User phil isett - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:25:35Z http://mathoverflow.net/feeds/user/7193 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29991/fourier-transforms-of-compactly-supported-functions Fourier transforms of compactly supported functions Phil Isett 2010-06-30T00:43:22Z 2012-11-20T06:58:14Z <p>One manifestation of the uncertainty principle is the fact that a compactly supported function $f$ cannot have a Fourier transform which vanishes on an open set. As stated, this phenomenon applies when $f$ lives on the integers, or on Euclidean space, but it is false for, say, the p-adic rationals, on which the characteristic function of the p-adic integers provides a counterexample.</p> <p>One normally proves this phenomenon on Euclidean space through complex analysis, although it does follow from the real-variable argument on this post of Tao on <a href="http://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/" rel="nofollow">Hardy's Uncertainty Principle</a>.</p> <p>For $f$ on the integers, it's even easier because only the zero polynomial has infinitely many roots (and this appears to me to be more or less the kind of input which goes into the other proofs as well).</p> <p>What I want to know is whether there is a proof of this phenomenon -- namely, the refusal of a compactly supported functions F.T. to vanish on an open set -- which directly relates to the connectedness of the frequency space, if that is what is behind it.</p> <p>Here is a failed effort in this direction...</p> <p>You know that $f = f \chi$ for any function $\chi$ which is $1$ on the support of $f$. Then $\hat{f}$ should remain unchanged when convolved with the Fourier transform of $\chi$, but since $\hat{f}$ lives on the reals you would like to think that this convolution would partially fill up some open set connected to the boundary of the original support of $\hat{f}$ and thus enlarge the support of $\hat{f}$. This argument goes through as long as you don't get an absurd cancellation; but while $\hat{f}$ may be assumed positive, one has less freedom to renormalize $\chi$.</p> <p>Does anyone know if there is a version of this argument which actually goes through? As stated it's no different than saying "$\hat{f} \ast \hat{\chi} = 0$ whenever $\chi$ lives away from the support of $f$; how weird..." It's possible my intuition here is all wrong and it's really the regularity of $\hat{f}$ which is completely behind the phenomenon, but I am at a loss for other examples of connected, locally compact abelian groups (are there any?) and I don't know how connectedness behaves with respect to Pontrjagin duality.</p> http://mathoverflow.net/questions/103138/what-is-the-simplest-oscillatory-integral-for-which-sharp-bounds-are-unknown What is the simplest oscillatory integral for which sharp bounds are unknown? Phil Isett 2012-07-26T01:03:53Z 2012-08-02T19:48:52Z <p>I have either heard or read that sharp asymptotics and bounds for oscillatory integrals of the form</p> <p>$\int e^{i \lambda \Phi(x)} \psi(x) dx \quad \lambda \to \infty$</p> <p>are unknown when the critical points for the phase function are not isolated. If this impression is correct, what are the simplest / most important integrals of this form for which the optimal decay rate and asymptotic have not been proven? E.g. are there examples with $\Phi$ being a polynomial? (I would also appreciate recommendations of references for estimating multidimensional oscillatory integrals if anyone has them.)</p> http://mathoverflow.net/questions/79907/what-is-a-complex-inner-product-space-really/97157#97157 Answer by Phil Isett for What is a complex inner product space "really"? Phil Isett 2012-05-16T21:05:10Z 2012-05-16T21:05:10Z <p>This answer is not complete, but I think the points are important.</p> <p>A complex vector space is one in which every vector $v$ has a special, oriented plane going through it, where each $\alpha \in {\mathbb C}$ rests upon the vector $\alpha v$. With this interpretation it is easy to see why $(\alpha + \beta)v = \alpha v + \beta v$, but actually a bit tricky to see why $\alpha(v + w) = \alpha v + \alpha w$. </p> <p>A real inner product gives a vector space a notion of geometry in which angles can be computed. In a normed complex vector space (which suffices to provide a notion of distance), the geometry of each plane $\alpha v$ going through $v$ looks like a rescaling of the usual complex plane because $|(\alpha v - \beta v)| = |\alpha - \beta| |v|$.</p> <p>One basic thing that separates an inner product space (or Hilbert space, rather) from a normed vector space is the fact that for any (closed) subspace $W$ of $V$ the projection $P_W(v)$ onto $W$, which finds the unique element in $W$ of minimal distance to $v$, turns out to be a (complex!) linear map. That's something very special about the shape of ellipsoids (unit balls in inner product spaces) which is not common to other kinds of unit "balls" -- for other shapes of the unit ball, this minimizer won't even be unique. This fact about linear projections, when applied to the space $W = { \alpha v }$ can explain the bilinearity of the inner product. (I think the real analogue of this fact can be demonstrated, or at least interpreted, by ruler and compass construction.) For Hermitian inner product spaces, we need a complex normed vector space, so we also want the rotations $e^{i \theta} v$ to be length-preserving, so that's one explanation for why we want conjugate linearity in the second variable of a Hermitian inner product.</p> http://mathoverflow.net/questions/96338/integration-under-functional-sign/96354#96354 Answer by Phil Isett for Integration under functional sign Phil Isett 2012-05-08T16:03:42Z 2012-05-08T16:03:42Z <p>Davide's right. Neither integral makes sense because the function was only continuous. If you assume $f$ is a smooth test function, then a priori it's only clear when $\mu$ is a finite linear combination of point masses. Thus, you cannot avoid using a Riemann sums trick: approximate the more general measure with a sequence of finitely supported measures $\mu_n$. My impression is that the general theory of distributions cannot start without taking Riemann-type sums at some point and that any argument probably has this maneuver underlying it somewhere. </p> <p>For the right hand side, you need to prove that $\int f(x,y) d\mu_n \to \int f(x,y) d\mu(y)$ in some $C^k$ topology as functions of $x$ over the support of $L_x$. For the left hand side, check that $\mu_n \rightharpoonup \mu$ weakly and $L_x f(x,y)$ is continuous in $y$. This step again uses that $L_x$ is continuous with respect to $C^k$ convergence for some $k$.</p> <p>Note, if you establish this identity when $\mu$ is, say, an absolutely continuous measure with a smooth density function, then you can pass to the limit for a general finite measure by using a mollifying kernel (analogous to taking $L_{\epsilon, x}$ in Liviu's argument, but this is a mollification in the $y$ variable, and is just measure theoretic).</p> http://mathoverflow.net/questions/90096/an-iterated-tensor-product-integral/90098#90098 Answer by Phil Isett for An iterated tensor product integral Phil Isett 2012-03-03T01:55:39Z 2012-03-03T02:01:52Z <p>Let's just view the function $\dot{X}(t) = \frac{d}{dt} X_t$ as a bounded function taking values in the vector space $V$. The notation $dX_{u_1}$ means $\dot{X}(u_1) du_1$ Then $dX_{u_1} \otimes \cdots \otimes dX_{u_k} = \dot{X}(u_1) \otimes \cdots \otimes \dot{X}(u_k) du_1 \cdots du_k$, should be regarded as a bounded function (or measure) on $[0,t]^k$ taking values in the vector space $V^{\otimes k}$.</p> <p>Because the operation of projecting to a symmetric part is a linear operation from $V^{\otimes k}$ to itself, you can take it inside of the integral. Let's call this symmetric part $dX_{u_1} \cdots dX_{u_k}$. I'll use the symbol $u \cdot v$ to denote the symmetric product of of $u, v \in V$ -- that is, $u \cdot u \cdot u = u \otimes u \otimes u$, and the general product is defined by the polarization identity. For example $u \cdot v = ( u \otimes v + v \otimes u) / 2$. The resulting multiplication is commutative.</p> <p>Therefore your integral is equal to</p> <p>$\int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k}$</p> <p>Because the symmetric product is commutative, for any permutation $\sigma$ of ${ 1 \ldots k}$ you get the same value by integrating any of the permuted regions</p> <p>$\int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{0 \leq u_{\sigma(1)} \leq \ldots \leq u_{\sigma(k)} \leq t} dX_{u_1} \cdots dX_{u_k}$</p> <p>Observe also that the region $0 \leq u_1 \leq \ldots \leq u_k \leq t$ is a fundamental domain for the action of $S_k$ on the cube $[0,t]$ -- that is, you can get the whole cube by permuting the variables $u_i$, and none of these regions overlap. Therefore, since there are $k!$ such permutations, we sum over permutations to conclude</p> <p>$k! \int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{[0,t]^k} dX_{u_1} \cdots dX_{u_k}$</p> <p>But, by Fubini and the multilinearity of the symmetric product, the integral on the right is just $(X_t - X_0)^k$.</p> <p>I should remark that this same proof in an even simpler setting also produces the $\frac{1}{k!}$ that appears when you prove the Taylor expansion through iterative use of the fundamental theorem of calculus.</p> http://mathoverflow.net/questions/89504/quick-computation-of-the-pontryagin-dual-group-of-torus/89544#89544 Answer by Phil Isett for Quick computation of the Pontryagin dual group of torus Phil Isett 2012-02-26T02:39:43Z 2012-02-26T03:53:08Z <p>If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis.</p> <p>For the circle, one needs only note that the distribution $u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula.</p> <p>The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument).</p> <p>In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then, </p> <p>$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$</p> <p>but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.)</p> <p>If you like to think in terms of Stone Weierstrass, you can view this argument is saying that the map $Tf$ which takes a Fourier transform and then performs the inverse Fourier transform is the identity. It relies on the fact that $f(x) = (e^{2 \pi i x} - e^{2 \pi i x_0}) g(x)$ whenever $f$ vanishes at $x_0$ which is a strong sense in which trigonometric polynomials separate points. The conclusion that $T$ is a multiple of the identity then follows from $T$ being linear not only over ${\mathbb C}$ but also over the algebra of trigonometric polynomials.</p> <p>Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below. </p> <p>Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. The same kinds of mollifications or similar calculations are used in proofs of Stone Weierstrass theorem; by a similar reasoning, once you have enough characters to separate points, you know you have a complete set.</p> <p>But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution). </p> <p>But then once you have established that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".</p> http://mathoverflow.net/questions/88513/h-principle-and-pdes/88682#88682 Answer by Phil Isett for H-principle and PDE's Phil Isett 2012-02-17T03:13:33Z 2012-02-17T04:12:07Z <p>You asked about the $h$-principle, but I'll say something about convex integration instead.</p> <p>Here is a survey by DeLellis and Szekelyhidi about instances of the "h-Principle" where convex integration is used to construct low regularity solutions to many equations of fluid mechanics:</p> <p><a href="http://arxiv.org/abs/1111.2700" rel="nofollow">http://arxiv.org/abs/1111.2700</a></p> <p>These analytic results, however, are different in flavor to what you usually call the $h$-principle in topology and geometry. In topology a nontrivial instance of the $h$-principle might say something like "you can invert the sphere $S^2 \subseteq {\mathbb R}^3$ through a regular family of immersions"; what makes it non-trivial is that there could have been a topological obstruction to doing so (for instance, you can't invert $S^1 \subseteq {\mathbb R}^2$ because the inclusion map $i$ and $-i$ have different degrees). In these analytic results, you're not exactly interested in homotopies.</p> <p>You can use the method of convex integration (at least, basically the same kind of convexity argument -- Gromov himself prefers not to call this convex integration) to construct wild solutions to PDE. For example, there are bounded solutions to incompressible Euler which are in the energy space and can have any prescribed energy density $\frac{1}{2} |v|^2(x,t)$ (in particular, they can be compactly supported in space and time). This is a bit shocking because sufficiently regular solutions to Euler conserve energy. The fact that weak solutions need not conserve energy is tied to ideas regarding the theory of turbulence, which is the main motivation for all these studies. </p> <p>If you read Springer or Gromov you may not immediately recognize the similiarities between the analysis convex integration and the topology/geometry version (for example, sometimes Baire category arguments are used in analysis to simplify the technical arguments, sometimes at the expense of some regularity in the solution). But the arguments closely parallel Nash's proof that short maps can be approximated by $C^1$ isometric embeddings, which is where the story of convex integration begins. More recent developments regarding isometric embeddings can be found in the references to the survey linked above. One main challenge regarding both Euler and the isometric embedding problem is to find the degree of regularity at which there is a transition from flexibility to rigidity.</p> <p>Preceding the developments in fluid mechanics, convex integration was also used by Kirchheim, Muller and Sverak to exhibit elliptic systems coming from Euler Lagrange equations with solutions that are Lipschitz but nowhere $C^1$ -- this flexibility result contrasts the result of Evans that minimizers of the same kinds of functionals are smooth off a closed set of measure 0. There are also many related investigations in the calculus of variations tied to the stability of differential inclusions $\nabla u \in K$, especially regarding how they arise in the mathematical theory of materials. For example, James and Ball presented the idea that if $u : \Omega \subseteq {\mathbb R}^3 \to {\mathbb R}^3$ is the configuration of a crystal, its deformation gradient $\nabla u$ minimizes free energy $\int_\Omega W(\nabla u) dx$ by taking values pointwise in the set $K$ of critical points of $W$. Muller's book "Variational Models for Microstructures and Phase Transitions" has more on this topic (for example regarding how you can explain microstructures as patterns which are "trying to minimize" such a functional), but I think this is a bit more distant from the original question. The relevance is only that convex integration can be used to produce wild solutions to $\nabla u \in K$; but here $K$ might even be a finite set, and $u$ is only Lipschitz, so it's fairly different from the topological setting.</p> http://mathoverflow.net/questions/79868/what-does-mellin-inversion-really-mean/80799#80799 Answer by Phil Isett for What does Mellin inversion "really mean"? Phil Isett 2011-11-13T04:25:33Z 2012-02-13T21:45:24Z <p>As others have pointed out, the Mellin inversion theorem is just the Fourier inversion theorem in disguise for the particular group ${\mathbb R_+}$ with invariant measure $\frac{dx}{x}$. The goal of the Fourier transform is to express a general function as a linear combination (i.e. integral) of the characters of the group, so that in this basis the operations of translations and all commuting operations will be diagonalized. For the ${\mathbb R_+}$, these characters look like $x \mapsto x^{-s}$ (the minus sign because of the normalization you chose in the question), and they are unitary (take values in the circle) for imaginary $s$ -- the operation of multiplying characters is just addition in the $s$ variable, so in the inversion formula you have the measure $ds$. There's also this funny thing about how there are $s$ with positive real part -- this is because in the "physical space" ${\mathbb R_+}$ you're always talking about distributions which are compactly supported away from $0$ when you use this transform. Let's ignore that.</p> <p>Since Mellin inversion is a disguised Fourier inversion, the real question is: why is the Fourier inversion formula on ${\mathbb R}$ true? To me the most convincing answer is the following: we can decompose a general function $f(x) = \int f(y) \delta(x-y) dy$ (this is the definition of $\delta$ but you have to take approximate delta-functions to make this rigorously work like a decomposition), so if we want to express a general function as a combination of the characters $x \mapsto e^{2 \pi i \xi x}$, it suffices to consider the $\delta$ function</p> <p>$\delta(x) = \int u(\xi) e^{2 \pi i \xi x} d\xi$</p> <p>One interpretation of this formal idea is that the distributions $\delta(x-y)$ are just like your usual standard basis functions.</p> <p>Now, observe that because $\delta(x)$ is invariant under multiplication by $e^{2 \pi i \eta x}$ for any $\eta$, the distribution $u(\xi)$ is translation invariant, and therefore must be constant. After you find the constant, plugging in $\delta(x) = C \int e^{2 \pi i \xi x} d\xi$ into $f(x) = \int f(y) \delta(x-y) dy$ gives the Fourier inversion formula. Complete, rigorous proofs all follow more or less these lines, but there are many flavors of how you like to phrase it. Of course, we can write the whole argument with multiplicative characters as well.</p> <p>Edit: The above argument assumes uniqueness of the representation, but one can also remark that if there is even a single function $f(x)$ for which $\int f(x) dx \neq 0$ and which can be realized as a linear combination $\int \hat{f}(\xi) e^{2 \pi i \xi \cdot x} d\xi$, then by rescaling, renormalizing and taking a limit, we obtain $\delta(x) = C \lim_{\epsilon \to 0} \epsilon^{-1} f(x/\epsilon)$, leading formally to the formula $\delta(x) = C \int e^{2 \pi i \xi \cdot x} d\xi$. One common rigorous execution of this philosophy is performed by taking $f$ to be a Gaussian.</p> http://mathoverflow.net/questions/5892/what-is-convolution-intuitively/86040#86040 Answer by Phil Isett for What is convolution intuitively? Phil Isett 2012-01-18T22:07:49Z 2012-01-18T22:07:49Z <p>I like the answer you gave when you asked the question. More generally, the convolution of two measures $\mu$ and $\nu$ is the pushforward of $\mu \times \nu$ by multiplication. In probability, that means you independently draw from $\mu$ and $\nu$ and add the resulting random vector. It's something that you can visualize to a certain extent if you do think of measures as fuzzy versions of points (like Terry Tao said).</p> <p>One point of view of measures is that they are linear combinations of points (or limits of things you can get from linear combinations of points). If you take this point of view, then convolution is simply the extension of the addition law by linearity to the case of measures.</p> <p>Since you can translate functions as well as measures, you can convolve, say, a probability measure with a function by randomly translating the function, giving the averaged out function $\int f(x-y) d\mu(y)$ which generally looks like a smoothed out version of your function $f$ -- $\mu$ tells you which translations you use and how to average. Again, you can view this as the extension of the operation of translating functions by linearity/continuity to the case of measures.</p> <p>The Lebesgue measure allows you to identify functions with measures, $g \mapsto g(x) dx$, so you can also convolve functions with other functions, but you might think of this operation is a bit less basic.</p> <p>Actually, the process of convolution extends by continuity to more than just measures but also to distributions. For example, you can approximate a tangent vector at $0$ (giving the distribution $u(x) = \sum_i c^i \partial_i \delta_0$) by differences of point masses, so convolution extends to distributions as well, but you can even get differential operators this way (in this example, $u \ast f$ is the derivative of $f$ in the $u$ direction). The technical difference here is that the approximation is only valid when integrated against $C^k$ functions (rather than $C^0$ functions in the case of measures). But the principle is the same -- it's the extension of the addition law by linearity and limits.</p> http://mathoverflow.net/questions/72210/why-are-there-so-many-smooth-functions/72226#72226 Answer by Phil Isett for Why are there so many smooth functions? Phil Isett 2011-08-06T04:46:09Z 2012-01-05T00:39:14Z <p>I wondered about this question for a while myself, but I think by now I've become convinced <strong>why there really are so many smooth functions</strong>. Maybe the ideas that convinced me can convince you. I think you answered the question when you asked it: the mollification process is basically responsible. If I recall correctly, the business about smooth homotopies is also most easily proven via mollification.</p> <p>If you start with a <em>non-negative</em> smooth bump function $\eta$ in the unit ball of ${\mathbb R}^n$ normalized so that $\int \eta(x) dx = 1$, then you can regard the measure $\eta(x) dx$ as a smooth probability measure. Likewise the measure $\eta_\epsilon(x) dx = \eta(\frac{x}{\epsilon}) \frac{dx}{\epsilon^n}$ is another smooth, probability measure supported in the ball of radius $\epsilon$ (it's the pushforward of the first one by multiplication by $\epsilon$).</p> <p>So with this interpretation, the mollification $f_\epsilon(x) = \int f(y) \eta_\epsilon(x - y) dy$ is what happens when you randomly translate the function $f$ and replace its value at each point by the expected value after these random translations. Now, it should seem intuitive that even when the original function is singular, the resulting function looks a lot like the original one, but is a whole lot smoother because the singularities get spread out and hence diminished. Imagine, for example, what results when you do this to the characteristic function of an open set -- you get a smooth cutoff that looks like the rough one. Or if you want a smooth partition of unity, start with a rough one (characteristic functions of sets) and just mollify it to get a smooth one.</p> <p>So, if you believe in smooth bump functions you should believe there are a lot of smooth functions. By the way, in case you don't know this, there is more than one way to produce a smooth bump function. One way (say we're on the line) is to repeatedly convolve characteristic functions of intervals in such a way that the support remains bounded -- the regularity increases by one every time you do it.</p> <p>EDIT: I noticed that the question also asked about <strong>what makes smooth functions so useful</strong>, which is something I didn't address at all. </p> <p>One reason they are so useful is that they simply do not have the defects that non-smooth functions have, and so they are less likely to introduce issues which are irrelevant to your problem. For example, if you want an asymptotic count of the integer lattice points in a large ball, you want to use Poisson summation so that the main term is simply the contribution from the $0$ frequency and everything else can be treated as an error. Unfortunately, the Fourier transform of the characteristic function of a ball fails to decay well enough for this idea to work (although it decays better than you would expect thanks to the curvature of the sphere, a fact which ultimately improves the error term). The problem here is related to the uncertainty principle -- Fourier analysis cannot give a count of lattice points that is precisely sensitive to points along the boundary; the characteristic function of an open ball or its closure define the same $L^2$ function. Thus, you have to mollify the ball for this strategy to work. Essentially the same issue arises in proving the prime number theorem (and commonly arises in analytic number theory in general) -- many proofs have irrelevant technicalities, all related to failing to smooth out the count.</p> <p>Another instance: proving that the fundamental group of the 2-sphere is trivial. It's easy to show that a smooth (or Lipschitz) curve will miss some point on the $2$ sphere (since it can't even increase Hausdorff dimension), and therefore such a curve is homotopic to a constant. But continuous curves could cover the whole sphere, giving a problem related more to oscillations and a lack of regularity and you might say less to topology. A similar problem arises when considering the degree of a continuous map -- they can hit their values infinitely many times thanks to uncontrolled oscillations so it's harder to interpret the degree as a count and usually requires an approximation. On the other hand, the distinctions between topological and smooth manifolds (e.g. exotic spheres) makes a really interesting topic, so one can't always dismiss lack of smoothness as just being a nuisance.</p> <p>One more unifying reason smooth functions are so useful is the fact that singular functions can be approximated by smooth functions, so concepts which already have explicit meaning for smooth functions (e.g. degree, distribution theoretic derivatives, Fourier transform) have a natural extension to singular functions exactly when they are continuous with respect to that kind of approximation. For example, degree is continuous with respect to uniform approximation so continuous functions have a degree (more interesting is that degree extends to maps with approximations in BMO -- see surveys of Brezis). When you work with smooth functions you rarely worry that something is defined, and then you just have to check the estimates/continuity (e.g. Fourier Transform maps $L^2 \to L^2$) to make sure defining by approximation is OK.</p> <p>Finally, you can only use smooth functions when you are on a manifold, so many of the axioms that people write down in point set topology to make sure they are not looking at some atrocious, pathological space, have already been built (entirely?) into the machinery of smooth functions and partitions of unity. Smooth functions can tell the difference between two closed sets, therefore the space is normal, etc. Thus, using smooth functions in arguments helps you avoid unwanted pathologies of the space as well as the maps.</p> http://mathoverflow.net/questions/84032/short-course-suggestions-for-high-school-students/84066#84066 Answer by Phil Isett for Short Course Suggestions For High School Students Phil Isett 2011-12-22T01:40:52Z 2011-12-22T01:40:52Z <p>If you can teach game theory, that could be good. It's bread and butter for mathematical economics and political science (even ecologists learn it now) -- I think the subject illustrates the point that math is not limited in application to situations which involve numbers. In addition to being useful, it's very elementary to solve games (although the fundamental fact that mixed strategy Nash equilibria exist requires topology to prove, it doesn't provide an algorithm for finding them -- actually solving games is more combinatorial). Proving that sets of strategies are/are not Nash equilibria can introduce students to the concept of a formal mathematical proof in a setting which I think is straightforward. </p> <p>Unfortunately, I can't think of a textbook that would be good, but maybe someone else knows one.</p> http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/57621#57621 Answer by Phil Isett for Why should one still teach Riemann integration? Phil Isett 2011-03-07T00:24:27Z 2011-12-08T03:25:28Z <p>One available compromise is to just work with the following definition of a Riemann integral (which works fine in ${\mathbb R}^n$ as well:</p> <p>A bounded function $f$ on $[a,b]$ is Riemann integrable if and only if for every $\epsilon > 0$ there are step functions $\psi_1, \psi_2 : [a,b] \to {\mathbb R}$ such that</p> <p>$\psi_1 \leq f \leq \psi_2$</p> <p>and</p> <p>$\int_a^b \left( \psi_2(x) - \psi_1(x) \right) dx &lt; \epsilon$</p> <p>I think using this definition is easy and geometrically intuitive, and on the other hand working with this definition prepares you conceptually for the Lebesgue integral where you juggle "simple" functions instead of step functions. Thus, you already get a nice piece of the Lebesgue point of view. </p> <p>The definition also demonstrates the broad principle that to construct an object in real analysis which should be a real number, one often needs a good way to overestimate/underestimate the object you're going for, and there are plenty of examples of that in real analysis outside of integration (liminf and limsup being the simplest) -- after all, this is just one point of view on how the real numbers are constructed.</p> <p>To give an example of the ease of use: to prove the fundamental theorem, suppose $F$ is continuous with a Riemann integrable derivative $F' = f$ and let $\psi_2$ be a step function above $f$ which induces the partition $a \leq x_1 \leq x_2 \leq \ldots \leq x_{n-1} \leq x_n = b$. Then the total change of $F$ from $a$ to $b$ is the sum of the "small" changes $F(b) - F(a) = \sum_{k=1}^n (F(x_n) - F(x_{n-1}))$ which is less than or equal to $\int \psi_2(x) dx$ by the mean value theorem. Similarly for the other inequality.</p> <p>Thus, one can go pretty far with this definition, but the results which separate the Riemann integral from the Lebesgue integral (e.g. that $g(f)$ is Riemann integrable for $g$ continuous and $f$ Riemann integrable or the fact that Riemann sums converge to the Riemann integral), one needs to use the observation that step functions can't be close in terms of area without being uniformly close on all but a small set. You could think this feature is either a clarification or a disadvantage. One certain disadvantage is that it is not the right point of view for integrating vector-valued functions. So you might decide it's better to define the integral in terms of Riemann sums in the first place (giving more of a "metric space" point of view and less of an "ordered space" point of view). Or you might even decide to skip some of these other topics, depending on your point of view and time available.</p> http://mathoverflow.net/questions/10574/how-do-i-make-the-conceptual-transition-from-multivariable-calculus-to-differenti/82345#82345 Answer by Phil Isett for How do I make the conceptual transition from multivariable calculus to differential forms? Phil Isett 2011-12-01T02:42:53Z 2011-12-01T04:25:41Z <p>I'm not sure if this point of view is taken up in the many references which are named here, but I'll say something about an "elementary" way to discover the exterior derivative which sounds like ordinary calculus. Let's take on the point of view that a $k$-form is something you integrate over a $k$-dimensional submanifold. If you imagine $k$-dimensional submanifolds as being composed of a $k$-dimensional blanket of little $k$-parallelograms, then this is a geometrically natural point of view since the $k$-form will assign a (small) number to each of these parallelograms. To actually realize a submanifold as such a "blanket" is to give a parameterization. (These parallelograms are oriented; this picture is different from surface integration of scalar functions in Riemannian geometry where one simply imagines some distribution of mass on the manifold and the integral is completely measure-theoretic. There the paralellograms have a <strong>positive</strong> mass given by the $k$-dimensional volume determined by the inner product.)</p> <p>In one-variable calculus, when $f$ is a function, $df$ tells you the change in $f$ per small change in its input, and if you integrate it over a curve from $a$ to $b$, it expresses the total change in $f$ from $a$ to $b$. Now, a one form $\eta$ is integrated not over points but rather over curves. Still, you can ask, how does $\int_\gamma\eta$ change when you perturb $\gamma$? Well, if you deform a closed curve $\gamma_a$ into another curve $\gamma_b$, the difference between the integrals over $\gamma_b$ and $\gamma_a$ is some derivative we can call "$d\eta$" integrated over the surface swept between the two. </p> <p>Picturing the case where $\gamma_a$ and $\gamma_b$ bound an annulus is a good thing to consider here; this interpretation tells you how to orient the boundary of the annulus if you want to think of $\int_\Sigma d\eta = \int_{\gamma_b} \eta - \int_{\gamma_a} \eta$ as being $\int_{\partial \Sigma} \eta$. On the other hand, you can take the point of view that the orientation for $\Sigma$ is determined by the requirement that we start at $\gamma_a$ and go to $\gamma_b$ (much like the case for $df$ of a function). You can then contract the inner circle to a point to recover Stokes' theorem for a disk -- the integral over the inner circle will vanish in the limit by the linearity and continuity of the form (a similar thing will happen in higher dimensions but the linearity is needed for the cancellation over the inner, closed surface).</p> <p>It's not completely necessary that the curve (or $k$-dimensional submanifold) you deform is closed, but by rule the boundary should remain fixed during the deformation or you will miss out on part of the boundary.</p> <p>Using a specific example like a square/cube, we can get a coordinate representation for $d\eta$ through the fundamental theorem of calculus. (For $0$ forms, every point is closed, so we did not need to worry about the word "closed" before.)</p> <p>It is easy to see many properties. For example, let's take $\eta$ to be a $1$-form in $3$-space; then $d^2 \eta$ is clearly $0$. Let $\gamma$ be a circle, and let $\Sigma_a$ and $\Sigma_b$ be the upper and lower hemispheres of a ball $B$ whose equator is $\gamma$. Then $\int_{\Sigma_a} d \eta = \int_\gamma \eta = \int_{\Sigma_b} d \eta$ by Stokes' theorem for a disk. On the other hand, the integral of $d^2\eta$ over the ball $B$ is just $\int_{\Sigma_b} d \eta - \int_{\Sigma_a} d\eta = 0$ because you can sweep out $B$ by deforming $\Sigma_a$ to $\Sigma_b$ with the boundary fixed. Since $\int_B d^2 \eta = 0$ for every ball, $d^2 \eta$ is identically $0$. When you execute this proof for a square, you see that mixed partials commute.</p> <p>I would like to know if the product rule can easily be seen through this interpretation, but I have not thought enough about it to see it clearly yet.</p> http://mathoverflow.net/questions/74841/an-example-of-a-beautiful-proof-that-would-be-accessible-at-the-high-school-level/81355#81355 Answer by Phil Isett for An example of a beautiful proof that would be accessible at the high school level? Phil Isett 2011-11-19T16:52:16Z 2011-11-19T16:52:16Z <p>For someone in high school, I think it's good to prove that the sum of the interior angles of a triangle is $\pi$ if they don't know why. Personally, I was never shown why this fact is true, and I feel that it's generally a bad idea to not know why something in math is true, especially when the answer is pretty. My favorite proof is to think about how the normal vector changes as you walk around the triangle -- it's nice because it generalizes to other shapes (which may not even be polygons).</p> http://mathoverflow.net/questions/80724/taylors-theorem-and-the-symmetric-group/80782#80782 Answer by Phil Isett for Taylor's theorem and the symmetric group Phil Isett 2011-11-12T22:19:47Z 2011-11-12T22:19:47Z <p>I think the most obvious way to see the symmetric group appearing is the following. To calculate the difference between $f(x)$ and the constant $f(0)$</p> <p>$f(x) - f(0) = \int_0^x f'(s_1) ds_1$</p> <p>Often we do this move because $x$ is small, so we figure that a small change in the input to $f$ will result in the output changing depending on the derivative of $f$. But the same logic goes for $f'$, so applying the same formula to $f'$, we get:</p> <p>$f(x) = f(0) + f'(0) x + \int_{0 &lt; s_2 &lt; s_1 &lt; x} f''(s_2) ds_2 ds_1$</p> <p>Note $f'$ may not change significantly from its initial value if $f''$ is under control, but there can be oscillations at the level of the next derivative responsible for $f'$ experiencing a small change -- in this case Taylor expansion usually is not helping us understand the problem. In any case, here we have an integral over the region ${ 0 &lt; s_2 &lt; s_1 &lt; x }$. This is a fundamental domain for the action of $S_2$ on the square ${ 0 &lt; s_1, s_2 &lt; x}$, so if $f''$ is a constant $f''(0)$, you get the volume $x^2 / 2!$ times $f''(0)$.</p> <p>Even if $f''$ is not constant, you can certainly write $f''(s_2) = f''(0) + \int_0^{s_2} f''(s_3) ds_3$ and get $f''(0) x^2 / 2!$ plus an integral over ${ 0 &lt; s_3 &lt; s_2 &lt; s_1 &lt; x }$, which is again a fundamental domain for the action of $S_3$ therefore has volume $x^3/3!$. Similar considerations apply to the higher order terms.</p> <p>Normally we replace the use of Fubini's theorem here with an equivalent integration by parts so that we never see iterated integrals. (Fubini's theorem is one way to prove integration by parts, so it really is the same move.)</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/79230#79230 Answer by Phil Isett for Dimensional Analysis in Mathematics Phil Isett 2011-10-27T04:08:39Z 2011-10-28T04:14:41Z <p>People have mentioned so far how dimensional analysis is fundamental in many inequalities in analysis, especially estimates (and even formulas) which come from partial differential equations. I want to elaborate on this point. For example, in ${\mathbb R}^{n + 1}$ ($n$ spatial dimensions), look at an equation like the wave equation $(-\partial_t^2 + \Delta) u = f$ -- for which we have the formal units $U/T^2 \sim U/X^2 \sim f$ where $T$ is the unit for the time variable and $X$ is the unit for the space variable. or the Schrodinger equation $(i \partial_t + \Delta)u = f$ (for which $U/T \sim U / X^2 \sim f$). The rigorous meaning of these units for Schrodinger is that for $\lambda \neq 0$, we can rescale a given solution to construct another solution $u(t/\lambda^2, x/\lambda)$ whose forcing term is $\lambda^{-2} f(t/\lambda^2, x/\lambda)$ (here we are imagining $\lambda$ has the units of $X$). (Similarly, scaling in the $x$ variable alone, we can change the equation to remove or reinsert the physical constants which usually appear in these equations.) Consider a large forcing term, e.g. $f$ belongs some $L^p$ space or mixed space-time $L^p$ space; you use dimensional analysis to figure out to what space we could possibly guarantee the solution $u$ lives in -- the goal here is to make rigorous the idea that $u$ depends continuously on $f$, and thanks to some abstract facts of functional analysis, proving this continuous dependence for a linear equation can essentially be achieved by no other means than by proving a (Strichartz) estimate such as </p> <p>$(\int |u|^r dt dx)^{1/r} \leq C (\int |f|^p dt dx)^{1/p}$ </p> <p>for some constant $C$ independent of $f$. Together with linearity, this estimate when applied to the difference of nearby $f$'s, says, if true, that as $f$ varies in $L^p$, $u$ varies continuously in $L^r$. </p> <p>The function space norms themselves have units'': e.g. $||u||_{L^2} = (\int |u|^2 dt dx)^{1/2}$ has units of $u$ times $T^{1/2}X^{n/2}$ where $T$ are the time units and $X$ are the spatial units ($T \sim X$ for the wave equation, but $T \sim X^2$ for Schrodinger). One can use these "units" to figure out what kind of estimates might possibly be possible. E.g. for Schrodinger, </p> <p>$(\int |u|^r dt dx)^{1/r} \leq C (\int |f|^p dt dx)^{1/p}$ </p> <p>would be impossible if the units were compatible, the left hand side is like </p> <p>$U (TX^n)^{1/r} \sim UX^{(n+2)/r}$ </p> <p>whereas the term with $f \sim (U/X^2)$ has dimensions $(U/X^2)\cdot(TX^n)^{1/p} = U(X^{-2+(n+2)/p})$. </p> <p>So dimensional analysis tells us that we have no chance of proving this kind of estimate unless $(n+2)/p - 2 = (n+2)/r$; otherwise one could rescale any solution to produce a counterexample. Similar considerations apply to the problem of determining in what sense $u$ can depend continuously on its Cauchy data.</p> <p>The dimensional analysis is very far from a proof of the estimate, which is not always true and requires a real understanding of, say, the dispersion properties of the specific equation at hand (you could have been looking at a different equation with the same scaling properties but which is qualitatively very different). Denis Serre mentioned how these dispersion properties can be read off from the curvatures of the parabola (for Schrodinger) and cone (for wave) in frequency space, although the proofs also bring in complex interpolation theory to also take into account the energy, and to get the most sharp results you need an interpolation scheme which treats individual frequency scales differently. You also need to look at norms which are different in the space and time variables, because the equation also distinguishes these variables. (The space of functions whose "energy" or "mass" is bounded in time gives a particularly natural spacetime norm.) Suffice to say, dimensional analysis is far from enough to justify a bound.</p> <p>For nonlinear problems with some kind of scaling symmetry, the units of the solution and the time or space dimensions can be tied together. E.g. consider the incompressible Euler equation without force for an unknown velocity field $\partial_t u + (u \cdot \nabla) u = - \nabla p$, $\nabla \cdot u = 0$. Then formally $U/T \sim U^2/X \sim P/X$ or $U \sim X/T$ (which is good for a velocity'' field), which means that given a solution $u(t,x)$ you can rescale to obtain a new solution $\lambda^{b-a} u(t/\lambda^a, x/\lambda^b)$ with pressure $\lambda^{2(b - a)} p(t/\lambda,x/\lambda)$. After all, if $\lambda^b \sim X$ and $\lambda^a \sim T$, then this rescaled $u$ looks like has units of $U$ and $P$ appears to have units of energy $U^2$. Any estimate for the Euler equations has to be consistent with this scaling, so it has consequences for studying the equation (of course, not so many a priori estimates exist, which is another problem...). On the other hand, like Deane Yang said, it's very good for dummy-checks.</p> <p>I guess all I've said is that the equations of physics often have a scaling symmetry, and due to this symmetry dimensional analysis is of use for rigorous mathematical treatment of the equation. (But that's no surprise since everyone's talking about the same equation.) But in mathematics the idea that inequalities must be consistent with a scaling symmetry goes a long way. For example, consider the Hölder inequality</p> <p>$\int f(x) g(x) d\mu(x) \leq ( \int |f(x)|^p d\mu(x))^{1/p} (\int |g(x)|^q d\mu(x))^{1/q}$</p> <p>This inequality is valid for any measure $\mu$; so you could be integrating over a surface or just taking a discrete weighted sum or something and it's still true. In particular, it is true for $\mu$ and true for the renormalized measure $\mu/\lambda$. If you compare "units", the left hand side is like $f g \mu$ and the right hand side is like $f g \mu^{1/p + 1/q}$. Thus, for Hölder's inequality to be true for arbitrary measure spaces, we see we need $1/p + 1/q = 1$. Here I am talking about abstract measure spaces -- something which has nothing to do with a physical problem or Euclidean spaces. Terry Tao has harmonic analysis notes where he uses this symmetry to renormalize $\mu$ and reduce the proof of Hölder's inequality to the case where $g$ is not even present, which allows one to view Hölder's inequality as an interpolation statement. The point of view I've expressed in great part derives from and is elaborated in those notes, which can be found <a href="http://www.math.ucla.edu/~tao/247a.1.06f/" rel="nofollow">here</a>.</p> <p>But from a broader mathematical viewpoint, dimensional analysis is probably only one example of paying attention to a group of symmetries (not just scaling symmetries). But at the moment, I cannot think of a particularly good example to illustrate this point. Probably one can also be found in the linked notes.</p> http://mathoverflow.net/questions/38639/thinking-and-explaining/77983#77983 Answer by Phil Isett for Thinking and Explaining Phil Isett 2011-10-13T02:04:02Z 2011-10-13T02:10:08Z <p>People have mentioned examples which are hard to share due to some kind of prerequisites. Here's one: I learned PDE from a professor who, in his mind, was always thinking about distribution theory, but officially could not talk about it until after he covered the material relevant to the exams. In distribution theory, whenever you see an integral over a domain $\int_\Omega u(x) dx$ you actually picture the characteristic function $\int \chi_\Omega(x) u(x) dx$ or $\int H(f(x)) u(x) dx$ if $f$ is a defining function for $\Omega$ and $H$ is a heaviside function. From this point of view, you imagine that all functions are smooth and compactly supported (or you can imagine their approximations), so that if you integrate by parts on $\int \chi_\Omega \nabla u(x) dx = - \int \nabla \chi_\Omega u(x) dx = \int \delta(f(x)) \nabla f(x) u(x) dx$. The boundary terms come when the derivative hits the characteristic function. Same thing for Stokes' theorem, Gauss's divergence theorem. It's pretty handy to compute this way.</p> <p>For a little while this was all I understood until I later found out what was going on. The limit of difference quotients of $\chi_\Omega$ is clearly supported on the boundary of $\Omega$ and it's clear, especially if you picture an approximation, that $\nabla \chi_\Omega$ points in the direction of increase of $\chi_\Omega$ -- i.e. the inward normal. More simply: there are two points of view -- if you were to take difference quotients of $u$, you use a Lagrangian point of view in which the point at position $x$ moves in the direction $i$, and you observe a change in $u$ between those points; instead, you can take an Eulerian point of view, (where the adjoint difference quotients go on the characteristic function) and you can instead look at movement of the region with $u$ fixed. </p> <p>Until I understood this point of view in a simpler way, it would not really be sensible to explain it to others. But now I know that giving a watered down version of the same proof when "proving" the fundamental theorem of calculus / Gauss's divergence for a calculus class in fact does not lose any key ideas (except for technicalities like how you need the mean value theorem to ensure the difference quotients are bounded). Of course, I would also talk about characteristic functions to any math student, since it is a nice point of view.</p> <p>By the way, in the calculus of variations, when your $u(x) = L(x, \phi(x) )$ is a Lagrangian and $\phi(x)$ is a solution to the Euler-Lagrange equations, and you take difference quotients using the flow of a vector field whose flow preserves the Lagrangian (a "symmetry"), you end up with Noether's theorem through only this one variation (there are only boundary terms in what I called "the Lagrangian point of view" because you vary through a family of solutions except at the boundary). So it's also a nice way to prove conservation laws in one swoop.</p> <p>My point: for a little while, distribution theory seemed like a magical theory with prerequisites that made it unexplainable in everyday talk, but once I really understood the ideas I could usually discard the vocabulary (actually, the whole theory can often be replaced by cutoffs, partitions of unity, Taylor expansion, and changes of variable -- although I still think it's great to learn). I suspect that this phenomenon is not uncommon for elementary applications of "fancy" mathematical theories. I believe that often once one has a more basic understanding, one can throw away the new words but still fully reveal the ideas (but maybe that's completely due to my own background). People here have talked about Feynman -- he was good at doing this in the context of physics. If you watch his (outstanding) lectures on Project Tuva you will see more or less the proof of Noether's theorem about which I just wrote.</p> <p>A second point:</p> <p>Another thing I think happens to me is that I feel some pressure not to convey just how <strong>often</strong> I rely on geometric modes of thought, especially when they go against the usual way of explaining things, or the background of a typical student, and are not completely necessary.</p> <p>Example 1: When you row-reduce a matrix, you make a bunch of changes (most importantly some "transvections") in the basis of the image space until a few of your basis vectors (say $v_1 = T e_1, v_2 = T e_2$) span the image of the matrix $T$. When you picture the domain of $T$ foliated by level sets (which are parallel to the null space of $T$), you know that the remaining basis vectors $e_3, e_4, ...$ can be translated by some element in the span of $e_1, e_2$ (i.e. whichever one lies on the same level set) in order to obtain a basis for the null space. Now, this is how we visualize the situation, but is it how we compute and explain? Or do we just do the algebra, which at this point is quite easy? If the algebra is easy and the geometry takes a while to explain and is not "necessary" for the computation, why explain it? This is a dilemma because once algebra is sufficiently well-developed it's possible that the necessity of (completely equivalent) geometric thinking may become more and more rare; and algebra seems to be more "robust" in that you can explore things you can't see very well. But then, when students learn the implicit function theorem, somehow I feel like having relied on that kind of foliation much more often would help understand its geometric content. Still, even if it's in your head and very important, are you going to draw a foliation every time you do row operations? We know the geometry, know the algebra, but it would take a while to repeatedly explain how to rely on the geometry while executing computations.</p> <p>Example 2: (Things that aren't graphs) </p> <p>Another problem geometric thinking faces is that modern math often seems to regard pictures as not being proofs, even if they are more convincing, so there is a bias regarding how to choose to spend class time. Let's say you want to differentiate $x^3$. You can draw a cube, and a slightly larger cube, and then look at the difference of the cubes and subdivide it into a bunch of small regions, three larger slabs taking up most of the volume. Algebraically, this subdivision corresponds to multiplying out $(x+h)^3$; collecting the terms uses the commutativity, which corresponds to rotating the various identical pieces. It is no different to write this proof out algebraically, the difference is that the algebraic one is a "proof" but the geometric one is.. not? Even if it's more convincing. So it's like the picture is only there for culture.</p> <p>Maybe I have the lecture time to teach both, I will. But I would like to go farther than that. When I differentiate the cube root function, the same cube appears and I go through it again if I feel like it just to convince myself of the truth. Actually, every time I ever use the product rule I always picture the same rectangle with a slightly larger rectangle. My point of view is that one important "definition" of multiplication is in terms of areas, and that a linear function is <strong>not</strong> necessarily a graph. When you think of a linear function, you should also picture things like rectangles, sectors, similar triangles like the kind that come up when "proving" basic differentiation formulas. Differentiating the integral may seem like a magical trick, but it's really just a continuation of the point of view that multiplication can look like an area/volume and differentiation means taking a small change in the input.</p> <p>Now, I'd like that point of view to be absorbed, but it's not exactly in the textbook, or completely consistent with what students' other teachers taught them. It's hard to go against the idea that "you should think graphically" -- if I ever think about the sine or tangent function now, it might be the area of a triangle, it might be the length of some vertical line segment, but it's basically never using the graph, which contains basically no additional information. If I have more than one shot at it, I'll try to explain both, but is it really of service to go around saying all the time why graphs aren't the end-all-be-all?</p> <p>Also, while I can express the pictures in my head one at a time, the fact that I repeatedly, repeatedly see this pictures is something that I feel is harder to express. After all, can't you just do algebra and get through this stuff more quickly? The algebra is "easier" too; it takes up less space.</p> http://mathoverflow.net/questions/72431/functionals-continuous-with-respect-to-weak-convergence Functionals continuous with respect to weak convergence Phil Isett 2011-08-09T03:12:40Z 2011-10-03T14:34:44Z <p>It's well known that a functional of the form $u \mapsto \int f(u) dx$ is continuous with respect to weak convergence (say weak* convergence in $L^\infty$) if and only if the function $f$ is affine. An even more interesting thing to study is functionals which involve derivatives $u \mapsto \int f(du) dx$. The function $f$ needs to be quasiconvex to be lower semicontinuous.</p> <p>I'm interested in functionals which also depend on the $x$ variable, like $\int f(x, u, du) dx$. Can anyone tell me a good place to read about continuity and semicontinuity with respect to weak convergence for functionals of this form. For example, must the $f(x,u)$ in $\int f(x,u) dx$ be convex / affine in the $u$ for almost every point $x$?</p> <p>EDIT: I should clarify the last question. Dorian has given me a good reference regarding weak continuity/semicontinuity of functionals which involve derivatives and depend on the spatial variables. The question that remains is simpler: when you have a functional $\int f(x,u) dx$ that is, say, continuous with respect to weak-* convergence in $L^\infty$, is it necessarily the case that $f(x,\cdot)$ is affine in $u$ at almost every point $x$? This question is purely measure theoretic and has nothing to do with derivatives (so without loss of generality I am asking about the measure space $I = [0,1]$ since I am interested in the case without atoms). By using a sequence of the form $\chi_{E_k} u + (1 - \chi_{E_k}) v$ for suitable characteristic functions $E_k$, we know that $f(x, \theta u + (1- \theta) v) = \theta f(x, u) + (1-\theta) f(x,v)$ for all measurable functions $\theta : I \to [0,1]$ and all bounded, measurable $u, v$ on $I$. Probably, this is enough to imply $f(x, u)$ is affine in $u$ for almost every $x$. Similarly, lower semicontinuity should imply convexity at almost every point.</p> http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/74078#74078 Answer by Phil Isett for Proofs that require fundamentally new ways of thinking Phil Isett 2011-08-30T16:18:33Z 2011-08-30T16:25:22Z <p>The Lebesgue integral seems to have been a fundamentally new way of thinking about the integral. It's hard to prove the convergence theorems if you have the Riemann integral in mind. I suppose there are probably many instances where you can give a computer a very ineffective definition of something and ask that it prove theorems. Ask it to prove anything about the primes where you start with the converse of Wilson's theorem as the definition of a prime. Can the computer figure out that its definition is terrible? Can it figure out what a prime really "is"?</p> http://mathoverflow.net/questions/74012/is-point-to-set-distance-continuous/74022#74022 Answer by Phil Isett for Is point to set distance continuous? Phil Isett 2011-08-30T00:43:42Z 2011-08-30T00:43:42Z <p>Yes, it is.</p> <p>Being an infimum of a family of continuous functions, it's clear that the closed upper contour sets ${ x : \phi(x) \geq t }$ are closed. (This means the function $\phi$ is upper-semicontinuous. I always have to think of the characteristic function of a closed interval $[0,1]$ which is "above" the lowersemicontinuous characteristic function of $(0,1)$, to keep my head straight about which one's which.) </p> <p>To see that the closed, lower contour sets ${ x : \phi(x) \leq t }$ are closed, you can consider a convergent sequence $x_n \to x$ (convergent with respect to the Euclidean metric) with $\phi(x_n) \leq t$. Associated to this sequence, there are also points $y_n \in A$ which are a distance $d(x, y_n) \leq t + \delta$ of $x_n$. For $n$ large enough, $d(x, x_n) \leq \delta$ (because $\phi$ is continuous with respect to the Euclidean metric). Which means that $d(x, y_n) \leq t + 2 \delta$. Thus $d(x, A) \leq t + 2 \delta$ for all $\delta > 0$.</p> http://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/72534#72534 Answer by Phil Isett for Can distribution theory be developed Riemann-free? Phil Isett 2011-08-10T01:41:01Z 2011-08-10T01:41:01Z <p>Regarding the first example: there is essentially no way to get around the "Riemann integration". Often when you use it (for example, to characterize monotonic functions) the distributions in question are measures. In this case, you can use Fubini's theorem to interchange the order of integration. Otherwise the statement you're trying to prove is:</p> <p>$\int \int u(x) \phi(y - x) dx \psi(y) dy = \int u(x) (\int \phi(y) \psi(y + x) dy) dx$</p> <p>where the $dx$ integral is to be viewed in the sense of distributions. A priori from the definition of a distribution, this formula is only clear when $\phi$ is a delta function (in which case the convolution is just a translation) or a linear combination thereof. So to prove the general case you will have to use the continuity in the definition of a distribution to pass from the limit by approximating $\phi(x) dx$ with point masses. This is essentially an exercise which is often done in Riemann integration, although you have to keep track of the error and make sure the convergence is in $C^k$. </p> <p>In Rudin's book, this discrete approximation is actually how he constructs the Lebesgue measure from scratch, and it's interesting that it actually even works in the measure case because you are bypassing the use of Fubini's theorem. From that point of view it isn't completely a "Riemann integral" approach.</p> http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71548#71548 Answer by Phil Isett for What should be learned in an introductory analytic number theory course? Phil Isett 2011-07-29T04:05:02Z 2011-07-29T04:05:02Z <p>Besides Davenport's book, which basically everyone has already recommended, why not talk a little more about the circle method? If I recall correctly, the only application Davenport gives is for counting the prime number solutions to an equation, but the circle method is probably simpler to understand when you're just looking for integer solutions. One can use Davenport's other book "Analytic Methods for Diophantine Equations and Inequalities". At the end, he also does the Oppenheim conjecture for 5 variables, which is somewhat simpler than the rest of the book beyond Waring's problem.</p> http://mathoverflow.net/questions/48067/is-square-of-delta-function-defined-somewhere/70871#70871 Answer by Phil Isett for Is square of Delta function defined somewhere? Phil Isett 2011-07-21T02:46:25Z 2011-07-21T02:46:25Z <p>The theory of distributions and operations on them are generally only useful in so far as they extend the operations on smooth functions. If you look in Hörmander, there is a criterion in terms of wavefront sets which is very useful (mentioned by others), and you'll also notice that the wavefront sets of $\delta$ and $\delta$ collide. The reason you can't square the delta-function is that when you approximate it by smooth functions, there is no unique limit. If you wanted to restrict to a smaller space of test functions, you would clearly have to consider test functions which vanish at the origin in some way. But do you have a particular purpose in mind for this question?</p> <p>EDIT: Sorry -- this was supposed to be a comment, not an answer.</p> http://mathoverflow.net/questions/8846/proofs-without-words/69834#69834 Answer by Phil Isett for Proofs without words Phil Isett 2011-07-08T21:49:24Z 2011-07-08T21:49:24Z <p>Here's a proof of the area of a circle (or sector) which is different from the one posted previously.</p> <p><strong>EDIT:</strong> I was unable to embed the file, which is in pdf form. Here is a link:</p> <p><a href="http://wildpositron.files.wordpress.com/2011/04/sectorarea2.pdf" rel="nofollow">http://wildpositron.files.wordpress.com/2011/04/sectorarea2.pdf</a></p> <p>I discussed what goes into making the proof complete to show that the map preserves area on my blog here (it requires just another picture or two, but it's essentially still only a geometric argument):</p> <p><a href="http://wildpositron.wordpress.com/2011/04/05/calculating-the-area-of-a-sector/" rel="nofollow">http://wildpositron.wordpress.com/2011/04/05/calculating-the-area-of-a-sector/</a></p> http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/69224#69224 Answer by Phil Isett for Generalizing a problem to make it easier Phil Isett 2011-07-01T01:10:17Z 2011-07-04T21:44:17Z <p>I've already posted an answer on this thread, but I found another example I'd like to describe separately. Let $r > 0$ and consider the following problem, coming from compound interest or as one definition of $e^r$:</p> <p>Show that $f(n) = (1 + \frac{r}{n})^n$ increases with $n$.</p> <p>One generalize the problem strategy is to allow $n$ to be a continuous variable (probably this trick could have its own article). Now, see if you can prove that $f(n)$ still increases. If you take this mindset, it's natural to use the definition of $n$th power for $n \in {\mathbb R}$ and write</p> <p>$f(n) = e^{n \log(1 + \frac{r}{n})}$</p> <p>And the problem has reduced to showing that $x \log(1 + \frac{r}{x}) = \int_0^1 \frac{r}{(1 + \frac{sr}{x})} ds$ increases with $x$, which it clearly does. (Here we've used the integral definition of the logarithm, but written in a way typically helpful for analyzing such products.)</p> <p>Another problem that can be solved through allowing a discrete parameter to be continuous is to prove Stirling's approximation for $n!$ (although to make that proof very clean you can also use other labor saving tricks like Taylor expansion by integration by parts and the dominated convergence theorem).</p> <p>If you ran into this problem from compound interest, or you were hoping for something more elementary which did not use such a heavy understanding of the exponential function, then you probably want to find a different proof. But finding a different proof still seems to require "generalizing the problem", but in a different way.</p> <p>Another proof, goes as follows. Imagine that interest at a rate $r$ works as follows: once an amount of money is invested, the value of each unit after a time $t$ is given by $(1 + tr)$. That is, the value of the money grows linearly. Now imagine you had the opportunity to withdraw and immediately reinvest your money at a time of your choice. Having this ability would allow you to raise more money, because it would allow you to accrue interest on the interest you've already earned (hence the name "compound interest"). With this interpretation, the number $(1 + \frac{r}{n})^n$ is the value of each unit of money after time $1$ and $n$ regularly spaced compoundings.</p> <p>The proof now goes as follows: if you had a choice of when these compoundings would occur, then the more compoundings the better, and the best way to allocate $n$ compoundings is to have them occur at $n$ regularly spaced time intervals. That is, we interpret $(1 + \frac{r}{n})^n = \max \prod_{i=1}^n (1 + a_i r)$ under the constraint that $0 \leq a_i \leq 1$ with $\sum a_i = 1$.</p> <p>For example, it is better to have one compounding than to have none at all, because after withdrawing and reinvesting the money, now not only does the initial investment grow linearly, but also the interest you earned before the withdrawal grows linearly. For the same reason, given $a_1, \ldots, a_n$, the opportunity to compound once more during, say, $0 &lt; t &lt; a_1$, would allow you to increase the amount of money at all later times.</p> <p>The fact that the best choice of $(a_1, \ldots, a_n)$ is to have $a_1 = a_2 = \ldots = a_n = \frac{1}{n}$ is the principle that the largest product you can obtain when the sum of positive numbers is fixed is to have all the terms equal. This is easy to check with two variables: you can either find the largest rectangle to fit inside an isosceles triangle, or otherwise just note that if $a_1 \neq a_2$, then changing to $a_1' = \frac{(a_1 + a_2)}{2} = a_2'$ gives an improvement for $(1 + a_1 r)(1+ a_2 r) &lt; (1 + a_1' r) (1 + a_2' r)$. The case of $n$ variables actually follows from this observation. </p> <p>So if you really wanted some elementary solution to the problem, this one would do. It's an interesting example because you can see that either solution involves some kind of generalization, but the two generalizations are unrelated to each other. The first one does not need to / is unable to consider these non-even partitions. The second does not need to / is unable to consider fractional $n$.</p> <p>By the way, does anyone know how to prove in an elementary way (i.e. expanding) that $\prod_{i=1}^n (1 + a_ir)$ tends to $e^r = \sum \frac{r^k}{k!}$ as $\max |a_i| \to 0$ with $0 \leq a_i \leq 1$ and $\sum a_i = 1$? An easy solution goes by writing the product with the exponential function so that you get the exponential of $\sum \log(1 + a_i r) = \sum \int_0^1 \frac{a_i r}{(1 + s a_i r)} ds$. </p> <p>You can then integrate by parts (i.e. Taylor expand) to obtain $\sum a_i r - \sum \int_0^1 (1-s) \frac{(a_i r)^2}{(1 + s a_i r)^2} ds$. Now, $\sum a_i r = r$ is the main term. After you take $\max |a_i|$ to be less than $.5 / |r|$, the error term is bounded in absolute value by $C \sum (a_i r)^2 \leq \max { |a_i| } \cdot \sum a_i |r|^2$. I can, of course, move this question to a different thread.</p> <p><strong>EDIT:</strong> I realized later on that there is a completely elementary proof, and it is also completely obvious even though I didn't think of it. Namely, you expand $(1 + \frac{r}{n})^n$ into powers of $r$, and it is easy to see after a little algebra that each coefficient increases with $n$. I still find the other solutions interesting, but this turns out not to be a good demonstration of how generalizing can make a problem easier. By the way, the last question I had asked was answered in this thread:</p> <p><a href="http://mathoverflow.net/questions/69272/a-limiting-product-formula-for-the-exponential-function" rel="nofollow">http://mathoverflow.net/questions/69272/a-limiting-product-formula-for-the-exponential-function</a></p> http://mathoverflow.net/questions/69272/a-limiting-product-formula-for-the-exponential-function A limiting product formula for the exponential function Phil Isett 2011-07-01T15:56:38Z 2011-07-04T17:53:03Z <p>By the way, does anyone know how to prove in an elementary way (i.e. expanding) that $\prod_1^n (1+a_i r)$ tends to $e^r=\sum \frac{r^k}{k!}$ as you let $\max|a_i|\to 0$ with $0\leq a_i \leq 1$ and $\sum a_i = 1$? An easy solution goes by writing the product with the exponential function so that you get the exponential of $\sum \log(1+a_i r) = \sum \int_0^1 \frac{a_i r}{(1+s a_i r)} ds$.</p> <p>You can then integrate by parts (i.e. Taylor expand) to obtain $\sum a_ i r − \sum \int_0^1 (1−s)\frac{(a_i r)2}{(1+s a_i r)2}ds$. Now, $\sum a_i r = r$ is the main term. After you take $\max|a_i|$ to be less than $.5/|r|$, the error term is bounded in absolute value by $C \sum |a_i r|^2 \leq \max|a_i|\cdot \sum |a_i| |r|^2 \leq C |r|^2 \max |a_i|$.</p> <p>I was hoping to find an elementary proof of this convergence by expanding the product $\prod_1^n (1+a_i r)$ and gathering terms with a common power of $r$. In particular, it would be nice to prove the convergence of this limit without the exponential function, since then the limit could be considered a definition of $e^r$. The case when all of the $a_i$ are equal is done in Rudin's "Principles of Mathematical Analysis".</p> <p>The motivation for this problem comes from compound interest, which I described in a different thread here: <a href="http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/69224#69224" rel="nofollow">http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/69224#69224</a> .</p> http://mathoverflow.net/questions/69432/looking-for-ideas-concerning-the-teaching-of-lower-division-differential-equation/69435#69435 Answer by Phil Isett for Looking for ideas concerning the teaching of lower-division differential equation courses... Phil Isett 2011-07-04T02:02:16Z 2011-07-04T02:02:16Z <p>One thing you can try (and if you decide to do it, I'd like to hear how it goes), is to discuss the differential equation $\frac{dz}{dt} = \alpha z$ with $\alpha \in {\mathbb C}$. The initial condition $z(0) = 1$, of course, corresponds to $e^{\alpha t}$, but what's nice is that you can draw the flow lines and see without calculation what they look like and how they depend on the signs of the real and imaginary parts of $\alpha$. Of particular interest is the case of $e^{it}$ and to explain why it is obvious that $e^{i t} = \cos t + i \sin t$ and $e^{i(t_1 + t_2)} = e^{i t_1} e^{i t_2}$. There's a nice construction of $\pi$ using the symmetries of the vector field -- it suffices to reach $y - x = 0$, and you can prove from the equation satisfied by $y - x$ that this line is reached within time $1$. The uniqueness of solutions implies the rest of the solution can be obtained by reflections. (Even though this equation is "solved explicitly" it can be usefully regarded as the <strong>definition</strong> of $e^{it}$ and the trig functions.) I suggest it because I think the equation is fundamental, and also it gives the opportunity to visualize where the product $\alpha z$ is compared to $z$.</p> <p>Some typical exercises which don't (shouldn't) involve solving explicitly are asking asymptotic questions about equations of the form $\frac{dy}{dx} = f(y)$ where $f : {\mathbb R} \to {\mathbb R}$ is some explicit function like a polynomial with some sign changes.</p> http://mathoverflow.net/questions/69272/a-limiting-product-formula-for-the-exponential-function/69310#69310 Answer by Phil Isett for A limiting product formula for the exponential function Phil Isett 2011-07-02T02:45:59Z 2011-07-02T03:15:20Z <p>Thanks, Anthony, for finding this solution. I was completely at a loss for how to handle all the indices. If you don't mind, I would like to write down one version of the argument that you've given in full detail.</p> <p>Claim: Under the hypotheses of the question $1 = k! \sum_{i_1 &lt; \ldots &lt; i_k} a_{i_1} \cdots a_{i_k} + O(\max |a_i|)$ where the error is non-negative.</p> <p>The claim is true without an error when $k = 1$, and follows from induction. If we write $1 = (\sum a_i)^{k+1} = (\sum a_i) ( \sum a_i )^k$ The induction hypothesis allows us to write this product as $(\sum a_i)\cdot(k! \sum_{i_1 &lt; \ldots &lt; i_k} a_{i_1} + O(\max |a_i|) ) = (\sum a_i)\cdot (k! \sum_{i_1 &lt; \ldots &lt; i_k} a_{i_1} ) + O(\max |a_i| )$</p> <p>If we now distribute out the product, we get the term we want $(k+1)! \sum_{i_1 &lt; \ldots &lt; i_k &lt; i_{k+1} } a_{i_1} \cdots a_{i_{k+1} }$ from the products with no repeats and then an error coming from products with exactly one term repeated. Take whichever term is repeated and bound one copy of it in absolute value by $\max |a_i|$. Then the error is bounded by $\max |a_i| ( \sum |a_i| )^k = O(\max |a_i|)$.</p> <p>Having this claim established and looking slightly more carefully at the dependence of the error on $k$ (the constant in the big O only grows like $C^k$), we also have prove the convergence that I was looking for (and we don't need non-negativity of the terms; just that $\sum |a_i|$ is bounded). In the non-negative case we can just observe the error is non-negative, so that the dominated convergence theorem applies (with respect to the finite measure $\frac{|r|^k}{k!}$), giving a small shortcut and a soft way to see the convergence without a rate.</p> <p>All credit goes to Anthony Quas for the idea; I just thought the induction was a fairly clear way to get the details all down.</p> http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/69156#69156 Answer by Phil Isett for Generalizing a problem to make it easier Phil Isett 2011-06-30T01:24:13Z 2011-06-30T04:18:28Z <p>I have some examples, but I might have more words of warning than examples. Example 1. Let's say you have just the axioms for the real numbers, and you want to establish that $2$ is not zero, but all you know is that $1$ is not zero (perhaps because you want to divide by 2 for whatever reason). Knowing that there is a field with $2$ elements where $2= 0$, you discover you have no choice but to prove 2 is positive, even though it's not what you were going for. (This is an example where taking on a more general point of view does not give you something different to prove instead, but limits the approaches to what you're trying to prove.)</p> <p>So in order to prove 2 is positive, it suffices to prove 1 is positive. And here it somehow just makes sense to prove that the square of any nonzero number is positive -- from this point of view 1 is positive "because it's a square".</p> <p>Here's another example from real analysis. You want to prove things about the cube root function from first principles. Like the fact that it's well-defined, continuous, differentiable, concave, etc. Typically one approaches by proving general theorems about smooth, increasing functions, and applies these general theorems to the function x^3 to learn about its inverse.</p> <p>So while this seems like the kind of example you're going for, I personally don't like the example only because it suggests this is the "right" way to go. Often when one takes a "generalize" approach solving the problem can become easier, especially if you already know the general facts which end up doing the job. But for this example, defining the cube root is just a bit easier than using the general intermediate value theorem since you can define it as $f(x) = \sup { t : t^3 \leq x }$ and prove directly that $f(x)^3 = x$.</p> <p>Moreover, it's also a good example to use the implicit definition to prove things directly for this function just because it makes manipulations concrete. For example, since $f(x+h)^3 = x + h$, we have</p> <p>$(f(x+h)^3 - f(x)^3) = h$ </p> <p>is also equal to</p> <p>$(f(x+h) - f(x)) \cdot \int_0^1 3 (s f(x+h) + (1-s) f(x))^2 ds$</p> <p>If you look at this expression for a little while, you will be able to deduce things like monotonicity, uniform continuity on any interval $[\epsilon, \infty)$, differentiability on such intervals, concavity, so on. The point is that even when general technology works, it can be quite instructive to use "general methods", but execute them on explicit examples.</p> <p>You might argue that these direct proofs are the most insightful, but one is more likely to first find a general proof (especially if the appropriate general apparatus already exists), and then the direct proofs can only be found later indeed because they require a more direct understanding of the specific problem. I personally use the cube root of 7 (which nobody can name) in calculus lectures to motivate the concept of continuity, so I agree here it's a very natural use of the concept to define the cube root.</p> <p>Differential equations have, I think, a lot of examples. If you want to learn something about the exponential function and your point of view is that it is the unique solution to $\frac{df}{dx} = f$ (or $\frac{df}{dx} = i f )$ with data $f(0) = 1$, then in order to give a differential equation-type analysis to prove the basic properties (which is fun) you will often have to consider the same differential equation with more general data (and in particular prove uniqueness of the $0$ solution). In fact, you may be tempted to consider more general ODE's like $\frac{df}{dx} = F(x, f)$ because this point of view can inspire some of the techniques. It also helps greatly (say if you want to prove $e^{it}$ is periodic) to just know what a vector field is and have that point of view.</p> <p>I think the problem is that the general theorems can easily end up being forced to consider cases which are more complicated than the problem at hand. Even if you find a proof, that doesn't necessarily mean you should settle for it. If you look at the function $f(x) = 1/k$ where $k$ is the smallest positive integer such that $kx \in {\mathbb Z}$, and you want to show $f$ is Riemann integrable, you can of course prove a theorem that characterizes Riemann integrable functions in terms of the measure of their sets of discontinuity. Of course, if you know this theorem off the top of your head, it is "easy" to prove $f$ is Riemann integrable. But this proof is inefficient because it invokes a theorem that takes care of the nastiest examples of Riemann integrable functions (which this $f$ is not).</p> <p>One more example: it's very easy to find a faulty proof of the chain rule if your point of view is not general enough. The first thing many people try to do to analyze the difference quotients for $f(g(x))$ is to multiply and divide by $g(x+h) - g(x)$. But then you get into these hairy problems where the number by which you divide may be zero. If you want to find the correct proof, you should take the point of view that the chain rule is a statement which should be true for maps between Euclidean spaces of any dimension. It just says the linearization of the composition is the composition of the linearizations. With this point of view in mind, you should not dare to try dividing, because dividing by $g(x+h) - g(x)$ does not even make sense in this generality.</p> <p>Here's another: prove that $\int_{\mathbb R} e^{i \xi x} e^{- x^4 } dx$ is bounded by $\frac{C}{(1 + |\xi|)^2}$. Somehow you have to recognize the key features of $e^{-x^4}$ are that it will cancel against something very oscillatory because it is so smooth. Similar example problem: prove the decay for large $\xi$ of $\int_{\mathbb R} \log(1 + .5 \sin(\xi x) ) e^{- x^4 } dx$. It seems like this "look for a more general setting" trick really needs to be drilled into you for it to work because... it may require some imagination or experience to know what the right general setting is. </p> <p>What some examples show is that finding the proof may be much more difficult, maybe impossible, without a willingness to work in this direction.</p> http://mathoverflow.net/questions/69129/who-is-the-weak-sequential-closure-of-the-set-of-finitely-supported-measures-on/69145#69145 Answer by Phil Isett for Who is the weak* sequential closure of the set of finitely supported measures on the integers? Phil Isett 2011-06-29T20:14:01Z 2011-06-29T20:14:01Z <p>Interesting question. I'd like to know if there is a more elementary answer to it. Here's a candidate start of an answer:</p> <p>Converging weak* in $l^\infty({\mathbb Z})^*$ implies in particular that $a_n$ is a convergent sequence for all $n$. Let me use the notation $f(a_n)$ to denote the dual pairing between $f \in l^\infty$ and $a_n \in l^1$. From the existence of a weak limit, we see there is a pointwise limit $a = (a_k)$, and hence $f(a_n) \to f(a)$ for all compactly supported $f$. To see that $a$ is in $l^1$, observe that we have the bound $|f(a_n)| \leq C$ for all $||f||_{l^\infty} \leq 1$ and all $n$ because $f(a_n) \to A[f]$ for some bounded, linear functional $A \in (l^\infty)^*$. With this estimate in hand, we can conclude that $f(a_n) \to f(a)$ for all $||f||_{l^\infty} \leq 1$ which vanish at infinity, and therefore $|f(a)| \leq C$ for all such $f$. It then follows that $a \in l^1$ by just the Lebesgue-theoretic definition of an infinite sum of positive numbers as a supremum of finite sums (just choose the $f$ correctly).</p> <p>What this argument has not shown is that $f(a_n) \to f(a)$ for all $f$ in $l^\infty$. It seems like the only thing that could have gone wrong is if somehow the mass of $|a_n - a|$ could have gone off towards infinity in ${\mathbb Z}$. But if this were the case, it seems like you could cook up an $f \in l^\infty$ so that $f(a_n)$ oscillates, which would contradict the weak convergence of the sequence.</p> http://mathoverflow.net/questions/103138/what-is-the-simplest-oscillatory-integral-for-which-sharp-bounds-are-unknown/103299#103299 Comment by Phil Isett Phil Isett 2012-08-03T05:30:34Z 2012-08-03T05:30:34Z Thanks! Does their condition b'' actually include all homogeneous polynomials? That sounds very general. I think the second example you wrote (which solves the wave equation) has well-known decay properties. http://mathoverflow.net/questions/103138/what-is-the-simplest-oscillatory-integral-for-which-sharp-bounds-are-unknown/103814#103814 Comment by Phil Isett Phil Isett 2012-08-03T05:22:54Z 2012-08-03T05:22:54Z Wow, thank you so much for the response and all the references! I will have to look into them. http://mathoverflow.net/questions/103138/what-is-the-simplest-oscillatory-integral-for-which-sharp-bounds-are-unknown/103145#103145 Comment by Phil Isett Phil Isett 2012-07-27T12:12:28Z 2012-07-27T12:12:28Z Thanks for the references. I had actually never thought about oscillatory integrals from the point of view of numerical approximation. Right now what I'm mostly interested in is knowing which oscillatory integrals do not even have theoretical bounds that are sharp up to a constant. But I find these interesting. http://mathoverflow.net/questions/96338/integration-under-functional-sign/96354#96354 Comment by Phil Isett Phil Isett 2012-05-08T20:35:07Z 2012-05-08T20:35:07Z (It looks like there is a coding error in the comment.) I think I'm starting to see the relationship between these two arguments. On the one hand, your argument appeals to Fubini to get started; the argument that I suggested actually avoids Fubini's theorem by taking Riemann sums. I actually thought that you couldn't get away without this move (for the reason in my post), but it looks like you can just get by with the continuity in the $y$ variable? Thoughts? http://mathoverflow.net/questions/62364/solving-x-partial-x-f-0-over-distributions/67287#67287 Comment by Phil Isett Phil Isett 2012-04-22T03:56:23Z 2012-04-22T03:56:23Z This answer is correct. The fact that $xu=0$ has only the functions $c_1\delta_0$ as solutions follows from how every smooth test function $\phi(x)$ with $\phi(0)=0$ can be written as $\phi(x)=x\psi(x)$ where $\psi(x)$ is another test function. You can see this by Taylor expanding: $\phi(x)=0+x\int^1_0\phi'(tx)dt$ -- in higher dimensions you should put in a cutoff to make sure this $\psi$ has compact support. Finally, to show $v'(x)=c_1\delta_0$ has only Heaviside solutions, observe that the only solutions to $v'(x)=0$ are constants, which can be proven by mollifying the distribution $v$. http://mathoverflow.net/questions/90521/spectrum-of-banach-algebras/90529#90529 Comment by Phil Isett Phil Isett 2012-03-08T04:45:45Z 2012-03-08T04:45:45Z Just to add: the fact that $0$ compactifies this space is the Riemann Lebesgue lemma. http://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula Comment by Phil Isett Phil Isett 2012-02-26T06:48:25Z 2012-02-26T06:48:25Z Poisson summation is the Fourier inversion formula for the circle in disguise. See <a href="http://mathoverflow.net/questions/89504/quick-computation-of-the-pontryagin-dual-group-of-torus" rel="nofollow" title="quick computation of the pontryagin dual group of torus">mathoverflow.net/questions/89504/&hellip;</a> and Darsh Ranjan's answer below. http://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula/14580#14580 Comment by Phil Isett Phil Isett 2012-02-26T06:44:32Z 2012-02-26T06:44:32Z One should be careful to note that being supported on ${\mathbb Z}$ is not enough to conclude $C(x) = \sum_n a_n \delta(x)$; for example, take $E(x) = \sum_n \frac{d}{dx} \delta(x - n)$. On the other hand, the property that $e^{2 \pi i \xi} \hat{C}(\xi) = \hat{C}(\xi)$ does imply the desired representation. Similarly, the integer periodicity of $\hat{C}$ follows from how $e^{2 \pi i x} C(x) = C(x)$. Also, instead of convolving, using the fact that $&lt; \hat{f}, C &gt; = &lt;f, \hat{C}&gt;$ is slightly more direct ($C$ and $\hat{C}$ are real and even). http://mathoverflow.net/questions/89462/recovering-a-measure-from-its-moments Comment by Phil Isett Phil Isett 2012-02-25T17:29:34Z 2012-02-25T17:29:34Z What do you mean by recover the measure? If you simply want to integrate a continuous function with respect to the measure, then the question reduces to approximating your desired continuous functions by polynomials effectively. You need a bound on the total variation to bound the error -- if your measure's positive, that's just the 0'th moment. In the positive case, you can also approximate plenty of discontinuous from above and below with continuous functions. http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces Comment by Phil Isett Phil Isett 2012-02-19T03:19:08Z 2012-02-19T03:19:08Z Maybe part of the conceptual issue is that you can define the &quot;tangent space&quot; of a set $K \subseteq {\mathbb R}^n$ (not a submanifold), but that different notions aren't equivalent. For example, you can try to think of smooth functions on ${\mathbb Q}^n$ as those which are restrictions of smooth functions on ${\mathbb R}^n$, then you will be able to discuss differential operators (as limits of difference quotients), but you won't have smooth curves into the space. On a related note, you might think that the $C^k$ tangent space is different from the $C^{k+1}$ tangent space. http://mathoverflow.net/questions/86344/interpolation-of-derivatives/86353#86353 Comment by Phil Isett Phil Isett 2012-01-22T04:31:00Z 2012-01-22T04:31:00Z This argument also shows that $f'$ is of H&#246;lder class $C^{0, 1/2}$ and in particular is continuous. So the conclusion is that $f$ belongs to $C^{1, 1/2}$, and in particular is continuously differentiable. http://mathoverflow.net/questions/41563/teaching-proofs-in-the-era-of-google/41568#41568 Comment by Phil Isett Phil Isett 2011-12-05T03:54:48Z 2011-12-05T03:54:48Z I think that one should have some sympathy for these people. I do not believe they are all really so aware that their habits actually hurt them (besides being extremely irritating to an educated observer). I think they often feel a lack of confidence that their thinking about math really will amount to anything. It's easy to have this attitude about math when you're a mathematician, but then how many of us were really so sincere when we endured courses in other, less appealing subjects? And at any rate, isn't it better to try to help rather than to foster a disdain? http://mathoverflow.net/questions/60457/elementaryshortuseful/82561#82561 Comment by Phil Isett Phil Isett 2011-12-04T04:22:50Z 2011-12-04T04:22:50Z You should definitely give the integral form of the remainder. This form somehow seems less well known despite being at least seven hundred thirty times as useful. http://mathoverflow.net/questions/81342/elementary-results-with-p-adic-numbers Comment by Phil Isett Phil Isett 2011-11-19T19:28:39Z 2011-11-19T19:28:39Z @Chris Wuthrich -- I think that proving that $({\mathbb Z}/p^n {\mathbb Z})^\times$ is cyclic is really neat and does motivate the p-adic thinking. (One should note that for p=2 this fact fails.) I'm aware you can easily see that the image of the homomorphism $x \mapsto 1 + px + p^2 x^2/2! + \ldots$ (defined for p≥3), ${\mathbb Z} \to ({\mathbb Z}/p^n {\mathbb Z})^\times$ consists of all those elements congruent to 1 mod p. It's not quite using the p-adics yet. How do you use the logarithm? Is it simpler? http://mathoverflow.net/questions/80724/taylors-theorem-and-the-symmetric-group/80725#80725 Comment by Phil Isett Phil Isett 2011-11-12T22:46:47Z 2011-11-12T22:46:47Z I was not familiar with this. Are there any places you would recommend for reading a systematic treatment of it?