User pietro majer - MathOverflow

I know that you know...

2012-11-23T14:57:55Z

A bit unsure if the following vague question has enough mathematical content to be suitable upon here. In the case, please feel free to close it.

In several circumstances of competition, a particular situation of partial information occurs, usually described as "I know that you know that I know... something". We may distinguish a whole hierarchy of more and more complicate situations closer and closer to a complete information. E.g. :

$I_0$: I know $X$, but you don't know that I know.
$I_1$: I know $X$, you know that I know, but I don't know that you know that I know.
$I_1$: I know $X$, you know that I know, I know that you know that I know, but you don't know that I know that you know that I know.
.... &c.

For small values of $k$, I can imagine simple situations where passing from $I_k$ to $I_ {k+1}$ really makes a difference (for instance: you are Grandma Duck, and $X$ is : "you left a cherry pie to cool on the window ledge". Clearly, $I_0$ is quite agreeable position; $I_1$ may lead to an unpleasant end (for me); $I_2$ leaves me some hope, if I behave well, and so on). But, I can't imagine how passing from $I_6$ to $I_7$ may affect my strategy, or Grandma's.

Are there situations, real or factitious, concrete or abstract, where $I_k$ implies a different strategy than $I_{k+1}$ for the competitors? What about $I_{\omega}$ and, more generally, $I_\alpha$ for an ordinal $\alpha$ (suitably defined by induction)? How these situations are modeled mathematically?

Answer by Pietro Majer for A New Analytic Inequality

2013-05-15T09:34:38Z

Yes, indeed $$\frac{1}{2\pi}\int_0^{2\pi} f(e^{it})\overline{f(e^{-it})}dt= |f(0)|^2,$$ as you can check applying the Cauchy formula to the holomorphic function $f(z)\overline{f(\bar z)}$.

Answer by Pietro Majer for radon-nikodým property of $\ell^\infty$

2013-05-07T17:51:33Z

For dual spaces, there is an important characterization: $X^*$ has the Radon-Nikodym property if and only if $X$ is Asplund (its separable subspaces have separable duals). Of course, $\ell_1$ is not Asplund.

Answer by Pietro Majer for Lagrangian submanifolds

2013-04-23T07:32:52Z

It is very elementary; the graph representation refers to the cartesian product $P\times iP$ coming from the real vector space direct sum decomposition $\mathbb{C}^n = P\oplus iP\sim P\times iP$.

Here the direct sum decomposition is possible because $\dim _ \mathbb{R}(P)=n $ and $P\cap iP=0$, from the definition of Lagrangian subspace (thus any $z$ writes uniquely as $z=x+iy$ with $x$ and $y$ in $P$).

The second projection $\pi_2: P\times iP \to iP$, restricted to a subspace $Q$, has kernel $Q\cap iP$, so it is injective exactly for $Q\in U_P$, in which case $Q=\operatorname{graph}\ \pi_1({\pi_2}{|Q})^{-1}$.

For other details check e.g. the introductory pages of Hofer-Zehnder's book, Symplectic Invariants and Halmiltonian Dynamics.

Answer by Pietro Majer for Cauchy's left endpoint integral (1823)

2013-04-16T07:33:44Z

The following argument may not be the most direct one, but follows as a quick consequence of the characterization of Riemann integrability, which is perhaps the main result of the theory.

For a bounded function $f:[a,b]\to\mathbb{R}$ and for $\lambda >0$ let's denote $$J _ \lambda :=\{ x\in [a,b]\ : \ \limsup _{y\to x}f(y)-\liminf _{y\to x}f(y) \ge \lambda\}\ .$$ The key point is that for any $\epsilon > 0$ there are two partitions $P^*$ and $P _ *$ of the interval $[a,b]$, both with mesh less than $\epsilon$, such that the corresponding left sums $s_L(f,P^*)$ and $s_L(f,P _ * )$ differ for more than $\lambda\operatorname{meas} (J _ \lambda)-\epsilon$ (to construct $P^*$ and $P _ *$, start from a covering of $J _ \lambda$ by a finite collection of disjoint intervals of length less than $\epsilon$, and whose left endpoints belongs to $J_\lambda$) . As a consequence $$\limsup _ {|P|\to0} s_L(f,P) - \liminf _ {|P|\to0} s_L(f,P) \ge \lambda\operatorname{meas} (J _ \lambda)\ , $$ and we conclude that if $f$ is left integrable in the Cauchy sense, then $\operatorname{meas} (J _ \lambda)=0$ for all $\lambda >0$, that is, it is continuous a.e., hence Riemann integrable by the characterization.

Answer by Pietro Majer for Derivative of log determinant and inverse.

2013-04-10T21:47:18Z

[modified according to the clarification given in comments].

In general, for an invertible square matrix $\Sigma=\Sigma(\rho)$, differentiably depending on the real variable $\rho$, we have: $(\Sigma^{-1})'=-\Sigma^{-1} \Sigma' \Sigma^{-1}$, and $\big(\det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')\det(\Sigma)$, so that $\big(\log \det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')$.

Here, $\Sigma'$ is the matrix with entries $-h_{ij}e^{-\rho h_{ij}}$, the Hadamard product of the matrices $-(h_{ij})_{ij}$ and $\Sigma$; and since both $\Sigma^{-1}$ and $\Sigma'$ are symmetric, $\operatorname{tr} (\Sigma^{-1} \Sigma')$ is just their Frobenius scalar product. I do not see other simplifications, unfortunately.

Answer by Pietro Majer for Does $h$ have infinitely many isolated zeros?

2013-04-10T19:06:10Z

I assume $f$ was a real-valued analytic function on $(0,1)$, otherwise I do not understand the notation for $h$. But then no zero of $h$ can be isolated. Indeed, the zero set of $h$ is the union of the zero sets of its $r+1$ factors, and all of them vanish on some $r$ dimensional submanifold of $(0,1)^{r+1}$.

Answer by Pietro Majer for Weierstrass Approximation Theorem with an additional condition

2013-04-01T09:44:39Z

Sure. Assume by simplicity $a=0, b=1$. If $p$ is any uniform approximation of $f$ (say with $p(b) \ge f(b) )$ just add $x^N/N$ to $p$, with $N$ large enough.

Integral representation of higher order derivatives

2011-05-17T18:40:50Z

I'm quite curious about the following phenomena, that still puzzle me although I have a proof, and I'd be really glad if someone may shred some light, showing an interpretation or a generalization. I will sketch the computations at request, which consist on a manipulation of the integral formula of the remainder of the Taylor expansion.

1. Let $v\in C^\infty(\mathbb{R})$, vanishing at $0$ with some order $p\in\mathbb{N} _ +$ . In other words, the formal Taylor series of $v$ at $0$ belongs to the ideal $x^p\mathbb{R}[[x]]$ . Then, the function $w(x):=v(x)/x^p$ is also (extensible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x$ (say $x\ge0$) in terms of the derivatives of order $k+p$ of $v$ on $[0,x]$ as follows:

$$\frac{w^{(k)}(x)}{k!}=\frac{\int_0^x (x-s)^{p-1}s^k\ \frac{v^{(k+p)}(s)}{(k+p)!}ds}{\int_0^x (x-s)^{p-1}s^k\ ds} \ . $$ In other terms, the $k-$th Taylor coefficient of $w$ at $x$ is an integral mean of the $(k+p)-$th Taylor coefficients of $v$, weighted with a Beta distribution on $[0,x]$. This is quite clear if $x=0$ and $v$ is analytic there, and not immediately obvious in general, but has it a special meaning, or is it an instance of a more general principle?

2. Let $u\in C^\infty(\mathbb{R})$, and assume that the formal Taylor series of $u$ at $0$ belongs to $\mathbb{R}[[x^2]]$ . Then, the function $w(x):=u( \sqrt { x } ) $ is also (extendible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x^2$ in terms of the $2k-$ order derivatives of $v$ on $[0,x]$ as follows:

$$w^{(k)}(x^2)=\frac{\ (2x)^{-2k+1}}{(k-1)!\ }\ \int_0^x (x^2-t^2)^{k-1}u^{(2k)}(t) dt\ $$ (this may also be written as an equality relating Taylor coefficients by means of an integral mean).

3. There is also a more general statement for a function $w(x):=u(x^{1/p})$ for $p\in\mathbb{N} _ +$, assuming that the formal Taylor series of $u$ is in $\mathbb{R}[[x^p]]$; the $k-$th Taylor coefficient of $w$ at $x^p$ is then an integral mean of the $kp-$th Taylor coefficients of $v$, supported on $[0,x]$, with certain densities depending on $p$ and $n$ recursively defined. Is there a more general statement connecting analogously operations in $\mathbb{R}[[x]]$ and $C^\infty$ functions via integral means of their Taylor formal series?

Answer by Pietro Majer for Need help understanding Riesz representation theorem for Reproducing Kernel Hilbert Spaces

2010-08-25T12:41:59Z

This was initially meant to be a comment, but became too long. So, firstly, I'd warmly suggest you to adopt and follow a good book on the subject, rather than using on-line material: which is good as a reference for single proofs, but not often enough orgainzed into a whole theory. The basic material on Hilbert spaces is quite easy and intuitive, but needs several little results, that fit better in a book. You may like Rudin's exposition in Real and Complex analysis (chapt. 4), or Akhiezer & Glazman's Theory of linear operators in Hilbert space (chapt. 1), or Halmos' books Introduction to Hilbert Space and A Hilbert space problem book,... &c: there are of course several very good elementary books on the subject: the best is visiting a library and choose yours. That said, as to the Riesz duality theorem, you may follow this path:

Two non-zero linear functionals on a vector space have the same kernel if and only if they are scalar multiple of each other. This is easy linear algebra.
Then, in order to represent a bounded linear functional $f$ on a Hilbert space via scalar product with a vector $u$, that is $f=\phi_u$ where $\phi_u(x):=(x\cdot u)$, the key point is to find $v\in H$ such that $\ker f = \ker \phi_v$ (so the representing vector $u$ will be some scalar multiple of $v$). But $\ker\phi_v$ is by definition the orthogonal of $v$, so if such a vector $v$ exists, it has to be a generator of the line $(\ker f)^\perp$ because of the (nontrivial) duality relation $V=(V^\perp)^\perp$ for closed linear subspaces.
From the above you are led to consider the orthogonal decomposition you mentioned, and easily conclude.

However, orthogonal decompositions and projectors are a topic which is important in its own, and deserve a separate little study (see Zen Harper's comment above). You may start from the concept of metric projection on a convex set C of a Hilbert space H (here the completeness and the uniform convexity of the Hilbert norm ensure existence and uniqueness of the point of C minimal distance from a given point of H). Particularizing the convex to be a linear closed space $V$ you'll see (here is the magic of the Hilbert structure) that the corresponding metric projector on $V$ is a bounded linear operator, and gives you the orthogonal decomposition $H=V\oplus V^{\perp}$, and the above mentioned relation).

Answer by Pietro Majer for Second order difference implies differentiability

2013-02-26T19:19:09Z

Some remarks on the issue of weakening the continuity assumptions on $f$ to locally integrable, and about generalizing to other exponents of $|h|$.

Keeping track of the big O term in the first part of Terry Tao's proof, we may state it as an a priori bound on the Hölder norm of $f'$: Let $f\in C^1(\mathrm{R})$:

If $f$ satisfies, for some $0< \alpha \le 1$ and for all $x$ and $h$ $$|f(x+2h)-2f(x+h)+f(x)|\le C|h|^{1+\alpha} \qquad \qquad(1)$$

then its derivative is $\alpha$-Hölder, and in fact

$$ |f'(y)-f'(x)|\le \frac{2^\alpha}{2^\alpha -1 } C|h|^{\alpha}\ .\ \qquad \qquad(2) $$

The same conclusion holds if we only assume $f\in L^1_{loc}(\mathbb{R})$. Indeed, we may consider the standard approximation of $f$ by convolution, $f_\epsilon:=f*\phi_\epsilon$ with $\phi_\epsilon(x):=(1/\epsilon)\phi(x/\epsilon)$, for $\phi\in C^\infty_c(\mathbb{R}) _ + $ with $\int_\mathbb{R}\phi\ dx=1\ .$ Then the $f_\epsilon$ are in $C^{1,\alpha}(\mathbb{R})$ and satisfy the above hypothesis (1) with the same $C$, so the $f _\epsilon'$ are equicontinuous. By the Ascoli-Arzelà theorem, since $f _\epsilon \to f$ locally uniformly, this is sufficient to conclude that $f$ is also in $C^{1,\alpha}(\mathbb{R})$, with the same bound on the Hölder norm.

For continuous (or just $ L^1_{loc}$, as before) functions $f$, the above hypotesis with $\alpha=0$ defines the Zygmund class. It is well known that a function in this class is "quasi-Lipschitz", i.e. it has modulus of continuity of the form $Kt(\log |t| +1)$, but may fail to be of bounded variation, and may be differentiable at no point. An example is the Hardy-Weierstrass function, or also the Takagi or blancmange function (see e.g. this MO question). Note that, for $\sigma=1$ and $C=1$, the Takagi function is the largest $f\in C^0$ in the pointwise order, with $\mathrm{supp}(f)\subset[0,1]$ satisfing (1). Hence, it produces a (local) modulus of continuity for functions in the Zygmund class, which implies the "quasi-Lipschitz" property of these functions, since the Takagi function itself is dominated by a function $Kt (\log|t| +1)$.

For completeness: the condition (1) for $\alpha > 1$ became trivial: a (locally integrable) $f$ satisfying it is then linear.

Answer by Pietro Majer for Who invented the expression "pairwise different" and what is its advantage over "different"

2013-02-04T09:11:57Z

However "distinct" may have the weaker meaning of not all coinciding. So, in case I would therefore use pairwise, for clarity (see e.g. here), like in the other situations you listed.

The fact is that, in lack of a standard agreement on a definition or a notation, people is led to use more specific forms than needed. For instance: some people use $\subset$ for inculsion, some for strict inclusion. Result: some use $\subseteq $ and $\subsetneq $, to avoid any doubt. (Or, I once heard somebody -maybe myself, using the expression, for a topology not stronger than another, weakly weaker ).

Answer by Pietro Majer for Limit of functions and asymptotic behaviour

2013-01-22T15:10:45Z

Note that the quantity $\lambda p$ depending on the parameter $x$, as you define it for polynomials $p(t):=\sum_{k=0}^m a_k t^k$, that is $$(\lambda p )(x)=\sum_{k=0}^m a_k \frac{1}{1+k/x}\ , $$ can be extended to a positive, bounded linear operator $$\lambda :C^0([0,1])\to C^0_b([0,\infty))$$ taking the function $f\in C^0(I)$ to
$$(\lambda f)(x):=\int_0^1 xt^{x-1}f(t)dt\ .$$ In particular, $(\lambda f)(x)=f(1)+o(1)$ as $x\to+\infty$. For $f\in C^1([0,1])$ we also have, integrating by parts $$(\lambda f)(x)=f(1)-\int_0^1 t^x f'(t)dt=f(1)-f'(1)/x+o(x^{-1}),\qquad (\mathrm{as}\ x\to+\infty) \ .$$

$$*$$

You can also write, changing variable in the integral, ($t=e^{-\tau}$) $$(\lambda f)(x):=x \int_0^{+\infty} e^{-\tau x}f(e^{-\tau})d\tau\ .$$
Thus your $\lambda f$ is the Laplace transform of $f(e^{-\tau})$, times $x$. Thus you can profit of the asymptotic theory for Laplace transform of functions, developed in most textbooks on the subject.

Answer by Pietro Majer for Proving a sequence of integrals increases (iterated minimax distributions)

2011-11-12T12:23:54Z

Let $c>1$ be any fixed real exponent and let us denote, for any $x\in I:=[0,1]$, $f(x):=(1-x)^c$, and $g(x):=1-x^{\frac{1}{c}}$. So $f$ is a strictly decreasing homeomorphism of $I$ into itself, with inverse map $g$ and with the interior fixed point $0 < L< 1/2$. Also, since all even-order iterated of $f$ are strictly increasing, for any $x\in I$ and for any $n\in\mathbb{N}$, we have $$x\le L\quad\mathrm{iff}\quad f(x)\ge x\quad\mathrm{iff}\quad F_1(x)\le x \quad\mathrm{iff}\quad F_{n+1}(x)\le F_n(x)\ .$$ The sets $$A:=\{(x,y)\in I^2\ : \ F_{n+1}(x)\le y \le F_n(x)\}$$ and $$B:=\{(x,y)\in I^2\ : \ F_{n+1}(x)\ge y \ge F_n(x)\}$$ are therefore included in the squares $[0,L]^2$, respectively $[L,1]^2$, so we have $$-\int_0^L(F_{n+1}-F_n)dx=|A|$$ and $$\int_L^1(F_{n+1}-F_n)dx=|B|\ .$$ Moreover, $(x,y)\in A$, that is $ F_{n+1}(x)\le y \le F_n(x)$, if and only if $$F _ {n+1}(f(x))=f(F _ {n+1}(x))\ge f(y) \ge f(F_ n(x))=F_ n(f(x))\ ,$$ that is, $(f\times f)(x,y):=(f(x), f(y))\in B$. So $A=(g\times g)(B)\ .$ The map $g\times g$ has Jacobian determinant $g'(x)g'(y)=\frac{1}{c^2}(xy)^{\frac{1}{c}-1}\ ,$ and since $x\ge L$ and $y\ge L$ in $B$ $$|A|=|(g\times g)(B)|=\int_{B} g'(x)g'(y)dxdy\le \left( \frac{L^{\frac{1}{c}-1}}{c}\right)^2|B|\ . $$

Let's prove that the factor in front of $|B|$ is less than or equal to $1$. So we wish to show that $$L^{\frac{1}{c}-1}\le c\ . $$ Recalling that $L=(1-L)^c$, that is, $c=\frac{\log L}{\log (1-L)}$, this reduces to

$$\frac{(1-L)\log(1- L)} {L\log L}\le 1 $$

which is indeed true exactly for $L\le 1/2$. So we have $$\int_0^1 F_{n+1}dx -\int_0^1 F_n dx = \int_0^L(F_{n+1}-F_n)dx + \int_L^1(F_{n+1}-F_n)dx=$$ $$= -|A|+|B|\ge 0\, $$ as required.

Answer by Pietro Majer for Probability density that minimizes the sample range

2013-01-28T16:22:07Z

For $n$ i.i.d. random variables $X_1,\dots,X_n$ with (cumulative) distribution function $ F(x)= \int _ 0 ^ x f(t)dt\ $ we have $\mathbb{P}(\max_ {1\le i \le n} X _i \le x) = F(x)^n \ $, $\mathbb{P}(\min _ {1\le i \le n} X _ i \le x) = 1 - \big ( 1- F(x)\big) ^n \ . $ Therefore the random variables $\max _ {1\le i \le n} X _ i$ and $\max _ {1\le i \le n} X _i$ have density functions respectively $\big( F(x)^n\big)'$ and $\big(1 - \big (1- F(x)\big )^n\big)'\ .$ Therefore

$$ \mathbb{E} \big ( \max _ {1\le i \le n} X _ i - \min _ {1\le i \le n} X _ i\big ) =\mathbb{E} \big (\max_ {1\le i \le n} X _i \big )- \mathbb{E} \big (\min _ {1\le i \le n} X _ i\big ) = $$ $$\int_0^1 x\bigg( F ^ n + \big( 1-F \big)^ n \bigg)'dx = 1 - \int_0^1 \bigg( F ^ n + \big( 1-F \big)^ n \bigg)dx =\int _ 0 ^ 1 p\big(F(x)\big)dx \ , $$ after integration by parts; here $p(t)$ is the strictly concave polynomial function $$p(t):= 1- t^n -(1-t)^n\ .$$

$$*$$

We have therefore to minimize the strictly concave functional $$J(f):=\int _ 0 ^ 1 p\Big( \int _ 0 ^ x f dt\Big) dx \ , $$ on the convex compact subset (see 1 below) $K\subset L^1([0,1])$ of all positive, concave, norm-one functions $f$. A continuous strictly concave functional defined on a compact convex set $K$ attains its infimum on the extreme points of $K$. Here (see 2 below), the extreme points of $K$ are exactly the family of piecewise linear functions, for $\{f_c \}_ {0\le c\le 1} $,

$$ f_c(x):= \frac{2x}{c}\mathrm{\quad if\quad } 0\le x\le c\ , $$ $$ f_c(x):= \frac{2(1-x)}{1-c} \mathrm{\quad if\quad } c\le x\le 1\ .$$ The minimization problem is therefore reduced to the computation of the one-variable maximum: $$\min _ {f\in K} J(f )=\min _ {0\le c \le 1} J(f_c) =1-\max _ {0\le c \le 1} g(c)\ ,$$

where $g(c)= 1-J(f_c)$ is $$g(c) =\int_0^1 \bigg( c^{n+1}x^{2n} + (1-c)^{n+1}x^{2n} + c( 1- cx^2 )^n +(1-c)\big( 1- (1-c)x^2 \big)^n \bigg) dx \ ;$$ so $g$ is a polynomial in $c$ symmetrical w.r.to $c=1/2$. It should be true that $g$ is concave in $c$, so that one eventually finds $$\min _ {f\in K} J(f )=1 -\frac{1}{2^n(2n+1)} - \int_0^1 \bigg( 1- \frac{x^2}{2} \bigg)^n dx \ .$$

For $n\to +\infty$ this is (see 3 below) $$ 1- \sqrt{\frac{\pi}{2n}} + o\Big(\frac{1}{n}\Big)\ . $$

Plots of the polynomials $g(c)$, for $n=2,\dots,12$, from top to bottom.

$$*$$

More details.

1. (Compactness of $K$). Note that by definition of $K$, each $f\in K$ satisfies $$ 0 \le f(x) \le 2\ ,$$ for all $x\in [0,1]$ and, for any $\epsilon > 0$, $$\mathrm{Lip}\big(f,\ [\epsilon, 1-\epsilon]\big) \le \frac{2}{\epsilon} \ ,$$ which is sufficient to conclude that $K$ is compact in $L^1$ by the Ascoli-Arzelà and the dominated convergence theorems.

2. (Extreme points of $K$). Note that for any $f\in K$ and $c\in [0,1]$, we have $f=f_c$ if and only if $f(c)\ge 2 $. This implies that all these $f_c$'s are extreme points of $K$ (if $f_c$ is an arithmetic mean of two elements of $K$, at least one of them has value at $c$ not less than $2$, hence it is $f_c$ itself). Conversely, if $f\in K$ has $\|f\|_\infty < 2$, then one can find a non-zero function $h$ with $\int_0^1 hdx =0$ and such that $f\pm h $ are concave and non-negative, hence in $K$, so that $f =\frac{1}{2}\big((f+h)+(f-h)\big)$ is not extremal.

3. (Asymptotics). Incidentally, a quick asymptotics for $\int_0^1 \big( 1- \frac{x^2}{2} \big) ^ n dx$ as $n\to\infty$ comes from the dominated convergence theorem, recalling that $0\le (1+t/n)^n\le e^t$ as soon as $1+t/n\ge0$. Changing variable, $x:=\xi/\sqrt n$,

$$\int_0^1 \bigg( 1- \frac{x^2}{2} \bigg) ^ n dx = \sqrt{\frac{1}{n}} \int _ 0 ^{+\infty} \bigg( 1- \frac {\xi^2}{2n} \bigg) ^ n \chi _ {[0,\sqrt n]}(\xi) d\xi $$

$$= \sqrt{\frac{1}{n}} \int _ 0 ^{+\infty} e^ { -\xi^2/2 } d\xi \ \big(1+ o(1)\big) = \sqrt{\frac{\pi}{2n}}+ o(\frac{1}{n})\ .$$

Answer by Pietro Majer for Metrization of weak convergence of signed measures

2013-01-30T11:01:14Z

If $X$ is an infinite dimensional separable Banach space (like $C^0(\Omega)$, for a compact metric space $\Omega$ ), and $\{y _ j \} _ {j\ge 1}$ is a dense sequence in its unit ball, one considers the norm on $X^*$ defined by $$ ||| u |||:=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u, y _ j \rangle |\ ,$$ which is weaker than the dual norm $\| \cdot \|$, since $ |||u|||\le \| u\|$. On the space $ X ^ * $, the $|||\cdot|||$ norm topology and the $w^*$ topology differ, because the latter is not metrizable. However, they induce the same topology on any (dual norm) bounded subset.

rmk. If $Y$ denotes the dense linear subspace spanned by $\{y _ j \} _ {j\ge 1}$ we may also consider the weak topology $\sigma(X ^ *,Y)$ on $ X ^ * $: that is, the smallest TVS topology that makes continuous any evaluation map at points $y\in Y$, that is $ X ^ * \ni u\mapsto \langle u, y\rangle $. With this topology, $X ^ * $ is a locally convex, Hausdorff and first countable space, though metrizable, yet not normable as not locally bounded. So this is strictly weaker than the above $|||\cdot|||$ norm-topology; it is also strictly weaker than the weak $^*$ topology $\sigma(X ^ *,X)$, since their dual spaces are the evaluations at points of $Y$, respectively of $X$. Again, on $\|\cdot\|$-bounded subsets they induce the same topology: indeed, if $u_n\to u$ in $\sigma(X ^ *,Y)$ with $\|u_n\|\le R$, then $ ||| u - u _ n |||=\sum _ {j=1} ^\infty 2 ^ {-j} |\langle u - u_n , y _ j \rangle |$ also converges to $0$, since the terms of the series converge to $0$ while dominated by the series $ \sum _ {j=1} ^\infty 2 ^ {-j} (2R) $. Moreover, it is also true that $u_n\to u$ in $\sigma(X ^ *,X)$, because for any $x\in X$ and $y\in Y$ $|\langle u - u_n , x \rangle |\le |\langle u - u_n , y \rangle | + 2R\| x - y \|$, whence $\limsup _ { n \to \infty} |\langle u - u _ n , x \rangle | \le 2R\| x - y \| $, and the RHS can be made arbitrarily small.

Answer by Pietro Majer for Good examples of random variables whose image is not a measurable set?

2013-01-29T18:45:10Z

I like this example, which is as natural as can be an example with sets that are not Lebesgue measurable. Start from the Cantor function $f:[0,1]\rightarrow \mathbb{R}$, and consider $h(x):= x+f(x)$, which is a homeomorphism $[0,1]\rightarrow[0,2]$. On each interval on the complement of the Cantor set $C$ this functions is a translation. Therefore $|h([0,1]\setminus C)|=|[0,1]\setminus C|=1$. Thus $|h(C)|=|[0,2]\setminus h([0,1]\setminus C)|=1$. So there exists a non measurable subset $V$ of $ h(C)$; let $W$ be $h^{-1}(V)\subset C$. Finally, the homeomorphism $h$ maps this Lebesgue measurable set $W$ into the non-measurable set $V$.

Also note that any Lebesgue, non Borel set in $h(C)$ is mapped by the homeomorphism $h^{-1}$ into a Lebesgue, non Borel subset of $C$.

Answer by Pietro Majer for What is the difference between a function and a morphism?

2013-01-26T08:35:39Z

A main point is that considering objects and morphism (more generally than sets with some structure and functions that preserve it), is both a clarifying and fruitful abstraction. There are situations in which a category may be represented by a sub-category of the Set category (though it can be more complicated than what you suggested. If you are interested, you may start from the Yoneda lemma). But, in order to understand universal properties and to make categorical constructions it could even be counter-productive, like e.g. insisting in metrization with topological spaces, or coordinates with vector spaces.

Answer by Pietro Majer for When is a solution to an ODE determined by its average value?

2013-01-25T15:21:15Z

For scalar equations, $$\dot x= v(t,x),\qquad x(t)\in\mathbb{R}$$ that is $n=1$, the problem is well-posed, provided (1) the solutions of Cauchy initial value problem (IVP) at $t=0$ are defined at least in the interval $[0,1]$, (which is ensured, for instance, assuming a linear growth: $|v(t,x)|\le a|x|+b\ $), and (2) there are unique solutions to any IVP (this is the case, for instance, in the Lipschitz hypotheses, of course true for a smooth $v$). The reason is simply the fact that solution are totally ordered (a direct consequence of the uniqueness: they can't cross), that makes their mean on $[0,1]$ depend bi-continuously on the initial value $x(0)$.

If we drop the uniqueness of solutions of IVP's, and consider a scalar equation with only continuous $v$, like in the classical example $\dot x= |x|^{1/2}$, then there can be several solutions with the same mean value, e.g. on $[-1,1]$ (for instance, in the above example, there is a continuum of odd solutions, hence with zero mean value).

Incidentally, an interesting fact here is that for scalar equation with continuous nonlinearity, the mean on the interval can be used to parametrize an arc of solutions connecting two given solutions of the same IVB (a set which is called the "Peano brush", emanating from $x_0$ at $t=t_0$). In other words, the Peano brush is not only connected (which is true for any $n$), but also arc-wise connected, (which is in general false for $n > 1$). Also note that if we change the above example into the equation $\dot x= |x|^{1/2}\mathrm{sgn} x$, it still lacks uniqueness of IVP, but now it has a backward uniqueness, the reason being that $|x|^{1/2}\mathrm{sgn}$ is a monotone function. As a consequence, prescribing the mean value on $[0,1]$ does produce a well-posed problem for this equation.

For systems (i.e. $n > 1$) these problems (solutions of ODE with assigned value of a given functional $L$, like the average on $[0,1]$) have been widely studied I think in the $70's$. For special classes of continuous nonlinearities, such conditions may give well-posed problems where the usual uniqueness of IVP fails. For equations where the uniqueness of IVP is already ensured, like in your case, there can be good motivations for these problems (as I guess you already have). In both situations I suspect there is no general result, but interesting results may arise from particular equations. A first nice category for systems comes again from monotonicity. The corresponding theorem should be (I should check) :

Let $v:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be continuous, monotone (i.e., $(v(x)-v(y))(x-y)\ge0$ for all $x$ and $y$, e.g. the gradient of a convex function), with linear growth. Then, for any $p\in \mathbb{R}^n$ there is a unique solution to $\dot x=v(x)$ with $\int_0^1 x(t)dt=p$.

Answer by Pietro Majer for What is the characteristic property of surjective submersions?

2013-01-24T22:38:19Z

The reverse implication, as it is, is not true, for quite an obvious reason (though I think a local version of it should be true).

Start by any smoothly final map $f_0:X_0\rightarrow Y$ (e.g. any surjective submersion), and a smooth map $f_1:X_1 \rightarrow Y$ which is not a submersion. Then, the disjoint union $f:=f_0\sqcup f_1: X_0\sqcup X_1 \rightarrow Y$ is not a submersion, nevertheless it is still smoothly final. ( Indeed, for any smooth manifold $Z$ and any map $g:Y\rightarrow Z$, if $g\circ (f_0\sqcup f_1)=(g\circ f_0)\sqcup (g\circ f_1) $ is smooth, so is $g\circ f_0$, hence $g$ because $f_0$ is smoothly final.

It is true that a smoothly final map $f:X\rightarrow Y$ is necessarily surjective (note e.g. that the above construction $f_0\sqcup f_1$ was surjective). In fact, for any $y\in Y$ there exists a map $g:Y\rightarrow\mathbb{R}$ differentiable in $Y\setminus\{y\}$ and not in $y$ (e.g., a map supported in the domain of a local chart at $y$, that in a local chart is $\|\cdot\|$ near $0$). Then, clearly, if $f:X\rightarrow Y$ is not surjective, say because there is $y\in Y\setminus f(X)$, then $g\circ f$ is smooth though $g$ is not, so $f$ is not smoothly final.

Answer by Pietro Majer for Differential equations and axiom of choice

2012-11-12T21:46:33Z

At least for scalar equations $\dot x(t)=f(t,x(t))$, that is with a nonlinearity $f\in C^0(\Omega,\mathbb{R})$, defined on an open set $\Omega\subset \mathbb{R}^2$, the Zorn's lemma is not necessary: the order structure of $\mathbb{R}$ allows to select a preferred solution (actually, two)

Any IVP admits an upper and a lower solution, whose domains are maximal intervals. For $(t_0,x_0)\in\Omega$, define, for $t\in\mathbb{R}$ $$x ^ * (t):=\sup\{x(t)\ : x\in C^1(\mathrm{co}(t_0,t),\ \mathbb{R}),\ \mathrm{graph}(x)\subset\Omega, x(t_0)=x_0, \dot x(s)=f(s,x(s)) \}\ ,$$ (where of course $\sup\emptyset=-\infty$). Define $\mathrm{dom}(x ^ *)$ to be the connected component of $t_0$ in the set $\{ t: x ^ *(t) \in\mathbb{R} \}$. Then, $x ^ *$ is a solution of the ODE with IVC $x(t_0)=x_0$, maximally defined on the interval $\mathrm{dom}(x ^ *)$.

Asymptotic growth of a certain integer sequence

2010-08-28T20:16:50Z

Some time ago, while putting my nose in the Sloane's Online Encyclopedia of Integer Sequences, I came over the sequence A019568 defined as follows:

$a(n):=$ the smallest positive integer $k$ such that the set $\{1^n, 2^n, 3^n, ..., k^n\}$ can be partitioned into two sets with equal sum.

In other words, $a(n)$ is the smallest $k$ such that there is a choice of signs + or - in the expression $$1^n\pm2^n\pm\dots\pm k^n \qquad\qquad(1) $$ that makes it vanish. In order to show that this $a(n)$ is a well-defined integer (that is: that at least one such $k$ does exist), a simple observation gives in fact a bound $$a(n)<2^{n+1}.$$ Reason: $(1-x)^{n+1}$ divides the polynomial $$Q(x):=(1-x)(1-x^2)(1-x^4)\dots(1-x^{2^n})=+1-x-x^2+x^3-\dots +(-1)^n x^{2^{n+1}-1},$$ therefore, if $S$ is the shift operator on sequences, the operator $Q(S)$ has the $(n+1)$-th discrete difference $(I-S)^{n+1}$ as factor, hence annihilates any sequence that is polynomial of degree not greater than $n$. In particular, the algebraic sum (1) with the signs of the coefficients of $Q(x)$ vanishes (incidentally, the sequence of signs is the so called Thue-Morse sequence A106400, $+--+-++--++-+--+\dots$.

However, looking at the reported values of $a(n)$ for $n$ from $0$ to $12:$

$$2,\ 3,\ 7,\ 12,\ 16,\ 24, \ 31,\ 39,\ 47,\ 44,\ 60,\ 71,\ 79,$$

it looks like the growth of $a(n)$ is much below $2^{n+1}$ (I have a weakness for sequences that grow slowly, here's possibly the main motivation of this question).

Question: Does anybody have a reference for the above sequence? Can you see how to prove an asymptotics, or a more realistic bound than $a(n)<2^{n+1}$?

Answer by Pietro Majer for Examples of "exotic" induction

2013-01-22T13:19:52Z

The following example is not completely elementary, in that it requires the notion of sum of a family of non-negative real numbers (defined as supremum of the finite sub-sums), and some topology of the real line. But it allows a nice picture, and leads to a visualization of ordinal numbers as subsets of $\mathbb{R}$, so that it could be understood and appreciated by well-motivated high-school students, if you take the time to explain the problem, possibly skipping the technicality, and solving particular cases first. Moreover, it is close to the birth-place of the ordinal numbers.

The additivity of the length on the family of all left-closed, right-open intervals of the real line.

Let $P$ be the family of all left-closed, right-open intervals of the real line. Suppose $I=:[a,b)\in P$ is partitioned into a family elements $J_x:=[x,x')$ of $P$, that we may indicize by the left end-point $x\in S$, that is, $I=\cup_{x\in S} J_x $. Then,
$$|I|=\sum_{x\in S} |J_x|\ .$$

Proof: You may first consider and solve the case of finitely many intervals, and the case of a sequence of intervals accumulating at $b$, that gives rise to a telescopic sum. For the general case, the key point is that $S$ is well-ordered by the natural order of $\mathbb{R}$. This allows to prove
$$\Big|\bigcup_{x\in S\atop x \le u} J_x \Big|= \sum_{x\in S\atop x \le u} |J_x| \ .$$ by transfinite induction on $u\in S$.

Answer by Pietro Majer for Does this poset have a unique minimal element?

2012-12-25T18:15:28Z

[edit 01.15.2013] The following proof is still incomplete, but the main ideas should be useful.

[edit 01.17.2013] I filled the lacking point in the case 2, small but subtle, completing the proof, so I wrote it (even if in the meanwhile a complete proof has been posted).

Let me start with some general notions, that I believe are known, for a tree $T=(V,E)$ with finite, nonempty vertex set $V$ and edge set $E$. I will assume that $T$ is a minimal element of $\mathcal{AFT}$ only in the end.

For a path of length $n$ (number of edges) , $ (v_0 \dots v_n) $ in $ T $, let's define the centre of the path as the set $ \big\{v _ {\lfloor\frac{n }{2}\rfloor},v _ {\lceil \frac{n }{2}\rceil } \big\}$, consisting of one or two vertices (thus, either the middle vertex, if $n$ is even, or the middle edge, if $n$ is odd). Given two paths, there is a third path including the centres of both, and one endpoint of each. As a consequence, all paths of maximum length in a tree share the same centre, that we can therefore refer to as centre of the tree, $C(T):=\{v,v'\}$ (so this notation allows that $C(T)=\{v\}=\{v'\}$, a singleton, precisely whenever the diameter of $T$ is an even number, as remarked).

Since the image of a maximum length path via an automorphism of $T$ is still a maximum length path, whose center is the image of the center of the path, the set $C(T)$ is invariant for any automorphism $f$ of $T$ (thus, it is either a fixed point, or a couple of fixed points , or a 2-periodic orbit of $f$).

The centre determines a natural genealogy order in $T$; in particular, we can attach to any vertex $v$ its progeny, $\Gamma(v,T)$, the set of all vertices $x$ such that the minimal path from $x$ to the centre passes by $v$. Thus, e.g. this reduces to $\{v\}$ if and only is $v$ is a leaf; if $ C(T)$ is a singleton $\{v\}$, $\Gamma(v,T)$ is the whole vertex set $V$; if $ C(T)$ is an edge $vv'$, $\Gamma(v,T)$ and $\Gamma(v',T)$ are the components of $(V, E\setminus\{C(T)\}$.

For a vertex $x$, denote $( x^0 \dots x^n )$ the unique minimal path in $ T $ connecting $x$ to the center: $x^0\in C(T)$, $x^n=x$; here $n$ is the path distance from $C(T)$. It is also convenient to consider the nested sequence $ \Gamma(x^i,T) $, and the vector $\gamma(x,T):=(\gamma_0,\dots,\gamma_n)\in\mathbb{N}^{n+1}$ whose $i$-th entry is the cardinality $\gamma_i:=|\Gamma (x^i,T)|$ of each of these sets. Note that, since the center of a tree is automorphism-invariant, any automorphism of $T$ satisfy $\gamma(f(x),T)=\gamma(x,T)$. Among all leaves, consider those with minimum $\gamma(x,T)$ in the lexicographic order (with leading coefficient $\gamma_0$ ); we may shortly call them minimal leaves. For instance, the three leaves of the tree $E_7$ have labels $(3,2,1)$, $(4,1)$, and $(4,3,1)$, in increasing lexicographic order.

Let $x$ be a leaf of $T$, with father $x'=x^{n-1}$. We may denote $ T_x:=(V_x,E_x)$ the tree obtained deleting the leaf $x$ and the edge $xx'$. For a minimal leaf $x$ we may distinguish the following alternative:

1. $\mathrm{diam}(T_x)=\mathrm{diam}(T)$. This means that $T$ and $T_x$ share a maximum length path, so they also have the same center. Thus, for any $v\in V_x$ we have $\Gamma(v,T_x)=\Gamma(v,T)\setminus\{x\}$, and in particular the entries of $\gamma(x',T_x)$ are simply $\gamma_i(x',T_x) = \gamma_i(x,T) -1$ for $i=0,\dots,n-1$. As a consequence, any automorphism $f$ of $T_x$ fixes the whole path connecting $x'$ to $C(T_x)=C(T)$ (this follows by induction on $i$, arguing on the cardinality of the connected components $\Gamma (x^i,T_x)$: now $\Gamma (x^0,T_x)$ has strictly minimum cardinality among the components of $(V_x, E_x\setminus \{C(T)\})$, so $f(\Gamma (x^0,T_x))=\Gamma(x^0,T_x)$ and $x^0$ is fixed; then $x^1$ is fixed because $\Gamma(x^1,T_x)$ has strictly minimum cardinality among the components of the sons of $x^0$ in $\Gamma (x^0,T_x)$, and so on ). Therefore, $f$ extends to an automorphism of $T$ that fixes $x$. Clearly, this is not the case if $T$ is a minimal element of $\mathcal{AFT}$.

2. $\mathrm{diam}(T_x)=\mathrm{diam}(T)-1$. This means that $x$ is an end of every maximal length path of $T$.

Now, assume $T$ is a minimal element in $\mathcal{AFT}$, so that we are in case 2. Then, $C(T)$ is an edge, i.e. $\mathrm{diam}(T)$ is an odd number $2n+1$, and no vertex of the minimal path $(x^0,\dots, x^{n})$ connecting $x$ to $C(T)$ is a branching point. Proof: consider first the case of odd diameter of $T$, where $C(T)$ is an edge. Assume by contradiction that $\Gamma(x^0, T)$ is not a single path. Then, there are in it leaves $y\neq x$. Take among them the one with minimum vector $\gamma(y,T)$ in the lexicographic order. Now, since $y\neq x$, we have $\mathrm{diam}(T_y)=\mathrm{diam}(T)$, and we can argue with $y$ like in the previous case 1. The automorphism $f_y$ of $T_y$ fixes all $x^i$ because $( f_y(x^0),\dots,f_y(x^n) )$ are an end of a maximum lenght path in $T$, so they must end at $x$, which implies $f_y(x^i)=x^i$ for $0\le i \le n$. But then, $f_y$ also fixes the path $y^i$, for the same inductive argument used in point $1$ (start with the greater index $j$ such that $x^j=y^j$ and proceed looking at the cardinality of $\Gamma(y^{j+1} , T_y)$, observing that $f_ y (y ^ {j+1} ) \neq x^{j+1} =f_y(x^{j+1} ) $ because $ y^{j+1} \neq x^{i+1}$. This is a contradiction as usual, because $f_y$ does not fix the father of $y$, as already observed. For an analog reason, the case $C(T)$ is a vertex implies that $\Gamma(x,T)$, that is the whole $T$, has no branching vertices, that is, it is a path, which however is impossible because $T$ has no nontrivial automorphism.

Conclusion of the proof: Since $(x^0,\dots, x^{n-1})$ is part of a maximum length path in $T_x$, and $ \mathrm{diam}(T_x)=2n $ is even, the center of $T_x$ is a single vertex, namely the other endpoint $y^0$ of $C(T):=\{x^0,y^0\}$. If $f_x$ denote the unique nontrivial automorphism of $T_x$, we know that $f_x(y_0)=y_0$ (it's the center of $T_x$), while $y:=f_x(x^{n-1})\neq x^{n-1} $ (otherwise $f_x$ would extend to $T$). Therefore, $ (y^0, f_x(x^0),f_x(x^1),\dots,f_x(x^{n-1}))$ is the $n$-edges path connecting $y$ to $C(T)$, and since the $x^i$ for $i\ge0$ are not branching points, this path has no branching points too, with the possible exception of $y^0$. Actually, $y^0$ must be a branching point, otherwise the path $\xi:=(x^n,x^{n-1},\dots,x^0,y^0,y^1,\dots y^n)$, which has maximal length $2n+1$ in $T$, would have no branching point at all, and therefore would be $T$ itself, what however is impossible because $T$ has no nontrivial automorphism.

Next, we may consider the automorphism $f_y$ of $T_y$. As to $C(T_y)$, it is either $\{x^0\}$ (if $\xi$ is the unique maximum length path of $T$ and $ \mathrm{diam}(T_y)=2n $) , or $C(T_y)=C(T)$, (if there are other maximum length paths in $T$ and $ \mathrm{diam}(T_y)=2n+1 $). Therefore $f_y(y_0)$ is either $x^0$, or $x^1$, or $y^0$; however, only the last case is possible, because $y_0$ is a branching points and $x^0$, or $x^1$ are not. Thus, $( f_y(y^0), f_y(y^1),\dots, f_ y(y^{n-1}))$ is a path of length $n-1$ , starting from the branching point $y^0=f_ y(y^0)$, without other branching points. For the same reason, $T$ must contain a family of paths emanating from $y^0$, with no branching points, of all lengths between $1$ and $n$; in particular, a leaf $z$ attached to $y^0$ (and possibly other matter). The unique involution $f$ of $T_z$ exchanges the endpoints of $C(T_z)=C(T)$ (otherwise it would be extensible to a nontrivial automorphism of $T$), therefore bijects the whole $\Gamma(x^0, T)=\Gamma(x^0, T_z)$ with $\Gamma(y^0, T_z)$. This proves that $n=2$ and $T$ is $E_7$.

Answer by Pietro Majer for Is it an integer for all positive integer n ?

2013-01-17T17:12:34Z

More generally, $\frac{(nm)!}{n!(m!)^n}$ counts the partitions of a set of $nm$ elements into $n$ classes of $m$ elements.

Answer by Pietro Majer for Totally Geodesic Submanifolds

2013-01-16T23:09:36Z

What about $M$ an Euclidean sphere, and $N$ a great circle minus a point?

Answer by Pietro Majer for real symmetric matrix has real eigenvalues - elementary proof

2013-01-11T17:38:27Z

Another elementary proof, based on the order structure of symmetric matrices. Let me first recall the basic definitions and facts to avoid misunderstandings: we define $A\ge B$ iff $(A-B)x\cdot x\ge0$ for all $x\in\mathbb{R}^n$). Also, a lemma:

A symmetric matrix $A$, which is positive and invertible, is also definite positive (that is, $A\ge \epsilon I$ for some $\epsilon > 0\ $).

We may say, equivalently: if $A$ is positive but, for any $\epsilon >0$, the matrix $A-\epsilon I$ is not, then $A$ is not invertible. (A quick proof passes through the square root of $A$: $(Ax\cdot x)=\|A^{1/2} x\|^2 \ge \|A^{-1/2}\|^{-2} \| x\|^2$; one has to construct $A^{1/2}$ before, without diagonalization, of course).

As a consequence, $\alpha^*:=\sup_{|x|=1}(Ax \cdot x)$ is an eigenvalue of $A$, because $ \alpha^*I-A$ is positive and $\alpha^*I-A-\epsilon I$ is not (and $\alpha _ *:=\inf _ {|x|=1}(Ax \cdot x)$ too, for analogous reasons).

The complete diagonalization is then performed inductively, as in other proofs.

Answer by Pietro Majer for Series of quotients with perturbed denominator

2013-01-11T16:30:29Z

Since $\sum_ {n=1}^\infty \frac{a_n}{b_n } < \infty$ and $0 \le \frac{a_n}{b_n + \sigma/N}\le \frac{a_n}{b_n} $, we have that $\sum_{n=1}^N \frac{a_n}{b_n + \sigma/N} \to \sum_ {n=1}^\infty \frac{a_n}{b_n }$ as $N\to\infty$, just by dominated convergence.

Answer by Pietro Majer for Lebesgue constant as condition number of polynomial interpolation

2013-01-09T18:35:31Z

A direct consequence of the definition, yes. Note that $X_T$ is in fact defined on $C(T)$, with the same norm $\Lambda _ T$ w.r.to the infinity norm $\|u\|$ (viewing the vector $u:=(u_0,\dots,u_n)$ as an element of $C(T)$, taking $x _ j$ to $ u _ j$). Then $u=p{|_ T}$ so $p=X_T u$ by the uniqueness; and $\hat p = X_T \hat u $ by definition. Therefore $\|p\|\ge \|p{|_ T}\|=\|u\|$ and $\|p-\hat p\| = \|X_T (u-\hat u)\| \le\Lambda _ T \|u-\hat u\|$.

Answer by Pietro Majer for Volume of intersection of a convex polytope with an affine space.

2012-12-21T14:37:05Z

Following Andreas Blass's hints, one possibility to show the continuity is the following. Let $d:=n-k$ be the dimension of the cutting affine spaces.

First, let's say that a face of a polytope is a bad face if it spans an affine space containing a $d$ dimensional subspace parallel to the cutting space; otherwise, call it a nice face. If all faces of a polytope are nice, let's say it's a nice polytope, after all.

For a nice simplex, one checks that the function $f$ is continuous on the whole $a$-space.

By genericity, any polytope can be subdivided into nice simplexes, plus a number of simplexes whose bad faces are included into some bad faces of the polytope.

In other words, a generic triangulation produces a simplicial complex where no bad faces are introduced. This shows that the points of discontinuity of the function $f$ of the polytope are exactly the projection in the $a$-space of its bad faces.

Comment by Pietro Majer

2013-05-26T05:49:49Z

Dear Gerardo, this site is devoted to research matter. Please check the FAQ about what to ask here; you can also find links to other maths sites more suitable for elementary questions.

Comment by Pietro Majer

2013-05-14T18:57:41Z

Just think of diagonal operators on $l_2$, to get any sequence accumulating to 0 as spectrum, 0 being an eigenvalue or not.

Comment by Pietro Majer

2013-05-12T21:30:23Z

Of course a union of sets is a suitable direct limit in Set. Conversely, for some concrete categories it happens that the underlying set structure of a direct limit is the limit of the underlying structures in Set: in other words, the forgetful functor U:C$\to$Set preserves the limit. This is true when U is a left adjoint. These are elementary notions of the categorical language.

Comment by Pietro Majer

2013-05-10T10:18:52Z

Please check the FAQ about the scope of this site, which is devoted to research questions and not elementary topics (for your question: approximate $h:=\chi_{[0,+\infty)}$ by convolution with a mollifier $\phi _ \epsilon$, and compute the limit of $(f*h*\phi _ \epsilon)'$ in two ways) .

Comment by Pietro Majer

2013-05-08T15:44:07Z

(more details on Ricky Demer's hint: Sublevel sets of ||F|| can't be all nowhere dense subsets of X. )

Comment by Pietro Majer

2013-05-07T17:52:53Z

(I see the update just after posting)

Comment by Pietro Majer

2013-05-03T21:59:35Z

or $x\sin x$

Comment by Pietro Majer

2013-04-29T19:52:01Z

discrete valuation ring I suppose, en.wikipedia.org/wiki/Discrete_valuation_ring

Comment by Pietro Majer

2013-04-28T11:01:48Z

What will you do with this distance? Will you use it to define a topology (how?), or maybe you want to modify it so as to get a true distance out of it? In any case, if I think of a point with a positive distance from itself, the name that occurs naturally to my mind for this distance is, in analogy, psychodistance ;)

Comment by Pietro Majer

2013-04-27T10:04:49Z

This question is not clear. Weak or strong convergence in $l_2$ refer to a sequence of elements of $l_2$ (hence a family of double indices data), where are they?

Comment by Pietro Majer

2013-04-24T17:30:47Z

Actually, for a continuous strictly decreasing $f$ and for positive integers $n$ and $m$ one always has $m x_m < nm x_{nm}$.

Comment by Pietro Majer

2013-04-22T16:43:22Z

yes, the reasoning is right, but the question is off topic.

Comment by Pietro Majer

2013-04-22T16:19:04Z

The Dirichlet function, $\chi_{[0,1]\cap\mathbb{Q}}$, is Lebesgue integrable and continuous nowhere. However, it is equal a.e. to a continuous function (the zero function). For the sake of a stronger example, there are Lebesgue integrable functions that are continuous nowhere, even if we allow to modify them on a null set. For instance, the characteristic function of a measurable set $C\subset (0,1)$ such that $0 < |C \cup A| < |A|$ for any nonempty open set $A\subset (0,1)$ (there exists such a set $C$).

Comment by Pietro Majer

2013-04-21T16:28:11Z

sorry, would you clarify the question?

Comment by Pietro Majer

2013-04-20T09:42:33Z

Well, as stated, without prescribing e.g. initial conditions, it's not true that trajectories form a compact set in $C^0$, because of course there are solutions with any $x(0)\in\mathbb{R}^n$. On the other hand, any subset of solutions with $\|x(t_0)\|\le C$ for some $t_0$ and $C$ is relatively compact by the Ascoli-Arzelà theorem. So, aren't you missing some constraint or condition on the trajectories you are studying?