User connor mooney - MathOverflow

Answer by Connor Mooney for analysis question related to $L^p$ type inequalities

2013-05-23T17:52:11Z

Yes, I think it's true. Say we follow the line $(x,\alpha x)$ for $x > 0$ and $0 < \alpha < 1$. Both sides of the desired inequality have no linear part at $0$, so we examine the second derivatives. Keeping only dependence of the coefficients on $\alpha$, the second derivative of the LHS goes like $$(1-\alpha)(1+x)^{p-2}$$ and for the RHS goes like $$(1-\alpha)(1+x^{p-2}).$$ It is clear that the left is controlled by a constant independent of $\alpha$ times the right for all $x > 0$. The other regions can probably be taken care of similarly.

Answer by Connor Mooney for Sharpness of the Sobolev embedding theorem

2013-04-12T15:23:23Z

This response is closely related to my answer Here.

For the case $W^{2,n}$ we automatically get Holder regularity by applying Sobolev and then Morrey. However, we don't get Lipschitz. The following example "integrates" the counterexample to boundedness of $W^{1,n}$ functions.

Take a function $\psi$ supported on $B_2$ with $\psi$ linear and nonconstant on $B_1$ and $|\nabla \psi| < c$, and add dyadic rescalings together as follows. Consider $$u(x) = \sum_{i=1}^{\infty} h_i\psi(2^ix)$$ for some $h_i$ we will choose to get bounded $W^{2,n}$ norm but unbounded derivative. Note that $|D^2(h_i\psi(2^ix))|$ grows like $h_i2^{2i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{2,n}$ norm we want $$\sum_i h_i^n2^{in} < C.$$

To give unboundedness of the derivative we want $$\sum_i h_i2^{i} = \infty.$$

The natural choice for $h_i$ is $\frac{2^{-i}}{i}$, which gives the counterexample.

Remark: This function has size $~ 2^{-k}\sum_{i=1}^k \frac{1}{i}$ ~ $2^{-k}|\log\log(2^{-k})|$ at $|x| = 2^{-k},$ so a more explicit example might look something like $|x||\log\log(|x|)|$, which looks almost like a cone away from $0$ but the slope gets unboundedly high near $0$.

Answer by Connor Mooney for What goes wrong for the Sobolev embeddings at $k=n/p$?

2013-03-12T16:59:15Z

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. My short answer is that under rescaling by factor $\lambda$, derivatives scale by $\lambda$ and volumes by $\lambda^{-n}$, so integrating derivatives to the $n$ won't change under rescaling. The following examples illustrate how this affects embeddings.

As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)

As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.

Answer by Connor Mooney for a diameter-perimeter-area inequality for convex figures

2013-02-02T00:39:36Z

Here's a variation on Dmitri's idea that works as a counterexample: Take the rhombus with long diagonal $2$ and short diagonal $2\epsilon$. Then the area (LHS) grows like $\epsilon$, but the perimeter minus twice diameter goes like $$4\sqrt{1+\epsilon^2}-4$$ which grows like $\epsilon^2$.

The idea is that by moving out $\epsilon$ in the "center" rather than the edges will give quadratic growth of the perimeter rather than linear.

Answer by Connor Mooney for Alternate definitions of $C^{1,\alpha}$ and $C^{1,\alpha}(\bar{D})$ maps

2012-11-13T02:16:10Z

It is a good exercise to show that for a function $u: \mathbb{R}^n \rightarrow \mathbb{R}$, $C^{1,\alpha}$ regularity is equivalent to the following: there exists a linear function $l_x$ such that $$\|u-l_x\|_{L^{\infty}(B_r(x))} \leq Cr^{1+\alpha}$$ where $C$ is uniform. The forward direction is clear; for the other direction, observe that for points distance $r$ apart, the linear approximations differ by $Cr^{1+\alpha}$ nearby these points, so (up to changing $C$ by constants depending only on $n$ and $\alpha$) their slopes can differ by at most $Cr^{\alpha}$.

Actually, in the proof above I only needed that I can choose $C$ uniform on neighborhoods. This is crucial; as Pietro notes, we can get this property at a point (say 0) by taking an arbitrary bounded function and multiplying by $|x|^{1+\alpha}$, which clearly doesn't give $C^{1,\alpha}$ regularity away from $0$.

The equivalence of "pointwise $C^{1,\alpha}$ regularity" with uniform constant and $C^{1,\alpha}$ is the basis of most approaches I know to proving $C^{1,\alpha}$ estimates in PDE. This is usually done by estimating a solution with a linear function, rescaling, and improving the estimate geometrically, i.e. finding $r_k \rightarrow 0$ and $l_k$ so that $$\|u-l_k\|_{L^{\infty}(B_r(x))} \leq r_k^{1+\alpha}.$$

One can analogously define "pointwise $C^{k,\alpha}$" by approximation with $k^{th}$ order polynomials to get Holder estimates on higher derivatives.

Answer by Connor Mooney for Elliptic Differential Equations with rough boundary data

2012-10-13T13:02:13Z

For constant coefficient linear equations, the question of regularity and existence is nicely answered by representation formulas such as the Poisson kernel, as you mentioned. These formulas tell us that on the interior, solutions are nice and smooth and we have interior derivative estimates which get worse as we approach the boundary; in fact, if we only have continuous boundary data we expect the $k^{th}$ derivatives of harmonic functions to blow up at worst like $dist(x,\partial\Omega)^{-k}$. Furthermore, if we have some regularity of boundary and boundary data, we can only expect to have this much regularity (plus 2 derivatives) for $u$ at the boundary.

I'll try to convince you that this phenomenon holds for a large class of equations:

Another approach is to separate existence theory from regularity theory. A natural method for proving existence of viscosity solutions to fully nonlinear elliptic equations $F(D^2u,Du,u,x)$ with merely continuous boundary data is Perron's method, which "algorithmically" produces a solution continuous up to the boundary by taking higher and higher subsolutions. Perron's method will work (roughly) for any nondivergence equation with a maximum principle. Of course, F must satisfy some structure conditions (probably uniform ellipticity, some regularity in $Du$ like Lipschitz, monotonicity in u and some regularity in x).

Once a solution is produced via Perron's method, if we have continuous boundary data again we can only hope for some interior estimates which get worse as we approach the boundary. In general, the best regularity one can hope for is $C^{1,\alpha}$ on the interior, which follows from the Krylov-Safonov Harnack inequality. If $F$ satisfies some structure conditions, we can do better; This is exactly the case for concave uniformly elliptic equations. The Evans-Krylov theorem gives an interior $C^{2,\alpha}$ estimate $\|u\|_{C^{2,\alpha}(B_1)} \leq C\|u\|_{L^{\infty}(B_2)}$ for viscosity solutions. One sees that this estimate gets worse near the boundary by rescaling it in smaller balls near the boundary. I believe Nadirashvili has produced a counterexample to $C^{1,1}$ regularity for general uniformly elliptic equations.

So, we can always produce a $C^{1,\alpha}$ solution to a uniformly elliptic fully nonlinear equation continuous up to the boundary. The next question is, if the boundary and boundary data have some regularity, does $u$ inherit this regularity up to the boundary? To answer this question, I only know the method of continuity: establish a $C^{2,\alpha}$ apriori estimate up the the boundary and solve the equation by first solving a simpler equation and perturbing towards the equation of interest, using the apriori estimate to make sure I can perturb the whole way. For concave equations, with $C^3$ boundary and $C^3$ boundary data, one can do this with the help of Evans-Krylov and a boundary Harnack inequality (see Fully Nonlinear Elliptic Equations by Caffarelli-Cabre for details).

It is very interesting that $C^3$ boundary data is actually optimal for the Monge-Ampere equation $\det(D^2u) = f$; there are examples of solutions with $C^{2,1}$ boundary data whose normal second derivatives blow up near the boundary!

Hope this helps!

Answer by Connor Mooney for Alexandrov-Bakelmann-Pucci maximum principle

2012-09-04T02:06:29Z

In my view, the key point the ABP estimate is that it ties pointwise information (the PDE) to information in measure (the contact set). This is crucial to regularity theory for non-divergence equations; it enables the proof of a Harnack inequality (due to Krylov and Safonov), and $C^{1,\alpha}$ regularity for viscosity solutions of fully nonlinear elliptic equations. For divergence equations, whose solutions are already defined by integrals, it is a bit easier to obtain regularity.

Note that the technique of looking at a local maximum, we only use the equation at a single point. In the proof of ABP that I know, we use the equation at many points. Here's the theorem I'd like to discuss:

If $a^{ij}(x)u_{ij} \leq 1$ in $B_1$ (uniformly elliptic)and $u|_{\partial B_1} \geq 0$ then $|inf_{B_1}u| \leq C |A|^{1/n}$ where $A$ is the set where $u$ agrees with its convex envelope. (The contact set).

Another way to describe $A$ is the collection of points in $B_1$ for which the graph of $u$ can be touched below by a supporting hyperplane. It is reasonable to believe that, given a supersolution, it doesn't have "cone-like" behavior, i.e. few points in $A$, roughly because the Hessian would be too large and contradict the equation at these points. This observation is made rigorous by the Area Formula; all the planes of slope $c|inf_{B_1}u|$ touch $u$ by below somewhere in $B_1$, so the derivative mapping $Du(B_1)$ contains a ball of radius $c|\inf_{B_1}u|$. The Jacobian determinant of this mapping is $detD^2u$, which tells us locally "how much we curved", i.e. the measure of the set of slopes of supporting hyperplanes nearby. The PDE tells us that at each of these points in $A$, $det D^2u$ is not too large, completing the proof.

Prof. Ovidiu Savin uses a similar technique in his paper "Small Perturbation Solutions of Elliptic Equations" to prove an ABP-type estimate; in this version, we slide paraboloids up from below the solution until they touch and show that the set of contact points has measure controlled below by the set of vertices of these paraboloids. This gives information in measure that can be exploited to prove a Harnack inequality and $C^{\alpha}$ regularity in the setting of the paper.

I hope this helps!

Answer by Connor Mooney for Question About Harmonic Function Theory

2012-08-06T20:52:57Z

This is actually pretty cool. Superharmonic functions bounded below in $\mathbb{R}^2$ are constant, while there are nonconstant superharmonic functions bounded below in $\mathbb{R}^n$ for $n \geq 3$. Here is a proof that doesn't use complex analysis, and only uses that the fundamental solution in $\mathbb{R}^2$ ($\log(|x|)$) is unbounded from above and below, and the maximum principle.

Slide $u$ so that its minimum on $\partial B_1$ is $0$. Take the fundamental solution $f(x) = -\log|x|$, which is $0$ on $\partial B_1$. Since $u$ is bounded below and log is unbounded, $\epsilon f(x) < u(x)$ for $|x|$ sufficiently large (depending on $\epsilon$). By the maximum principle, $u(x) \geq \epsilon f(x)$ in $\mathbb{R}^2 - B_1$ for all $\epsilon$. Taking $\epsilon$ to $0$, we see that $u \geq 0$ outside $B_1$. But then, we see that $u$ takes its minimum in $\bar{B_1}$, and by the mean value inequality any superharmonic function with an interior minimum must be constant!

This result is false in higher dimensions. For a counterexample, just take the fundamental solution $|x|^{2-n}$ and cap it off above in $B_1$ by a paraboloid and smooth it out.

Answer by Connor Mooney for compactly supported harmonic functions

2012-08-06T20:32:50Z

The only such functions are $0$.

Compact support implies $$\int_{\Omega} \Delta u = 0.$$

This along with the subsolution hypothesis means that $\Delta u = 0$. Any compactly supported harmonic function is identically zero by analyticity.

Answer by Connor Mooney for Derivable functions & Sobolev spaces

2012-08-02T04:55:50Z

No. Consider $x^{3/2}$ on $(-1,1)$. It has Holder continuous but not Lipschitz derivative. See Gilbarg-Trudinger Ch. 4 for some related exercises (e.g. a C^1 function with derivative continuous but not Holder continuous for any $\alpha$).

Answer by Connor Mooney for Moser regularity proof avoiding John-Nirenberg lemma

2012-03-25T18:07:06Z

For the homogeneous equation, I have seen a proof of $C^{\alpha}$ regularity using an oscillation estimate based only on local boundedness and a Poincare-Sobolev inequality. Specifically:

Let u be a subsolution in $B_2$ satisfying $|(u \leq 0) \cap B_1| \geq \frac{1}{2}|B_1|$. Then $\sup_{B_{1/2}}u^{+} \leq \gamma \sup_{B_1}u^{+}$, where $\gamma < 1$ depends only on the ellipticity constants and $n$. (|.| denotes Lebesgue measure).

From there, one concludes that the oscillation of a solution decays by a fixed proportion each time we localize, which gives Holder regularity.

Answer by Connor Mooney for PDEs as a tool in other domains in mathematics

2011-12-10T23:00:05Z

Reilly used PDEs to give a very elegant proof that spheres are the only embedded hypersurfaces of constant mean curvature in $\mathbb{R}^n$.

Let $\Sigma$ be such a hypersurface, bounding a region $\Omega$. He showed that any solutions to the PDE $\Delta u = -1$ in $\Omega$ with $u=0$ on $\Sigma$ must be a second order polynomials with leading term proportional to $|x|^2$. One sees that level sets of this function are spheres by completing the square.

Answer by Connor Mooney for Roadmap to learning about Ricci Flow?

2011-08-10T03:22:14Z

In my view, the best place to start learning about Ricci flow is Hamilton's famous 1982 paper "Three-manifolds with positive Ricci curvature," modulo the short-time existence section. (DeTurck later came up with an easier way to prove short-time existence of solutions).

I like Hamilton's paper because it introduces the reader to the intense tensor computations involved in Ricci flow theory and requires only basic Riemannian geometry: Riemannian metrics, the Levi-Civita connection, covariant differentiation of tensor fields, parallel transport, geodesics, the exponential map, normal coordinates, curvature, the Hopf-Rinow theorem, variations of energy and Myers' theorem come to mind. Moreover, Hamilton proves the tensor maximum principle and illustrates the power of maximum principle techniques.

From there, you should be equipped to handle expository work on the Ricci flow. All of the sources mentioned above are great; I particularly like Simon Brendle's book "Ricci Flow and the Sphere Theorem" as a reference for convergence theory.

Applications of Berger's Curvature Estimate

2011-08-09T03:23:15Z

I'm interested in applications of the following estimate of Berger on the Riemann curvature tensor: Let $(M,g)$ be a Riemannian manifold of dimension $n \geq 4$, let $p \in M$, and assume that the sectional curvature $K(\pi)$ at $p$ lies in $ [\lambda,\Lambda]$ for all $2$-dimensional subspaces $\pi \subset T_pM$. Then for any orthonormal collection ${e_1,e_2,e_3,e_4}$ in $T_pM$, the Riemann curvature tensor $R(.,.,.,.)$ satisfies

$|R(e_1,e_2,e_3,e_4)| \leq \frac{2}{3}(\Lambda-\lambda).$

This is obtained by using the symmetries of the curvature tensor and the first Bianchi identity. One nice application is that pointwise strict quarter-pinching of sectional curvature implies positive isotropic curvature.

Does anyone know of other striking applications?

Answer by Connor Mooney for What are your favorite instructional counterexamples?

2011-07-30T16:38:19Z

A nice counterexample to the statement "$L^p$ convergence to $0$ implies pointwise a.e. convergence to $0$" is obtained by taking characteristic functions of length $\frac{1}{n}$ wrapping around the interval $[0,1]$. These integrate to $\frac{1}{n}$, but converge nowhere to $0$ because the harmonic series diverges.

A counterexample to the converse is easier: just take $f_n = n(n+1)\chi_{[\frac{1}{n+1},\frac{1}{n}]}$. These integrate to $1$ and converge everywhere to $0$.

Comment by Connor Mooney

2013-05-17T12:12:12Z

@Peter: So is the LHS along the diagonal.

Comment by Connor Mooney

2013-04-12T14:51:37Z

In the case $W^{1,n}$ how do you interpret $C^{-1,1}$?

Comment by Connor Mooney

2013-02-02T01:16:34Z

Yes, in restrospect I should have posted this as a comment on your answer- cheers!

Comment by Connor Mooney

2012-12-22T18:31:36Z

Interesting question! It seems to me that if we start with a convex obstacle (like $f = |x|^2$) and smoothly deform it to an obstacle with "2 humps" something would be irregular; the set where $\phi$ agrees with $f$ will start out as a sphere and pinch off into 2 spheres.

Comment by Connor Mooney

2012-11-13T04:11:49Z

@Analysis Now: Yes, $l_x(z) = u(x) + \langle Du(x), z-x \rangle$. Let $L = l_x-l_y$ with $|x-y| = r$. Then $L$ is linear, and $|L| \leq |l_x-u| + |u-l_y| \leq Cr^{1+\alpha}$ on say $B_{2r}(x)$. The oscillation of a linear function on a ball of radius $r$ is $r$ times the slope, hence $L$ must have slope bounded by $Cr^{\alpha}$, giving $Du(x)-Du(y)$ differing by at most $Cr^{\alpha}$.

Comment by Connor Mooney

2012-10-14T01:06:55Z

Yes, I agree. But I think an easier way to solve it might be to take a limit of solutions with mollified boundary data. The $C^{2,\alpha}$ interior estimate guarantees that the limit is a solution, and if one has (for instance) only $C^2$ boundary data then we at least have a gradient estimate up to the boundary that guarantees that the limit is continuous up to the boundary.

Comment by Connor Mooney

2012-10-03T04:43:01Z

The numerator is a harmonic function $u$ with $C^{1,\alpha}$ boundary data, so we do expect $C^{1,\alpha}$ regularity from the boundary; in particular, if we subtract the linear part of $u$ at $a$, we expect that the quotient tends to $0$. However, in the expression you give we haven't subtracted the gradient plane, which in general has a normal component (as in the nice examples of Alexandre below). Since $u$ and its tangential derivative at $a$ are $0$, the quotient will tend to $0$ if we travel along the boundary but will be nonzero if we approach from the interior.

Comment by Connor Mooney

2012-08-29T12:31:13Z

(In the above comment I mean solve the Dirichlet problem for Laplace with solution continuous up to the boundary).

Comment by Connor Mooney

2012-08-29T12:28:19Z

A quick remark on the solvability of Laplace's equation: We need $n = 2$ for the segment endpoint criterion to work, I think because complex analysis gives a nice way to construct a barrier. In higher dimensions, an exterior cone condition suffices; for a weaker condition see "Wiener criterion" in e.g. Gilbarg-Trudinger.

Comment by Connor Mooney

2012-08-06T21:58:08Z

@Yemon: Yes, what timur said. One way to show this is to use the interior gradient estimates for harmonic functions (which in turn follow from the mean value property, which holds in all dimensions).

Comment by Connor Mooney

2012-08-02T12:10:43Z

I should also mention here that $L^p$ convergence of $\{f_k\}$ to 0 implies that a subsequence converges pointwise a.e. to zero. To see this take a subsequence with $\int |f_k|^p < 2^{-k}$ (or any summable series) and use the monotone convergence theorem to conclude that $\int \sum |f_k|^p < \infty$.

Comment by Connor Mooney

2012-03-26T15:20:34Z

I hadn't seen the paper before, but that's exactly the proof I was thinking of. Thanks for the reference!