User daniel spector - MathOverflow

Characterizing the Dual of $W_0^{s,p}$

2013-02-14T10:21:57Z

I am interested in literature/results characterizing the dual of the fractional Sobolev space $W^{s,p}(\Omega)$, where $\Omega \subset \mathbb{R}^N$ is open, bounded, and smooth, $0< s<1$, and $1< p<\infty$.

For example, we can characterize the dual of $W_0^{1,2}(\Omega)$ as follows. If $f \in W^{1,2}(\Omega)$ then there exist $f_i \in L^2(\Omega)$ such that (for $v \in W^{1,2}_0(\Omega)$)

$\langle f,v \rangle = \int_\Omega f_0v+\sum_{i=1}^N f_iv_{x_i}\;dx$,

and so we write $f = f_0 - \sum_{i=1}^N (f_i)_{x_i}$.

A similar characterization holds for the dual of $W_0^{1,p}(\Omega)$, where instead the $f_i$ are in $L^p(\Omega)$ (though I do not have a reference for this - the $W_0^{1,2}$ case can be bound in Evan's PDE book).

My question is do we have such a representation $f=f_0 + (-\Delta)^\frac{s}{2} f_1$, for $f_0,f_1 \in L^p(\Omega)$, or something like this?

Optimality of p-Lebesgue Differentiation Theorem for Sobolev Functions

2013-04-19T07:35:59Z

This is the third question in a series whose purpose has been to flesh out an example of the optimality of the p-Lebesgue differentiation theorem for Sobolev functions. This theorem says that for $f \in W^{1,p}_{loc}(\mathbb{R}^N;\mathbb{R})$,

$\lim_{r\to0} \frac{1}{r^{N+p}}\int_{B(0,r)} |f(x+h)-f(x)-\nabla f(x)h|^p\;dx = 0$

for $\mathcal{L}^N$ almost every $x\in \mathbb{R}^N$.

EDIT: The previously mentioned lower bound is true, but it is not infinite, and thus does not serve as a counterexample to the question at hand. Therefore, the real question is about the optimality of the p-Lebesgue differentiation theorem for Sobolev functions. Can one find $f \in W^{1,1}$ or $f \in W^{1,p-\epsilon}$ (and $f \notin W^{1,p}$) such that the above limit is plus infinity on a set of positive measure?

Finding a good ordering of $\mathbb{Q}$

2013-04-17T11:30:33Z

Oftentimes in density arguments we let $\{x_n\}$ be a dense sequence and this is sufficient to imply the desired result.

From a research question I am working on I have simplified the example/counterexample to the following problem, which I believe is perhaps a nice exercise in choice (and yet, I cannot make a good one).

Precisely, can one choose a "good" ordering of a dense set $\{x_n\} \subset (-1,1)$ and a "good" sequence $r_n>0$ such that

$\sum_n r_n <\infty$

and

$|\{x \in (-1,1):x\in B(x_n,r_n) \text{ for infinitely many } n\}|>0$

where I have used $|\cdot|$ to denote the Lebesgue measure on $\mathbb{R}$.

Answer by Daniel Spector for Interior regularity for elliptic equations

2013-03-17T07:45:59Z

Ok. So I thought a little more carefully about it, though I still have not seen the Lions-Magenes book, but I think $C^\infty$ boundary is related to the above result holding for more general spaces (for example, $D^{n+\frac{1}{2}}$ for $n \in \mathbb{N}$. If this is the case, the smoothness is probably needed for an extension as I mentioned above.

Is your corollary straight out of the book? Since the logical conclusion of the theorem you mentioned is

1) If $u \in H^\frac{1}{2}$ and $\Delta u$ is in a funny space, then $u|_{\partial \Omega} \in L^2(\partial \Omega)$.

and

2) If $\Delta u$ is in a funny space and $u|_{\partial \Omega} \in L^2(\partial \Omega)$ then $u \in H^\frac{1}{2}$.

That is, from my reading of the theorem (and thinking about Craig's comment) is that you do need the boundary data. However, $C^\infty$ should not be needed for the boundary, only smooth enough to extend an $H^\frac{1}{2}$ function to all of $\mathbb{R}^N$. Hence, the cube should be sufficient.

Boundedness of the derivative of the trace of an H^1 function

2013-01-08T12:27:45Z

As a research preface, this question is linked to a problem of increasing magnetism in Ginzburg-Landau equations that I have distilled for the purpose of getting to the bottom of this technical matter.

Suppose $u_n\in H^1(B_1)$ (actually, $u_n$ are smooth), where $B_1$ is the unit ball in $\mathbb{R}^N$, and that we know $||u_n||_{H^1} \leq C$, so that up to a subsequence, $u_n \to u$ in $H^1(B_1)$ weakly. What can we say about the boundedness of the quantity

$\int_{\partial B_1} \nabla u_n(x) \cdot n(x)\;d\mathcal{H}^{N-1}(x)$,

where $n(x)$ is the unit normal to $\partial B_1$?

In particular, is this quantity always finite under these hypotheses, or is there a counterexample that shows this blows up for a bounded subset of $H^1$?

Answer by Daniel Spector for existence of a minimum for a convex functional on a non-reflexive space

2013-03-06T08:49:00Z

An equivalent formulation of your question is to prove that

$f:X^\prime \to \mathbb{R}$ be written as the supremum of affine functions, i.e.

$f(x^\prime)=\sup_{i \in \mathbb{N}} \langle x^\prime,x_i\rangle+c_i$.

This is equivalent to weak star lower semicontinuity (cf Ambrosio, Fusco, Pallara Functions of Bounded Variation and free discontinuity problems). Perhaps you have some representation that implies this? You don't need the full strength of weakly star lower semicontinuity then (though they are equivalent), since if $x_n^\prime \to x^\prime$ weak star, then we have \begin{align} \liminf_n f(x_n^\prime) &\geq \liminf_n \langle x_n^\prime,x_i\rangle+c_i \newline &=\langle x^\prime,x_i\rangle+c_i \end{align}

and taking the supremum on the right hand side we obtain sequential lower semicontinuity with respect to weak star convergence, which along with your subsequence implies existence of a minimizer. What does your functional look like?

Answer by Daniel Spector for Fourier Coefficients and Hölder Continuity

2013-01-30T17:37:49Z

I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An Introduction) which is one place to understand the subtleties of this issue.

One thing I will mention is that the Sobolev embedding theorem implies sufficient conditions for Holder continuity. If, for example, $n^2 |\hat{f}(n)|^2$ is summable ($f \in H^1$), then $f$ is $C^{0,\alpha}$ for $\alpha<\frac{1}{2}$. More generally, you can find conditions based on the following idea:

$|f|_\alpha \leq \sum_n |\hat{f}(n)| |n|^\alpha = \sum_n |n|^{\alpha+\frac{1}{2}+\epsilon}|\hat{f}(n)| \frac{1}{|n|^{\frac{1}{2}+\epsilon}} \leq \sum_n |n|^{2\alpha+1+2\epsilon}|\hat{f}(n)|^2 \sum_n \frac{1}{|n|^{1+2\epsilon}}$

Therefore if $|n|^{2\alpha+1}|\hat{f}(n)|^2$ is summable then $f$ is Holder continuous of any order strictly less than $\alpha$.

Answer by Daniel Spector for How to define Laplacian on $L_2$

2013-01-28T19:10:57Z

I can think of two differing viewpoints, and though many would prefer the latter due to its simplicity inherited from the Hilbert structure, I prefer the former for its connections to Sobolev spaces and PDE.

I. The Laplacian can be defined in the sense of distributions, since the Laplacian of an $L^2$ function is a distribution. In particular, for $u \in L^2(\Omega)$ define

$<\Delta u, \phi> := \int_\Omega u \Delta \phi\;dx$

for $\phi \in C^\infty_c(\Omega)$.

This would then say that in one dimension, for example, $\Delta |x| = 2\delta_0$, where I have used $\delta_0$ to denote the Dirac mass at zero, a distribution/measure.

II. For $u \in L^2(0,1)$, $sin(n\pi x)$ and $cos(n\pi x)$ form a basis and we can write

$u(x)= b_0 + \sum_n a_n sin(n\pi x) + b_n cos(n\pi x)$,

and we have $\sum_n a_n^2+b_n^2 <\infty$

Then we can define, formally, $\Delta u := \sum_n (n\pi)^2(-a_n sin(n\pi x) - b_n cos(n\pi x))$. In general, $\Delta u$ will not make sense pointwise, unless we know that $\sum_n n^4(a_n^2+b_n^2) <\infty$ (but as above, we can write $<\Delta u, \phi> = \int u \Delta \phi\;dx = \sum_n (n\pi)^2 (-a_n a_n^\prime-b_nb_n^\prime)$

where $a_n^\prime$ and $b_n^\prime$ are the Fourier coefficients of $\phi$. Now this makes sense for any $u \in L^2(0,1)$ if $\phi$ satisfies $\sum_n n^4((a_n^\prime)^2+(b^\prime_n)^2)<\infty$.

Of course, II can be done in higher dimensions - I have only chosen one dimension to illustrate with a simple example the basis functions. I do comment that I is more general, since it does not require the Hilbert structure.

Answer by Daniel Spector for Norm of differential operator between Sobolev spaces

2013-01-20T08:38:37Z

Indeed, the norm is one.

To see this, fix a cutoff function $\phi \in C^\infty_c(U)$ (which we only need if $U$ is unbounded, to make sure the constructed functions are integrable) and define

$f_n(x):=\phi(x) \frac{sin(nx_1)}{n^m}$.

Then $f_n \to 0$ strongly in $W^{m-1,p}(U)$ and

$||\nabla^m f_n||_{L^p}= ||\phi(x) sin(nx_1)||_{L^p} + o(\frac{1}{n})$.

Therefore, $||\partial^\alpha f||_{W^{m-|\alpha|,p}} = ||\phi(x) sin(nx_1)||_{L^p} + o(\frac{1}{n})$,

and so in the computation of

$\sup \frac{||\partial^\alpha f||}{||f||}$,

plugging in $f_n$ we have the following lower bound

$\sup \frac{||\partial^\alpha f||}{||f||} \geq \frac{||\phi(x) sin(nx_1)||_{L^p}+o(\frac{1}{n})}{||\phi(x) sin(nx_1)||_{L^p}+o(\frac{1}{n})}$.

Combining this with the upper bound mentioned in the question, we obtain that the norm is in fact one.

Answer by Daniel Spector for Solution to differential equation

2012-12-29T18:04:56Z

The result you want is true because of existence and Uniqueness for second order non-linear ODE, for example see Boyce and Diprima.

Writing $y^{\prime\prime}=f(t,y,y^\prime)$ and verifying that $f$, $f_y$, and $f_{y^\prime}$ are continuous you can guarantee existence and uniqueness in a small interval of the initial condition. Here you need to see that the functions you have in the problem ($coth$ and $sinh$) are smooth.

Answer by Daniel Spector for Classical Derivative, Weak Derivative and Integration by Parts

2012-12-17T20:17:22Z

Suppose $f \in W^{1,1}_{loc}(U)$. Then no, since for such an $f$, we have that $Df$ exists and the approximate limit

$ap\lim_{y\to x} \frac{f(x)-f(y)-Df(x)(x-y)}{|x-y|} = 0$

exists for almost every $x$, while from assuming classical differentiability we have

$\lim_{y\to x} \frac{f(x)-f(y)-\nabla f(x)(x-y)}{|x-y|} = 0$

exists for every $x \in U$. In particular, the classical differential is a candidate for the approximate differential, and so $Df=\nabla f$ wherever the two exist, and hence in $U$ up to a set of measure zero.

http://www.encyclopediaofmath.org/index.php/Approximate_differentiability

Minkowski's Inequality for Integrals in Orlicz spaces

2012-12-12T11:50:21Z

EDIT: I have changed the question to have less parameters, fitting it into the context of Orlicz spaces.

Suppose $f:[0,\infty)\to[0,\infty)$ is convex and increasing, $f^{-1}:[0,\infty)\to[0,\infty)$ is concave and increasing and $\int_{B(0,\delta)} \rho^\delta(z)dz =1$.

Is it true that

$f^{-1}\left(\int_{B(0,h)} f\left(\int_{B(0,\delta)} |u(x,z)| \rho^\delta(z)dz\right)\;dx\right)\leq \int_{B(0,\delta)} f^{-1}\left(\int_{B(0,h)} f\left(|u(x,z)| \right)\;dx\right)\rho^\delta(z)dz$?

When $f(t)=t^p$, then $f^{-1}(t)=t^\frac{1}{p}$ and this is precisely Minkowski's Inequality for integrals, as the title suggests, and the proof uses duality in $L^p$. Does this theorem stretch to some class of $N$-functions $f$ ($\Delta_2$, etc.)?

Answer by Daniel Spector for Is there any known condition for the following property?

2012-12-07T09:12:50Z

There may be a name for this, but it seems like a strange condition. Such a function cannot take a constant value on any set of positive Lebesgue measure, otherwise the inverse image of that constant (having zero 1-D Hausdorff measure in the range) would have positive Lebesgue measure, and therefore infinite 1-D Hausdorff measure.

A good start might be to investigate the situation on maps $f:[0,1] \to \mathbb{R}$ with the Lebesgue measure in both places.

There is also a related notion, called Lusin's N property, which means $f$ takes sets of measure zero into sets of measure zero (as opposed to $f^{-1}$, as you desire). This is a quality of Lipschitz functions that Sobolev functions also inherit, and is necessary to satisfy the fundamental theorem of Calculus (along with being differentiable a.e., etc.).

Answer by Daniel Spector for ODE's without a Lipschitz condition

2012-12-05T18:57:12Z

The reason for a Lipschitz condition is to guarantee uniqueness, as a standard example of non-uniqueness when $f$ is not Lipschitz is

$f(x,y)=y^{1/3}$

with the initial condition $y(0)=0$.

Then $y=0$ and $y=cx^\frac{3}{2}$ can be checked as two solutions satisfying the initial condition for the appropriate choice of $c$.

Answer by Daniel Spector for A consequence of convexity

2012-11-29T14:02:00Z

In general, whenever you have a (separately) convex function $f :\mathbb{R}^N \to \mathbb{R}$, which means it is convex in each variable, this implies the function is locally Lipschitz. The fact that it is decreasing allows you to make some explicit calculations of the constant, as Brian Rushton does, but in general the inequality looks like this:

$|f(x)-f(y)| \leq \frac{N}{r} (\sup_{B_{2r}} f - \inf_{B_{2r}} f) |x-y|$

for all $x,y \in B_{r}$.

Answer by Daniel Spector for Divergence form Elliptic PDE Removable Singularity/Regularity Question

2012-11-28T12:31:37Z

I believe the following is a counterexample:

Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$,

and

$u^\prime(x) = 2H(x)-1$,

where $H$ is the Heaviside function, $H \in L^\infty \cap W^{1,2}_{loc}((-1,1)\setminus {0})$.

Thus $u$ solves $(\frac{1}{2} u^\prime)^\prime = H^\prime$,

which is the PDE in one dimension, with $H$ in the right space, $u$ is smooth away from the origin and $u \in W^{1,\infty}(-1,1)$, but $u^{\prime\prime}$ is only a measure and so not in $L^p(-1,1)$ for any $p>1$.

In particular, $u$ is not $C^1$, $u$ is not smooth at the origin, and $u^{\prime\prime} \notin L^p(-1,1)$.

More generally, the computation extends into more dimensions, as far as I can tell.

Answer by Daniel Spector for About a generalization of the Radon Nikodym Theorem

2012-11-25T09:10:40Z

If you assume you have proven the Radon-Nikodym theorem for measures $\mu,\nu$, $\nu <<\mu$, then for any $F$ in such a class, the measure $\mu_F,\nu_F$ defined by

$\mu_F(E):=\int_E F(x)\;d\mu$,

$\nu_F(E):=\int_E F(x)\;d\nu$,

then since $\nu_F << \mu_F$ (if your definition of absolute continuity is equivalent to $\nu(E)=0$ whenever $\mu(E)=0$), you can apply the Radon-Nikodym theorem for $\nu_F,\mu_F$ and achieve the desired result.

Answer by Daniel Spector for inequality for coupling of measures

2012-11-25T08:50:20Z

Here is what seems to me a good beginning, if the argument can be carried all the way.

We have that $dist(\mu_1,\mu_2):=\inf_c \sup_s f(s,c)$,

where $f(s,c):= c( { (\omega^1,\omega^2) \in X\times X : \omega^1_s \neq \omega^2_s } )$.

Then for any indices $ { s_i }$ for $i$ from $1$ to $r$, we have

$\sup_s f(s,c) \geq f(s_i,c)$, so that

$r \sup_s f(s,c) \geq \sum_{i=1}^r f(s_i,c)$,

and taking the infimum in $c$ we have

$r dist(\mu_1,\mu_2) \geq \inf_c \sum_{i=1}^r f(s_i,c)$.

But $c$ is a measure, and so we should be able to arrange the right hand side as $c$ of some set where the components differ in at most $r$ spots, the infimum of which can be bounded below by the difference of the measures.

For this to work, maybe the set in the definition of $rift$ should have ONLY $\omega^1_s \neq \omega^2_s$?

Answer by Daniel Spector for injection with sobolev space

2012-11-19T11:39:39Z

I think it would help to specify what you mean by subregion $\omega \subset \Omega$. In general, it is absolutely true for $\omega$ a lower dimensional object. For example, when $n=2$ and $u \in H^1(\Omega)$, you can expect that for most (Lebesgue almost every) slices the restriction $u|_\omega$ is absolutely continuous (and therefore continuous and bounded).

If you want a subregion of full measure, then the dimension and exponent are important. A nice example is a function like $|x|^a$, whose derivative is $a|x|^{a-1}$ and second derivative is $a(a-1)|x|^{a-2}$. Computing the $L^2$ norm of this second derivative near zero, we find that we require

$\int_0^1 r^{2(a-2)}\;r^{n-1}dr <\infty$

which is equivalent to saying $2(a-2)+n-1>-1$, so that we should have $a>2-\frac{n}{2}$. Therefore, if $n > 4$, the function $|x|^{-\epsilon}$ is in $H^2(B)$ but not in $L^\infty(B)$. Then, you can define

$u(x):= \sum_n \frac{1}{2^n} |x-x_n|^{-\epsilon}$,

where $(x_n)_n$ is a dense set and obtain a function which is unbounded in every open set and still in $H^2$, by the above calculation.

Also, YangMills comment is right, since we see when $n=4$ it should still be possible, but one should use logarithms.

Answer by Daniel Spector for Coupled first order pdes in two unknowns

2012-11-19T09:18:35Z

The only smooth solution on some smooth, open domain in $\mathbb{R}^2$ is the zero solution.

Consider the following:

If $f$ solving this equation were smooth, then we would require $f_{xy}=f_{yx}$, and so computing

$f_{xy} = xf+(x+y)xf_x$

and

$f_{yx} = yf+(x+y)yf_y$,

and equating these we would need

$xf+(x+y)^2x^2f = yf+(x+y)^2y^2f$,

and at every point for which $f \neq 0$, this would require

$x+(x+y)^2x^2 = y+(x+y)^2y^2$,

which is not true on any open set, and so the smoothness of $f$ implies $f \equiv 0$.

Answer by Daniel Spector for Showing a coercivity condition for this bilinear form

2012-11-17T19:14:11Z

I think the first thing to iron out is that $\Omega$ should be open, since compact would be closed and then even defining $H^1(\Omega)$ is not trivial. Maybe assume $\Omega \subset \mathbb{R}^N$ is open and $J,f$ being $C^1(\overline{\Omega})$.

However, this is not the issue you are interested in. You want to know the relationship between

$\int_\Omega \nabla u MM^t \nabla u\;dx$

and

$||\nabla u||_{L^2(\Omega) }$.

Given $M$, can you show that the eigenvalues of $MM^t$ are strictly positive? The ellipticity constant is just a statement on the eigenvalues of the matrix associated with the coefficients. So what you are looking to do is to prove that this matrix is strictly positive definite. I hope you can compute this successfully.

EDIT: The result you are looking for in this case is to show that $\xi MM^t\xi \geq c$ for all $\xi \in S^{N-1}$, which would imply the coercivity you are looking for. A priori, $MM^t$ positive definite means you can this inequality with $c=0$, but you need a little better if this is your approach. For example, if the eigenvalues of $MM^t$ are ${\lambda_i}_{i=1..N}$, with $\min_i \lambda_i = \tilde{\lambda}>0$, and these eigenvalues correspond to eigenvectors $x_i$ which form an orthogonal basis for $\mathbb{R}^N$, then this is true, as follows.

Any $x\in S^{N-1}$ can be represented as $x=\sum_{i=1}^N c_i x_i$, where $\sum_i c_i^2=1$, and we compute $MM^t x = \sum_i c_i MM^t x_i = \sum_i c_i \lambda_i x_i$, therefore $xMM^tx = \sum_i c_i^2 \lambda_i \geq \tilde{\lambda}$, where we have used $\sum_i c_i^2=1$.

This is only an example - you should verify that the eigenvalues correspond to orthonormal eigenvectors to do this, or adapt the proof to something which is close and true in your case.

Answer by Daniel Spector for Is there a good metric under which a sequence of compact sets can converge to an infinite dimensional set?

2012-11-15T16:42:42Z

Ok. Suppose $\mathcal{F}_n \subset L^2(\Omega)$ such that $\mathcal{F}_n$ is spanned by $n$ linearly independent "vectors" with bounded norm, which means $L^2(\Omega)$ functions such that $||f||\leq C$ for all $f \in \mathcal{F}_n$. Then being finite-dimensional subspaces these sets will be compact in the strong topology.

However, in general, the limit will almost never be compact, since in this case it will recover a ball in $L^2(\Omega)$ which is not compact in the strong topology. You are going to need to make some more assumptions on the relationship between the sets to recover some information on the limit.

Answer by Daniel Spector for the intersection of a sequence of measurable sets

2012-11-15T08:02:09Z

Perhaps if you consider the case where $\Omega_j$ are sets of finite perimeter you can recover the statement. A set $\Omega_j$ is of finite perimeter in $\Omega \subset \mathbb{R}^N$ if $\chi_{\Omega_j} \in L^1(\Omega)$ and $Per(\Omega_j,\Omega):=\sup_\phi \left\vert\int_\Omega \chi_{\Omega_j} div\; \phi\;dx \right\vert <\infty$ where the supremum is taken over $\phi \in C_c^1(\Omega;\mathbb{R}^N)$ such that $||\phi||_\infty \leq 1$.

In particular, balls are sets of finite perimeter, since we may integrate by parts to show that $\left|\int_\Omega \chi_B div\;\phi\;dx\right| = \left|\int_{\partial B} \phi \cdot n \;d\mathcal{H}^{N-1}\right| \leq \mathcal{H}^{N-1}(\partial B)$.

Then a general compactness result for sets of finite perimeter/$BV$ functions implies that if we have a sequence $\Omega_j$ with $\epsilon \leq |\Omega_j| \leq C$ and $Per(\Omega_j,\Omega) \leq C$ we may find a subsequence whose characteristic functions converge strongly in $L^1(\Omega)$, and using this subsequence it should be possible to choose another subsequence with the desired property (eventually this subsequence will converge to some limit $A$ with $|A|\geq \epsilon$ from the strong convergence in $L^1(\Omega)$, and therefore choosing $\Omega_{j_k}$ which are near the limit we can find a countable family with positive measure of the intersection.

Answer by Daniel Spector for A minimum problem of the CoV

2012-11-14T13:17:19Z

Perhaps you have simplified this from a multidimensional radial problem, which explains the constraint $\int_0^B u^k(t) t^{h-1}\;dt = C$, which if we view $u$ as a radial function in $h$ dimensions is equivalent to $||u||_{L^k(B_r)}=C$. Then this is equivalent to minimizing the functional

$E(u):= \frac{1}{|S^{h-1}|}\int_{B_R} \left(\sqrt{1+|\partial_r u(r)|^2} - a\right) u^{k-1}(r) r^{h-1}\;dr ds$

over radial functions in $W^{1,1}(B_r)$, and a priori you will not necessarily stay in this space, since there is a lack of compactness in $W^{1,1}$ (though later being radial may save things). Using the direct method, we can define

$C:= \inf E(u) = \lim_n E(u_n)$,

and so for $n$ large we have

$C + a \int_{B_r} u_n^{k-1}(r) r^{h-1}\;drds \geq \int_{B_r} |\partial_r u_n(r)| u_n^{k-1}(r) t^{h-1}\;drds $

But $\int_0^B u_n^{k-1}(t) t^{h-1}\;dt$ is bounded from the constraint and Jensen's inequality. This will give bounds on $u^k_n$ in $L^1$ and $\nabla (u_n^k)$ as well, so that $u^k_n$ converges, up to a subsequence, to some $u \in BV(B_r)$. Thus existence of a minimizer can be established, and then there are is the question of uniqueness and regularity to address. I hope this gives a good start to your question/encourages more thoughts on the subject.

Answer by Daniel Spector for Integral inequality

2012-11-14T12:39:13Z

Also, something should be attached to the "arbitrary" subsets. Do you mean measurable?

Comment by Daniel Spector

2013-04-29T11:05:02Z

Maybe. I need more characters, so I wrote this sentence. But I meant to say maybe to your vague question.

Comment by Daniel Spector

2013-04-25T06:37:06Z

Try taking the derivative of $f(c):=\int_\Omega |u-c|^p\;d\mu$, and then think about justifying it later (dominated convergence, etc). Then you can see why $c$ should be the average of $u$ when $p=2$, and what you might expect otherwise.

Comment by Daniel Spector

2013-04-24T07:37:06Z

In general, the $L^\infty$ norm can be controlled by the Sobolev norm within the right parameters, but the converse cannot be true. Sobolev functions have some nice properties of the derivatives, but $L^\infty$ (even continuous, H$\"o$lder continuous) can have pathologically bad derivatives.

Comment by Daniel Spector

2013-04-17T12:10:58Z

Thanks for the reference Sean. What I want to prove is false! Great to know now :)

Comment by Daniel Spector

2013-04-01T12:23:18Z

Symmetrization and ODE analysis.

Comment by Daniel Spector

2013-03-31T15:09:52Z

It seems you are right about needing a different constant - my apologies for sloppiness. See my edit for 2.

Comment by Daniel Spector

2013-03-16T16:38:53Z

Ok. Sorry for the delay. I read what you wrtoe more closely and apologize for missing the $0 \in \partial \Omega$. However, the question you mention is not unique, since in general, the solution $u=0$ is a regular solution. So maybe the question is not correct in asserting solvability in the right space versus the operator is bounded from one space to the other (and in fact, as you show, could be bounded on some functions not in this space).

Comment by Daniel Spector

2013-03-13T08:36:05Z

I guess I am saying that we do not have $\Delta u =0$ or $\Delta u_m=0$. In fact, we have $\Delta u = \delta_0$ and $\Delta u_m = \delta_{x_m}$, so the equation satisfied by $u,u_m$ is with a right hand side in $(C_0(\Omega))^\prime$, and therefore the standard estimates can not be used, and something else is needed.

Comment by Daniel Spector

2013-03-11T16:44:43Z

This is a sufficient condition, however, the applied mathematicians have demonstrated that it is not necessary.

Comment by Daniel Spector

2013-03-10T09:54:53Z

The (typical) definition of the norm in a dual space is $||x^\prime||_{X^\prime} := \sup_{x \in X, ||x||\leq 1} |\langle x^\prime,x\rangle|$, from which weak star lower semicontinuity is a consequence in general for norms in a dual space. A good reference for these ideas is Brezis Analyze Functionelle, which is available in English as "Functional Analysis, Sobolev Spaces, and Partial Differential Equations".

Comment by Daniel Spector

2013-03-06T13:44:57Z

I am not familiar with such a functional, so off-hand I cannot say. The norm on $\ell^1$ is weakly star lower semicontinuous, so if $I^T$ is continuous, then the answer is yes.

Comment by Daniel Spector

2013-03-05T14:24:54Z

Then again, for the function you mention, $\Delta u$ is not in $L^2(\Omega)$ either. The singularity at $0$ will mean the Laplacian exists in the sense of distributions and is a measure, which is not in the space of square integrable functions (i.e. $f=\delta_0$).

Comment by Daniel Spector

2013-03-04T11:40:04Z

@Andras There is no compact embedding on $\mathbb{R}^N$ but since the functions have zero boundary value they can be extended to a large ball where the embedding will hold, and therefore the restriction to $\Omega$ also embeds into $L^p$ compactly for $p<p^*$, as holds for smooth domains.

Comment by Daniel Spector

2013-03-03T19:56:09Z

My question seems dumb because it asks a basic question about solvability of the PDE for $P$ you mention. So there is the weak formulation. Are the hypothesis sufficient to apply Lax-Milgram for existence, or is something missing? (i.e. can one really apply the existence theory for Poisson's equation in this setting?)

Comment by Daniel Spector

2013-03-01T13:02:56Z

Ok. Now about solving this weak PDE with no boundary condition for $P$? What would you do?