5
$\begingroup$

Let $\Omega$ be a bounded domain in $\mathbb R^2$. By the Sobolev embedding theorem, if $k>\frac np$ (in this case $k>\frac 2p$) then

$u\in W^{k,p}(U) \implies u\in C^{k-[\frac 2 p]-1,\gamma}(U)$

for a certain $\gamma$; where $[\frac 2 p]$ is the integer part of $\frac 2p$.

If $k=1$ and $p=2$ then a function $u\in W^{1,2}(U)$ is not necessarily in $L^\infty(U)$, like the function $\log|\log|x^2+y^2||$ shows. (Let $U$ be the disk centered at the origin of radius $\frac 12$.)

What is an example that shows that if $k=2$ and $p=2$ then for a function $u\in W^{2,2}(U)$, $|\nabla u|$ is not necessarily in $L^\infty(U)$?

More generally, is there a standard way to construct such example from the previous one, that is $\log|\log|x^2+y^2||$?

$\endgroup$

2 Answers 2

5
$\begingroup$

Take $u$ in $\mathscr S'(\mathbb R^2)$ with $$ \hat u(\xi)=\frac{\mathbf 1(\vert\xi\vert\ge 2)}{\vert\xi\vert^3 \ln\vert\xi\vert},\quad \vert \xi\vert^2\hat u(\xi)=\frac{\mathbf 1(\vert\xi\vert\ge 2)}{\vert\xi\vert \ln\vert\xi\vert} $$ so that $u\in W^{2,2}$. However, $$ \vert\xi \hat u(\xi)\vert=\frac{\mathbf 1(\vert\xi\vert\ge 2)}{\vert\xi\vert^2 \ln\vert\xi\vert} $$ so that $\nabla u\notin L^\infty$.

Generally speaking in $n$ dimensions, you take $$ \hat v=\frac{\mathbf 1(\vert\xi\vert\ge 2)}{\vert\xi\vert^{n} \ln\vert\xi\vert},\quad\text{so that}\quad v\in W^{\frac n 2,2}\quad\text{but $v\notin L^\infty.$} $$ The first assertion is due to the convergence of $\int_2^{+\infty}\frac{dr}{r(\ln r)^2}$ and the second to the Fourier inversion formula and to the divergence of $$ \int_2^{+\infty}\frac{dr}{r \ln r}. $$

$\endgroup$
4
  • $\begingroup$ It may not be entirely obvious that $\sqrt{(-\triangle)}u \notin L^\infty \implies \nabla u \notin L^\infty$ $\endgroup$ Apr 26, 2012 at 13:09
  • $\begingroup$ OK, but you can modify the definition of $\hat u$ above by multiplying the numerator by $ \mathbf 1(\xi_n\ge \vert\xi\vert) $ so that $\partial_n u$ will not be bounded. $\endgroup$
    – Bazin
    Apr 26, 2012 at 16:08
  • $\begingroup$ Thanks. Do you know of a more explicit example or is this the only way? I am not familiar with Fourier inversion. $\endgroup$
    – Giuseppe
    Apr 26, 2012 at 19:10
  • $\begingroup$ Well, may I say that you should get familiar with Fourier analysis. For $u\in \mathscr S(\mathbb R^n)$, you have $$ \hat u(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx,\quad u(x)=\int e^{2i\pi x\cdot \xi} \hat u(\xi) d\xi, $$ and both formulas can be extended to temperate distributions, i.e. to the topological dual of $\mathscr S(\mathbb R^n)$. $\endgroup$
    – Bazin
    Apr 26, 2012 at 19:18
4
$\begingroup$

You can take as an example $$ u(x) = x_1^{k - 1} (\log \lvert x \rvert)^\beta: $$ if $\beta < 1 - \frac{1}{n}$, $u \in W^{k,n} (B_1)$ and if $\beta > 0$ then $D^{k - 1} u \not \in L^\infty (B_1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.