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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 $\log|\log|x^2+y^2||$?

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2 Answers 2

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 $$ \int_2^{+\infty}\frac{dr}{r \ln r}. $$

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It may not be entirely obvious that $\sqrt{(-\triangle)}u \notin L^\infty \implies \nabla u \notin L^\infty$ –  Willie Wong Apr 26 '12 at 13:09
    
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. –  Bazin Apr 26 '12 at 16:08
    
Thanks. Do you know of a more explicit example or is this the only way? I am not familiar with Fourier inversion. –  Giuseppe Apr 26 '12 at 19:10
    
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 $\mathscr S(\mathbb R^n)$. –  Bazin Apr 26 '12 at 19:18

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 $D^{k - 1} u \not \in L^\infty (B_1)$.

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