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I'm currently studying the properties of Sobolev Spaces in calculus of variations and functional analysis and was wondering if there is a function, that is $W^{1,p}$ but is unbounded on any open subset of $B_1(0)$.

Nhat
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  • Do you know an example of a function $f$ in $W^{1,p}$ that is unbounded in every neighborhood of a single point? – Umberto P. Dec 07 '15 at 18:41
  • How about $u(x)$ piecewise with $u(x) = \frac{1}{\vert x \vert ^\alpha}$ outside of $0$ and $u(0) = 0$. For certain $\alpha(n,p)$ we have $u \in W^{1,p}$ and its unbounded on any neighbourhood around $0$. – Nhat Dec 07 '15 at 18:46
  • Right. If you enumerate the rationals in $B_1(0)$ by ${q_k}$ and define $u_k(x) = 2^{-k} u(x-q_k)$, then $\sum |u_k|_{1,p}$ converges (as a real number), so that $\sum u_k$ converges in $W^{1,p}$. This function is unbounded at every rational point. This is the idea behind the answer given below. – Umberto P. Dec 07 '15 at 18:49
  • Thanks for the help, though I'm still trying to understand the function. What do you mean by "$\sum \Vert u_k \Vert_{1,p}$ converges as a real number"? $u_k$ is in $W_{1,p}$ so every summand is a real number. How does $\sum u_k$ then converge to something in $W^{1,p}$? Is it the $2^k$ in the denominater which gives us convergence? – Nhat Dec 07 '15 at 19:05
  • Never mind, I got it. Thanks! – Nhat Dec 07 '15 at 19:23

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One standard example is given in section 5.2.2 of Evans' book (p.260 in the second edition), and is $$u(x) = \sum_{k = 1}^{\infty} \frac{|x - r_k|^{-\alpha}}{2^k}$$ where $\{r_k\}_k$ is a countable, dense subset of your set $U$. This function is in $W^{1, p}(U)$ if $\alpha < (n - p) / p$.