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I was wondering whether it is possible to construct a function $\space f:\left( -1,1 \right) \rightarrow \mathbb{R} \space$ such that following holds:

(1) $\space f \in L^1 \left ( -1,1 \right )$

(2) $\space f \notin L^{\infty} \left ( -1,1 \right )$

(3) $\space$there is a weak derivative $\mathrm{D}f$ of the function $f$ on $ \left ( -1,1 \right )$

(4) $\space \mathrm{D}f \in L^1 \left ( -1,1 \right )$.

In other words: $\space f \space$ belongs to the Sobolev space $\space W^{1,1} \left ( -1,1 \right ) \space$ but doesn't belong to $L^{\infty} \left ( -1,1 \right )$.

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I know that this is not possible in terms of classical derivatives, but I'm curious whether it can be done weakly. Unfortunately my knowledge of Sobolev spaces is not deep enough to solve this problem.

I will appreciate any help.

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This situation can not happen in $1$D. This is because $W^{1,1}(I)$, where $I$ is an open interval, coincides with the space of absolutely continuous functions. As absolute continuity implies uniform continuity you can continuously extend your function to $\overline I$ by Heine-Cantor theorem, and easily infer the boundedness.

You can find a function with the desired properties in higher dimensions. Namely, you are looking for some set open, bounded $D$ with Lipschitz boundary such that $W^{1,1}(D)$ is not embedded in $L^\infty(D)$.

Sobolev embeddigs theorem say that, given $D\subset \mathbb{R}^n$

$$W^{k,p}(D) \hookrightarrow W^{h,q}(D)$$ if and only if $$\frac{kp-n}{p} \ge \frac{hq-n}{q}$$ with strict inequality if $q=\infty$ (unless $n=1$, in which the large inequality suffices).

In particular, as soon as $W^{k,p}$ magic number $(kp-n)/p$ is non positive, you do not have the embedding in $L^\infty$.

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  • $\begingroup$ It maybe is worth pointing out that the first part of your answer implies that the strict inequality required in the second paragraph for $q=\infty$ is not actually necessary when $n=1$. This is an exception though, as for $n>1$ there are functions that are in $W^{1,n}(\mathbb{R}^n)$ but not in $L^{\infty}(\mathbb{R}^n)$ $\endgroup$
    – Bananach
    Nov 21 '17 at 16:55
  • $\begingroup$ Ops, you are right. I started writing with in mind only the $n=1$ case, then got excited by the whole thing and went a bit deeper! $\endgroup$ Nov 21 '17 at 16:57
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No such functions exist, and this is a particular case of the Sobolev embedding theorem (https://en.wikipedia.org/wiki/Sobolev_inequality).

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