Sobolev embeddings; continuous representatives

Question

From various Sobolev inequalities, we know that for $N\geq 2$ if $\Omega=\mathbb{R}^N$ or if $\Omega\subset \mathbb{R}^N$ bounded with $C^1$ boundary, then for $p=N$ we have $W^{1,p}(\Omega)\subset L^{q}(\Omega)$ for all $q\in[p=N,\infty)$. However, for $N=1$, we have the result that if $u\in W^{1,p}(I)$ with $I$ being an interval, either bounded or unbounded, and $1\leq p\leq\infty$, then $u$ has a continuous representative, so essentially $W^{1,p}(I)\subset C^0(I)$ for all $1\leq p\leq\infty$. In particular, we have $W^{1,p}\subset C^0$ for $N=p=1$. But this is not true for $N=p=2,3,4...$? Is this correct?

Answer 1

Correct. You can find the example of a discontinuous function in $W^{1,2}$ for $N=2$ in this post log log

Answer 2

As pointed out by @Gio67, functions in $W^{1,n}(\mathbb R^n)$ in general are discontinuous and even unbounded. An example is $$ f(x) = \begin{cases} \log (1 - \log |x|) & \text{for } |x|<1, \\ 0 & \text{for } |x|>1, \end{cases} $$ which is unbounded, but it belongs to $W^{1,n}(\mathbb R^n)$ if only $n > 1$. It might be instructive to check it yourself and see where the assumption $n>1$ comes in.

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On the other hand, another generalization is true. Namely, functions in $W^{n,1}(\mathbb R^n)$ are continuous. Note that $W^{n,1}$ is a subspace of $W^{1,n}$ by Sobolev embedding.

And yet another generalization. If we assume that $\nabla f$ belongs to the smaller Lorentz space $L^{n,1}$ instead of $L^n = L^{n,n}$, continuity follows as well. See this short note by E. Stein.