For $\beta\in N_{+}$, $d\in N_{+}$ define the Sobolev space $\mathcal{W}^{\beta,\infty}([-1,1]^d)$ of functions $f:[-1,1]^d \to \mathbb{R}$ such that $f$ and its weak partial derivatives up to order $|\beta|$ are elements of $L^{\infty}$-the space of essentially bounded functions.
Sobolev Imbedding Theorems (such as Thrm 4.12 of Adams and Fournier, 2003) seem to indicate that functions in this Sobolev space are continuous, $\mathcal{W}^{\beta,\infty}([-1,1]^d) \subset C^0([-1,1]^d)$.
Question:
I have ran across what seems to be a counterexample to this result.
Let $\mathbb{Q}$ denote the rational numbers, and consider the function $\boldsymbol{1}_{\mathbb{Q}}(x)$ which is equal to one if $x\in \mathbb{Q}\cap [-1,1]$ and is equal to zero if $x\in [-1,1] \setminus \mathbb{Q}$. Clearly this function is bounded so it is in $L^{\infty}$ and it is well known that this function has a weak derivative $v(x)=0$.
It follows that this function should be an element of $\mathcal{W}^{1,\infty}([-1,1])$, however it is not continuous which seems to be a violation of $\mathcal{W}^{\beta,\infty}([-1,1]^d) \subset C^0([-1,1]^d)$.
Can someone explain to me what is going on here? Is this a counterexample or is one of my conclusions wrong?
(Additional references on this topic would also be much appreciated)
That is a fine technicality of $L^P$ ($p=\infty$ included) spaces. Technically, you are using equivalence classes of functions: $$ f \sim g \iff f=g \; \textrm{a.e.} $$ In your case $0 \sim 1_\mathbb{Q}$ and the Sobolev embedding theorem tells you that there is a continuous representative in this class of functions. In your case, the $0$ function.
From that perspective, you could take any $f \in C_c^{\infty}([0,1])$ and add $f+1_\mathbb{Q}$ to make it nowhere continuous, even though all your Sobolev embeddings apply.
It is therefore common to always refer to the "most regular representative" in an equivalence class of $L^p$-functions. See the comments to see what this means precisely.