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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)

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1 Answer 1

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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.

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    $\begingroup$ Nice! This doubt is common. The immersion $W^{k,p} \hookrightarrow L^p$ means that for all $u \in W^{k,p}$ there exists a function $u^* \in L^p$ such that $u=u^*$ a. e. $\endgroup$ Sep 13, 2022 at 22:30
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    $\begingroup$ Note that the most regular representative of an element of $L^1_{loc}$ is equal to $\lim_{\epsilon \to 0^+} \frac{1}{m(B_\epsilon(x))} \int_{B_\epsilon(x)} f(y) dm(y)$ whenever this limit exists. It doesn't really matter all that much what it is where this limit doesn't exist. $\endgroup$
    – Ian
    Sep 13, 2022 at 22:31
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    $\begingroup$ Yes, both of your comments give a nice definition what I called "most regular representative". And I had this doubt too when I was first introduced to Sobolev spaces. $\endgroup$
    – F. Conrad
    Sep 13, 2022 at 22:34
  • $\begingroup$ When writing the most regular representative as @Ian did does this mean for $f\in L^{1}_{loc}$ the most regular representative of $f$ is $g$ such that $g(x) = \lim_{\epsilon\to 0^{+}} \frac{1}{m(B_{\epsilon}(x))} \int_{B_{\epsilon}(x)}f(y)dm(y)$? $\endgroup$
    – Chad Brown
    Sep 13, 2022 at 23:02
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    $\begingroup$ @ChadBrown That is what I meant, yes. $\endgroup$
    – Ian
    Sep 13, 2022 at 23:46

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