I am currently learning about Sobolev spaces, and I am trying to build some intuition of weak derivatives. My current intuition is imagining the weak derivative of f as a function equal to f's derivative almost everywhere. However, this assumption assumes that f is differentiable almost everywhere. Is this always the case for weakly differentiable functions?


If $p>n$, then the function is differentiable a.e. and the derivative coincides with the weak derivative a.e.

If $p\leq n$ (or even $f\in BV$) the function is only approximately differentiable a.e.

Both results can be found in Evans & Gariepy, Measure theory and fine property of functions, section 6.1 (and if I recall correctly "a.e." can be replaced by "outside a set of zero $p$-capacity", which is slightly stronger).

To construct a counterexample to a.e. differentiability for $p$ strictly below $n$, consider a nonnegative function $\eta \in C^\infty (\mathbb{R}^n)$ with support in $B_1$ and with value $1$ on $B_{1/2}$, and enumerate the rationals as $\mathbb{Q}=\{q_i\}_{i\in \mathbb{N}}$. Choose a sequence $r_i\searrow 0$ to be specified later, and define $$f(x)=\sum_{i\in \mathbb{N}}\eta\left(\frac{x-q_i}{r_i}\right).$$ This is a dense sum of bumps with smaller and smaller support. By the scaling of the $L^p$ norms you can check that indeed $f\in W^{1,p}$, provided $\sum_{i\in \mathbb{N}}r_i^{\frac{n}{p}-1}<\infty$.

However, the support of $f$ is contained in $\bigcup_{i\in \mathbb{N}} B(q_i,r_i)$ which can be made as small as wanted by sending $r_i$ quickly to zero, therefore at most points the function is zero. On the other hand, $f$ has value at least $1$ on a dense set (and the same holds for any function in the same equivalence class), therefore it can not be differentiable where it attains value zero.

I couldn't come up with a similar counterexample for $f\in W^{1,n}(\mathbb{R}^n)$, but maybe a similar construction would work.

  • $\begingroup$ Isn't $1_\mathbb{Q}$ a counterexample? it's weak derivative is $0$, but even nowhere continuous. $\endgroup$ – David Lingard Jul 31 at 0:00

Here are 2 references on the topic: https://www.math.ucdavis.edu/~hunter/pdes/ch3.pdf https://en.wikipedia.org/wiki/Weak_derivative

  • $\begingroup$ the down vote is because this gives a counter example of: continuous+derivable a.e implies weakly derivable. $\endgroup$ – David Lingard Jul 30 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.