1
$\begingroup$

Let $f \in C^0([0,T])$ be such that $f'$ exists in the classical sense everywhere, but $f'$ may not be continuous. Is it true that $f'$ is the weak derivative of $f$ too, if it exists?

I know this is true if $f \in C^1$, but I don't have that..

$\endgroup$
  • $\begingroup$ Theorem 7.21 in Rudin's "Real and Complex Analysis" states that if $f$ is differentiable everywhere and $f' \in L^1$, then $f(x) - f(a) = \int_a^x f'(t) \, dt$ for all $x$, so that $f$ is absolutely continuous, which implies that $f$ is weakly differentiable. It is easy to generalize this to the case $f' \in L^1_{\rm loc}$. But if $f' \notin L^1_{\rm loc}$, then all bets are off. $\endgroup$ – PhoemueX Sep 26 '14 at 13:07
0
$\begingroup$

In order for $f'$ to be the weak derivative of $f$ (i.e., weak derivative in the sense of distributions) the integral $$ \int_0^1 \varphi(x)\, f'(x)\,dx, \quad \varphi\in C^\infty_0(0,1), $$ should make sense, and although $f'$ is not in general locally $L^1$, the above is alternatively defined to make sense as $$ -\int_0^1 \varphi'(x)\, f(x)\,dx, $$ which has the same value with the first integral if $f'$ is continuous.

However, $f'$ is not necessarily a strong derivative of $f$ (strong is weaker than classical but stronger than weak), as strong derivatives belong to $L^p$ spaces.

$\endgroup$
  • $\begingroup$ So if $f' \in L^2$ say, then $f'$ is also the weak derivative. $\endgroup$ – asdad Sep 26 '14 at 11:55
  • 1
    $\begingroup$ If $f\in L_{\mathrm{loc}}^1$, then $f'$ always defines a weak derivative. So what you say is true. $\endgroup$ – Yiorgos S. Smyrlis Sep 26 '14 at 11:59

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.