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..

  • $\begingroup$ 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

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.

  • $\begingroup$ So if $f' \in L^2$ say, then $f'$ is also the weak derivative. $\endgroup$ – asdad Sep 26 '14 at 11:55
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    $\begingroup$ 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

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