Let $f\colon\mathbb{R}\to\mathbb{R}$ be a continuous function with weak derivative (i.e. the derivative in the sense of distribution) in $C^1(\mathbb{R})$. Does this condition imply that $f$ is two times continuously differentiable (i.e. $f\in C^2(\mathbb{R}))$?
A sufficient condition for a function to be of class $C^2$ in the weak sense.
2 Answers
The answer is yes.
Lemma Assume that u \in \mathscr{D}'(X), where $X$ is an open interval in $\mathbb{R}$ and that u' = 0 (in the sense of distributions). Then $u$ is (can be identified with) a constant.
This result should be in most textbooks on distribution theory. I'll give a sketch of the proof for $X = \mathbb{R}$.
Step 1: Show that $\phi \in C^\infty_c$ is a derivative of a test function if and only if $\int_{-\infty}^{\infty} \phi\,dx = 0.$
Step 2: Choose a fixed $\psi \in C^\infty_c$ with $\int \psi = 1$, and let $C = \langle u, \psi\rangle$.
Step 3: Take an arbitrary $\phi \in C^\infty_c$ and choose $c$ so that $\phi - c\psi$ has integral $0$, i.e. $c = \int \phi$. Then \phi-c\psi = \chi' for some $\chi \in C^\infty_c$.
Put everything together: 0 = \langle u', \chi \rangle = - \langle u, \chi' \rangle = -\langle u, \phi \rangle + c\langle u, \psi\rangle = -\langle u, \phi \rangle + C\int \phi = -\langle u, \phi \rangle + \langle C, \phi\rangle.
for all $\phi \in C^\infty_c$. This shows that $u$ is in fact a constant $C$.
Returning to the original question. Let $g \in C^1$ be the distributional derivative of $f$ and define $ G(x) = \int_0^x g(t)\,dt $ Then $G \in C^2$, and it's clear that G' = g (both in classical sense, and in distributional sense). Hence G' = g = f' (in the sense of distributions), so by the lemma, $f = G + C$. In other words, $f$ can be identified with a $C^2$ function.
Let $g\in C^1(\mathbb{R})$ be a weak derivative of $f$. That means that we have \int_{\mathbb{R}}f\,\psi'=-\int_{\mathbb{R}}g\,\psi for all $\psi\in C^\infty_c(\mathbb{R})$ (the set of compactly supported $C^\infty$ functions). Since $g$ is continuous, the function $ h(t)=\int_0^t g $ is well-defined and satisfies h'(t)=g(t) for all $t\in\mathbb{R}$. So
\int_{\mathbb{R}}f\,\psi'=-\int_{\mathbb{R}}g\,\psi=-\int_{\mathbb{R}}h'\,\psi=\int_{\mathbb{R}}h\,\psi' for all $\psi\in C^\infty_c(\mathbb{R})$ ; as this set is invariant under taking derivatives and anti-derivatives, we get $ \int_{\mathbb{R}}(f-h)\psi=0 $ for all such $\psi$. Since $C^\infty_c(\mathbb{R})$ is dense in $L^2(\mathbb{R})$, $f=h$ almost everywhere; but both $f,h$ are continuous, so $f=h$. This shows that $g$ is the actual derivative of $f$, and thus $f$ is twice continuously differentiable.
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0Chris, you are right. I haven't thought about these such for such a long time. I'll change the answer to avoid approximations of the identity. – 2012-03-10