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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. $f\in C^2(\mathbb{R}))$?

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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.

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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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    Is it true that $C^\infty_0$ is invariant under derivatives? I thought that a necessary condition for the integral of a $C^\infty_0$ function to be $C^\infty_0$ is that the function integrate to zero. Otherwise when you define $\psi(t) = \int_{-\infty}^t \varphi(t)dt$, $\psi$ will not vanish at infinity. Maybe I'm missing something silly here.2012-03-10
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    I think you are right. What I'll do is change the proof to use C$^\infty$ functions of compact support. That class is invariant under both derivatives and anti-derivatives.2012-03-10
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    Martin, I think if the problem pops up for functions which vanish at infinity then a fortiroi it pops up for compactly supported functions. When you integrate a function with compact support, if the mass is positive then its integral will eventually be a constant function greater than zero (so it will stop growing eventually, but its support will not be compact). Take the usual normalized bump function as an example: the integral of that function is $0$ on $(-\infty,-1]$ and $1$ on $[1,\infty)$.2012-03-10
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    Chris, you are totally right. Now, if I'm not wrong (again!), my proof is only using derivatives, and that I'm fairly sure that derivatives of smooth compactly supported functions are again compactly supported smooth.2012-03-10
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    I think you need to show that approximations to the identity can be constructed as the derivatives of other $C^\infty_c$ functions to do it this way. I don't think that is the case since approximations to the identity are supposed to be essentially delta functions which need to integrate to 1 to make sense.2012-03-10
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    Chris, 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