I recently took an exam where I had to solve the following question:
Find the second distributional derivative of the function
\begin{equation} u(x) = \begin{cases} 1 - |x|, \quad |x| < 1, \\ 0, \quad \quad \quad \; \;|x| > 1 \end{cases} \end{equation}
Here's my solution:
Note that $u(x)$ is the tent function,
\begin{equation} u(x) = \begin{cases} 1 - x, \quad 0 \leq x < 1, \\ 1 + x, \quad -1 < x \leq 0, \\ 0 \quad \quad \quad \; |x| > 1 \end{cases} \end{equation}
We show that the second distributional derivative, $D^{2}u = 0$ almost everywhere. Let $\phi$ be a test function (compactly supported, continuous function) in $C^{\infty}_{c}(\mathbb{R})$. Since $u$ is a locally integrable function, $u$ defines a distribution by its action on the set of all test function. Hence,
\begin{align} (Du)(\phi) := \langle Du, \phi \rangle & = - \langle u , \phi' \rangle \\ & = - \int_{-\infty}^{\infty} u(x) \phi'(x) dx \\ & = - \Bigg [ \phi(x)u(x) \Bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u'(x) \phi(x) dx \Bigg ] \\ & = \int_{-\infty}^{\infty} u'(x) \phi(x) dx, \quad \quad \; \text{since} \; \phi(x) \in C_{c}^{\infty}(\mathbb{R}) \\ & = \int_{-1}^{0} (1) \phi(x) dx + \int_{0}^{1} (-1) \phi(x) dx \end{align} where, \begin{equation} f(x) = \begin{cases} -1, \quad \quad \; \; 0 \leq x < 1, \\ 1 \quad -1 \leq x < 0, \\ 0 \quad \quad \quad |x| > 1, \\ \end{cases} \end{equation} in the sense of distribution. This shows that $Du = f(x)$. By a similar integration by parts calculation, we can show that, $D^{2}u = D(Du) = Df = 0$ almost everywhere:
\begin{align} (D(Du))(\phi) := \langle Df, \phi \rangle & = - \langle f , \phi' \rangle \\ & = - \int_{-\infty}^{\infty} f(x) \phi'(x) dx \\ & = - \Bigg [ f(x)u(x) \Bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f'(x) \phi(x) dx \Bigg ] \\ & = \int_{-\infty}^{\infty} f'(x) \phi(x) dx, \quad \quad \; \text{since} \; \phi(x) \in C_{c}^{\infty}(\mathbb{R}) \\ & = \int_{-1}^{0} (0) \phi(x) dx + \int_{0}^{1} (0) \phi(x) dx = 0 = \langle Df , 0 \rangle \end{align}
This proves the claim.
I scored 4/10 on this problem, and I can't seem to figure out my mistake. Suggestions?
Yes, $D\phi=f$, where $f$ is what you said. It's clear that $f'=0$ almost everywhere, but that does not imply that the distributional derivative $Df$ vanishes! It doesn't. (It would be right if $Df$ were defined by a locally integrable function...)
Hint regarding a correct calculation of $Df$: Say $g=\chi_{[0,1]}$ (so $g'=0$ almost everywhere). If $\phi$ is a test function then $$-\int g\phi'=-\int_0^1 \phi'=\phi(0)-\phi(1).$$So $Dg=\delta_0-\delta_1$.