Fundamental solution to $\frac{\mathrm{d}^2}{\mathrm{d}x^2}=I(x)$

Question

Here, $I(x)$ is the indicator function on the interval ([0,1]) (i.e. $I(x)=1$ if $x\in [0,1]$ and $0$ otherwise).

According to Wikipedia, we can find the fundamental solution $F(x)=\frac{|x|}{2}$ and then take its convolution with $I(x)$ to find a solution $f(x)$. Thus far, I understand.

What I don't get is how to differentiate that convolution twice in order to get $I(x)$. Here's my attempt: \begin{align*}(I*F)(x)&=\int_{-\infty}^{\infty}F(x-y)I(y)\mathrm{d}y\\ &=\int_{0}^{1}\frac{|x-y|}{2}\mathrm{d}y\\ &=\left[\frac{y}{4}(2x-y)\mathrm{sign}(x-y)\right]_0^1\\ &=\frac{1}{4}(2x-1)\text{sign}(x-1). \end{align*}

Thing is, if I differentiate this twice, don't I just get $0$?

Answer 1

Not sure I understand the step where you integrate; instead, I'd do \begin{align*} \int_{0}^{1} \frac{|x - y|}{2}\, dy &= \frac{1}{2} \int_{0}^{1} |y - x|\, dy = \frac{1}{2} \int_{-x}^{1-x} |y|\, dy \\ &= \frac{y|y|}{4}\bigg|_{-x}^{1-x} = \frac{(1 - x)|1 - x| + x|x|}{4} = \frac{x|x| - (x - 1)|x - 1|}{4}, \end{align*} which checks.