Dirac delta function proof I got the function
$$f(x)=e^{-|x|}$$
I want to show that $f''(x)=f(x)-2\delta(x)$ where $\delta(x)$ is the Dirac delta function.
I know that I can solve it with a known theorem but can I prove it without using it?
 A: Here is a heuristic answer:
\begin{align*}
f''(x)
&= \bigl( e^{-|x|} \bigr)'' \\
&= \bigl( - \operatorname{sgn}(x)e^{-|x|} \bigr)' \\
&= -2\delta(x)e^{-|x|} + (\operatorname{sgn}(x))^2 e^{-|x|} \\
&= -2\delta(x) + f(x).
\end{align*}
For a more rigorous proof, let $\varphi \in \mathcal{D}(\mathbb{R})$ be any compactly supported smooth function. Since we are dealing with derivatives of distributions, we may invoke the weak derivative to reformulate the question. Indeed, it is equivalent to showing that
$$ \int_{\mathbb{R}} f(x)(\varphi(x) - \varphi''(x)) \, \mathrm{d}x = 2\varphi(0). $$
Now, by performing integration by parts twice,
\begin{align*}
\int_{\mathbb{R}} f(x) \varphi''(x) \, \mathrm{d}x
&= \int_{-\infty}^{0} e^{x} \varphi''(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi''(x) \, \mathrm{d}x \\
&= \underbrace{\left[ e^{x} \varphi'(x) \right]_{-\infty}^{0} + \left[ e^{-x} \varphi'(x) \right]_{0}^{\infty}}_{=0}
 - \int_{-\infty}^{0} e^{x} \varphi'(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi'(x) \, \mathrm{d}x \\
&= \left[ -e^{x} \varphi(x) \right]_{-\infty}^{0} + \left[ e^{-x} \varphi(x) \right]_{0}^{\infty}
 + \int_{-\infty}^{0} e^{x} \varphi(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi(x) \, \mathrm{d}x \\
&= -2\varphi(0) + \int_{\mathbb{R}} f(x) \varphi(x) \, \mathrm{d}x.
\end{align*}
Therefore the desired claim follows.
A: Let $f(x)=e^{-|x|}.$
$$\frac{d |x|}{dx}=\text{sgn}(x)=2
\theta(x)-1,$$
Where $$\theta(x)=1, x>0;~ 0, x\le 0.$$
Further, $$\frac{d \theta(x)}{dx}=\delta(x).$$
So $$f'(x)=\frac{d e^{-|x|}}{dx}=-e^{-|x|} \text{sig}(x)=-e^{-|x|}~[2\theta(x)-1]$$
$$f''(x)=e^{-|x|}~\text{sgn}^2(x)-e^{-|x|}\frac{d~\text{sgn}(x)}{dx}=f(x)-2f(x) \frac{\theta(x)}{dx}=f(x)-2\delta(x) f(x).$$
Note that $\text{sgn}^2(x)=1.$ Next, we can write $$f''(x)=f(x)-2\delta(x) f(0)=f(x)-2\delta(x)$$
A: We have that $f$ satisfies the following integral equation:
$$f'(x)-\int_{0}^x f=-\operatorname{sgn}(x)=-2u(t)+1$$
Which is equivalent to the given differential equation.
