How to compute the distributional derivative of $\operatorname{sgn}(x){\rm e}^{-\lvert x \rvert}$?

Question

Consider the function $\operatorname{f}(x) = \operatorname{sgn}(x){\rm e}^{-\lvert x \rvert}$ (viewed as a tempered distribution). Show that

$$\operatorname{f}'(x) = 2\delta(x) - {\rm e}^{-\left\vert x\right\vert}.$$

If I am not mistaken this boils down to computing

$$\int_{-\infty}^\infty -\operatorname{sgn}(x)\operatorname{e}^{-\lvert x \rvert} \varphi^\prime(x) \mathrm{d}x$$

for some Schwartz function $\varphi$. I tried to use partial integration, but it did not work. Could you please tell me if I am on the right track?

Answer 1

There's a simpler way. Consider $f$ as a piecewise function:

$$x > 0 \implies |x| = x, f(x) = e^{-x} \implies f'(x) = -e^{-x}$$ $$x < 0 \implies |x| = -x, f(x) = -e^{-(-x)} = -e^x \implies f'(x) = -e^{x}$$

Note that you can unite the two sides of the derivative as:

$$x \ne 0 \implies f'(x) = -e^{-|x|}$$

Now we just need to figure out how to treat the case of $x = 0$. Note that $f$ is discontinuous there.

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -e^{x} = -1$$ $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{-x} = 1$$

So, if a function's value jumps by 2 at $x = 0$, what does that mean for its distributional derivative?