The following question is related to this: Tempered distribution, Schwartz space, functional analysis
For $f:\mathbb{R}\to\mathbb{C}$, $f(x)=|x|$ and $h:\mathbb{R}\to\mathbb{C}$, $h(x)=\begin{cases} 1, ~\text{for}~ x\geq 0\\ -1,~\text{else}\end{cases}$
holds $(S_f)'=S_h$. Calculate the derivative of $S_h$ (therefore $(S_h)'$).
I am not sure what I have to do. I have to show, that the functions $(S_f)'$ and $S_h$ are equal. Hence for every $g\in\mathcal{S}(\mathbb{R})$ we have $(S_f(g))'=S_h(g)$. Right?
I tried to evaluate:
$S_h(g)=\int_{\mathbb{R}} h(x)g(x)\, dx=\int_0^\infty g(x)\, dx-\int_{-\infty}^0 g(x)\, dx$
Using integration by parts, I get to
$xg(x)\vert_{x=0}^{x=\infty}-\int_0^\infty xg'(x)\, dx-(xg(x)\vert_{x=0}^{x=\infty}-\int_{-\infty}^0 xg'(x)\, dx=-\int_{\mathbb{R}} xg'(x)\, dx$
But how do I calculate $(S_f(g))'=\left(\int_{\mathbb{R}} |x|g(x)\, dx\right)'$, since this is not even a function, and the derivative should be 0?
Thanks in advance for your comments.
$\displaystyle\int_{0}^{\infty}g(x)dx-\int_{-\infty}^{0}g(x)dx=-\int_{0}^{\infty}xg'(x)dx+\int_{-\infty}^{0}xg'(x)dx=-\int_{\bf{R}}xh(x)g'(x)dx$, note that $xh(x)=|x|$.
So $\left<g,S_{h}\right>=-\displaystyle\int_{\bf{R}}f(x)g'(x)dx=-\left<g',S_{f}\right>=\left<g,S_{f}'\right>$.
And $\left<g,S_{h}'\right>=-\left<g',S_{h}\right>=-\displaystyle\int_{0}^{\infty}g'(x)dx+\displaystyle\int_{-\infty}^{0}g'(x)dx=g(0)+g(0)=2g(0)$.