Let $g:\mathbb{R}\to\mathbb{C}$ be a bounded, continuously differentiable function with $g'\in L^p(\mathbb{R})$ for some $1\leq p\leq\infty$. $g$ and $g'$ define tempered distributions by $S_g(f)=\int_{\mathbb{R}} f(x)g(x)\, dx$
Show: For every $f\in\mathcal{S}(\mathbb{R})$ holds $S_{g'}(f)=-S_{g}(f')$
Using integration by parts, I get:
$S_{g'}(f)=\int_{\mathbb{R}} f(x)g'(x)\, dx= g(x)f(x)\vert_{-\infty}^\infty-\int_{\mathbb{R}} f'(x)g(x)\, dx=g(x)f(x)\vert_{-\infty}^\infty-S_g(f')$
I need to show, that $g(x)f(x)\vert_{-\infty}^\infty=0$. Then the formula holds.
How can I proceed?
$\lim_{a\to\infty} g(a)f(a)-g(0)f(0)$ and $\lim_{a\to -\infty} g(0)f(0)-g(a)f(a)$
Do you have a hint? Thanks in advance.
$|f(x)g(x)|\leq\|g\|_{L^{\infty}}|f(x)(1+|x|^{2})|\dfrac{1}{1+|x|^{2}}\leq\dfrac{1}{1+|x|^{2}}\|g\|_{L^{\infty}}\sup|(1+|x|^{2})f(x)|$