Product of a tempered distribution and bounded function is a tempered distribution? If $f$ is a tempered distribution and $g$ is bounded, then the product $fg$ is tempered distribution?
Thanks ins advance.

No. (Were you given more information about $g$ in the statement of the problem?) In fact the product $gu$ need not even exist:
Recall first that the definition of $gu$ is $$(gu)(\phi)=u(g\phi)\quad(\phi\in\mathcal S).$$
Define $u$ by $$u\phi=\phi'(0).$$. Let $$g(t)=\begin{cases}1,&(t\ge0),
\\-1,&(t<0).\end{cases}$$If there were  such a thing as $gu$ we would  have $$(gu)(\phi)=(g\phi)'(0).$$But  $(g\phi)'(0)$ need not exist for $\phi\in\mathcal S$.
Note assuming just that $g$ is infinitely differentiable is not enough; consider $u(\phi)=\int_0^\infty\phi$ and $g(t)=\sin\left(e^t\right)$. (Now $gu$ is a distribution, but not a tempered distribution.)
