Given a function $u(x) \in \mathscr{S}(\mathbb{R})$, I would like to establish if:
$\xi \cdot \hat{u}(\xi)$ is bounded in $\mathbb{R}$, where $\hat{u}(\xi)$ is the Fourier transform of $u$
$e^x \cdot u(x)$ is bounded in $\mathbb{R}$
but I don't know how to approach the problem.
For part 1, I tried to use that Fourier transform is always bounded with $\Vert{\hat{u}}\Vert_{\infty} \le \Vert{u}\Vert_{L^1} $ and in this case $u \in L^1(\mathbb{R})$ but multiplying any $u$ by $\xi$ the result is yet bounded? And the same doubts arose with part two.
Can anyone help me? Thank you in advance
If you know that the Fourier transform is actually an automorphism on the Schwartz space, then the first assertion follows pretty much by the definition (or a equivalent characterization) of a Schwartz function, that is, $\varphi$ is in $\mathcal{S}$ if and only if $$ \sup_{x\in\mathbb{R}} |x^m D^n\varphi(x)| < \infty \qquad \text{for all} \quad n,m \in \mathbb{N}_0,$$ where $D^n$ refers to the $n$-th derivative (so $D^n \varphi (x) = \frac{d^n}{dx^n}\varphi(x)$ and $D^0\varphi =\varphi$). If you don't know the fact about the Fourier transform being an automorphism on the Schwartz function, you can also get this first result "by hand" by using the following fact: If $\varphi$ is a Schwartz function then there exist for any positive integers $N$ and $m$ a constant $C_{N,m}$ such that $$ |D^m\varphi(x)| \leq C_{N,m} (1+|x|)^{-N}. $$ Indeed, this actually is another characterization of the Schwartz functions (one that I found to be quite useful). And it is this rapid decay to zero of any Schwartz function at infinity (among other things) that is used to show that $\widehat{\varphi}$ is again a Schwartz function.
This also relates a bit to the second questions, since it is at least true that any Schwartz functions decays to zero quicker than any polynomial.
Finally, with these tools at hand and the comments by reuns (below) and Mindlack one can also show the first assertion quite elegantly.