This might be a silly question, but how do I prove that $u \in S'$?
Specifically, I have $u=e^{-x^2+8x}\chi_{[0,1]}$ and its Fourier transform $\hat u$ (which is not to be calculated).
Among other things, I am asked if $\hat u \in S'$ and I believe the answer is yes, assuming $u \in S'$. I'm just not sure how to prove that it's in fact true.
Thanks.
First, let us distinguish the function $u$ from the associated distribution say $U$. The function is $u:\mathbb{R}\rightarrow\mathbb{R}, x\mapsto u(x)=e^{-x^2+8x}\chi_{[0,1]}$. The associated distribution is the linear map $U:\mathscr{S}(\mathbb{R})\rightarrow\mathbb{R}$ given by $$ U(\phi)=\int_{\mathbb{R}}\ u(x)\phi(x)\ dx\ . $$ The Schwartz space $\mathscr{S}(\mathbb{R})$ is the space of rapidly decaying smooth functions $\phi$, with the locally convex topology defined by the collection of seminorms $$ ||\phi||_{n,k}=\sup_{x\in\mathbb{R}}(1+x^2)^{\frac{k}{2}}|\phi^{(n)}(x)| $$ indexed by nonnegative integers $n,k$. Since $x^2-8x\ge -16$, $|u(x)|\le e^{16}$ and so for any $\phi$, $$ |U(\phi)|\le \int_{\mathbb{R}}\ |u(x)|\ |\phi(x)|\ dx\le e^{16}\int_{\mathbb{R}}\ |\phi(x)|\ dx $$ $$ = e^{16}\int_{\mathbb{R}}\ (1+x^2)^{-1}\ (1+x^2)|\phi(x)|\ dx \le ||\phi||_{0,2}\ e^{16} \times \pi\ . $$ So $U$ is continuous and is a distribution in $\mathscr{S}'(\mathbb{R})$. For the Fourier transform, just look up the definition of fourier transform of distributions in $\mathscr{S}'(\mathbb{R})$.