Uniform convergence and convergence in $S'(\mathbb{R}^n)$ Let
$$\hat{f_\epsilon}: \xi \mapsto \exp(-\epsilon |\xi|) \frac{\sin(|\xi|t)}{|\xi| t}$$
denote to the Fourier transform of $f$. How do I see


*

*$\hat{f_\epsilon}$ converges uniformly on $\mathbb{R}^n$ to $\hat{f}=\frac{\sin(|\xi|t)}{|\xi|t}$ as $\epsilon \to 0$ ?

*$\hat{f_\epsilon} \to \hat{f}$ in the tempered distributions $S'(\mathbb{R}^n)$

*${f_\epsilon} \to {f}$ in $S'(\mathbb{R}^n)$?
Who can help me?
 A: *

*Dealing with the cases $t>0$/ $t<0$, and making a substitution in the supremum, we have to show that $g_{\varepsilon}$ converges uniformly to $g$ on $\Bbb R_{\geq 0}$, where 
$$g_{\varepsilon}(x)=e^{-\varepsilon x}\frac{\sin x}x,\quad g(x)=\frac{\sin x}x.$$
To see that, write
\begin{align}
\sup_{x>0}|g_{\varepsilon}(x)-g(x)|&=\sup_{t>0}\left|e^{-t}\frac{\sin\frac t{\varepsilon}}{t/\varepsilon}-\frac{\sin\frac t{\varepsilon}}{t/\varepsilon}\right|\\
&=\varepsilon\sup_{t>0}(1-e^{-t})|\sin\frac t{\varepsilon}|\leq \varepsilon.
\end{align}

*Note that uniform convergence implies converges in the dual of Schwartz space. Indeed, if $g_n\to g$ uniformly on $\Bbb R^d$, and $\varphi\in\mathcal S(\Bbb R^d)$, then 
\begin{align}
|\langle g_n,\varphi\rangle-\langle g_n,\varphi\rangle|&=\left|\int_{\Bbb R^d}(f_n(x)-f(x))\varphi(x)dx\right|\\
&\leq\sup_{x\in\Bbb R^d}|f_n(\xi)-f(\xi)|\int_{\Bbb R^d}|\varphi(x)|dx, 
\end{align}
the last integral being convergent since $\varphi\in\mathcal S(\Bbb R^d)$.

*The Fourier transform is sequentially continuous, and so is the map on $\mathcal S'$, which maps the distirbution $S$ to a distribution $T$
$$\langle T,x\mapsto \varphi(x)\rangle:=\langle S,x\mapsto \varphi(-x)\rangle.$$

