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$$\hat{f_\epsilon}: \xi \mapsto \exp(-\epsilon |\xi|) \frac{\sin(|\xi|t)}{|\xi| t}$$

denote to the Fourier transform of $f$. How do I see

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

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

  3. ${f_\epsilon} \to {f}$ in $S'(\mathbb{R}^n)$?

Who can help me?

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up vote 2 down vote accepted
  1. 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}

  2. 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 $\varphi\in\mathcal S(\Bbb R^d)$.

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