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For $f\in L^1(\mathbb R^d)$ look at $f_n:\mathbb R^d \rightarrow \mathbb K, x\mapsto\int_\mathbb {R^d}\hat{f}(\xi)e^{2\pi i\langle\xi, x\rangle} e^{-\frac{\pi^2}{n}|\xi|^2}d\xi$.
Show $\lim\limits_{n\to\infty} \|f_n-f\|_1=0$
$\hat{f}(x)$ is the Fourier transformation $\hat{f}(\xi)=\int_{\mathbb R^d}f(x)e^{-2\pi i\langle x,\xi\rangle}dx$
How can I show this? I know the Fourier transformation of $g_n(x)=e^{-\frac{\pi^2}{n}|\xi|^2}$ is $\hat{g}_n(\xi)=\frac{n^{d/2}}{\pi^{d/2}}e^{-n|\xi|^2}$ with $\|g\|_1=1$. Maybe I can use Fubine or Lebesgue dominated convergence theorem?
Thanks for a hint.

andy
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    Take the integral $\int_\mathbb {R^d}\hat{f}(\xi)e^{2\pi i\langle\xi, x\rangle} e^{-\frac{\pi^2}{n}|\xi|^2}d\xi$, insert the definition of $\hat f$ and apply Fubini; you should get the convolution of $f$ with an approximate idenity. – David C. Ullrich Jul 04 '22 at 14:09
  • I get $\int_{\mathbb R^d} \int_{\mathbb R^d} f(y)e^{-2\pi i \langle y,\xi\rangle}e^{2\pi i\langle\xi, x\rangle}e^{-\frac{\pi^2}{n}|\xi|^2}dyd\xi$. Where is the convolution with $f$? – andy Jul 04 '22 at 15:25
  • @DavidC.Ullrich how can I simplify the integral? – andy Jul 04 '22 at 17:43
  • Let $\varphi_n$ be such that $\hat{\varphi_n}(\xi)=e^{-\frac{\pi^2}{n}|\xi|^2}$, then $f_n=(\hat f \hat{\varphi_n})^\check\ =f*\varphi_n$ and ${\varphi_n}$ is an approximation to the identity. – Feng Jul 05 '22 at 06:17

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