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I would like to know if my proof is correct. Define Fourier transform of $f\in L^1(\mathbb{R}^n)$ as $$\widehat{f}(x)=\int_{\mathbb{R}^n}e^{-ix\cdot y}f(y)\text{d}y.$$ I want to show that $\widehat{f}$ is continuous. So pick any sequence $x_n\rightarrow x$. Now $g_n(y)=e^{-ix_n\cdot y}f(y)$ is measurable for all $n\in\mathbb{N}$, $g_n(y)\rightarrow e^{-ix\cdot y}f(y)=g(y)$ for all $y$ and $|e^{-ix_n\cdot y}f(y)|=|f(y)|$ where $|f(y)|$ is integrable. So by dominated convergence theorem we have $$\lim_{n\rightarrow \infty}\widehat{f}(x_n)=\lim_{n\rightarrow \infty}\int_{\mathbb{R}^n}e^{-ix_n\cdot y}f(y)\text{d}y=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^n}g_n(y)\text{d}y=\int_{\mathbb{R}^n}g(y)\text{d}y=\int_{\mathbb{R}^n}e^{-ix\cdot y}f(y)\text{d}y=\widehat{f}(x).$$ Therefore $\hat{f}$ is continuous. Am I correct?

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Infinitebig
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  • Yes, you are correct. – uniquesolution Oct 06 '17 at 10:03
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    There is even a better result the fourrier transform is always uniformly continuous – Guy Fsone Oct 06 '17 at 10:03
  • @GuyFsone is that more difficult to prove or is it just slight modification of my proof? I guess DCT is needed there also – Infinitebig Oct 06 '17 at 10:08
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    In fact together with the continuity results in other to conclude the uniform continuity , we can the Riemann-Lebesgue Lemma which says that Fourier transform $\widehat{f}(x)\to 0$ as $|x|\to\infty$ – Guy Fsone Oct 06 '17 at 10:12
  • @GuyFsone so it's basically continuous on compact set, hence uniformly continuous, and outside that compact set all values are arbitrarily close to 0? – Infinitebig Oct 06 '17 at 10:16
  • https://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuous/2399462#2399462 – Guy Fsone Oct 06 '17 at 10:21
  • that's what I was thinking about, thanks. – Infinitebig Oct 06 '17 at 10:26
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    You can use the dominated convergence theorem, but note it is not hard to prove it directly from :

    $$|\hat{f}(x)-\hat{f}(x_0)| \le |\hat{f}(x)-\hat{f}m(x)|+|\hat{f}_m(x)-\hat{f}_m(x_0)|+|\hat{f}_m(x_0)-\hat{f}(x_0)|$$ where $\hat{f}_m(x) = \int{ |y| \le m}e^{-i x \cdot y}f(y)\text{d}y$ is clearly uniformly continuous

     – reuns Oct 06 '17 at 15:37

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