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$\textbf{Statement}: \text{ If } f \in L^1(\mathbb{R^n})$, then the Fourier transform of $f(x)$, that is $\hat{f}(w) \rightarrow 0$ as $|w| \rightarrow \infty$.

Proof: $|\hat{f}(w)|:= \int_{\mathbb{R^n}} |f(x) e^{-2\pi iwx}dx| \\=|\frac{1}{-2\pi i}|\int_{\mathbb{R^n}} |f'(x) e^{-2\pi iwx}dx|\\ \leq \frac{1}{2\pi |w|}||f'||_{L^1(\mathbb{R^n})} \rightarrow 0 $.

$\textbf{Question:}$ Does this implies that $f(x) \rightarrow 0 \text{ as }|x| \rightarrow \infty$

It is clear that $|\hat{f}(w)| \leq ||f(x)||_{L^1(\mathbb{R^n})}$ and $|w| \rightarrow \infty \implies f(x)=0$.

Atul Anurag Sharma
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  • Note that you assumed vanishing behavior of $f$ at infinity in your proof (along with differentiability). The proof involves working on a dense subspace such as compactly supported smooth functions, then doing an approximation argument. – cmk Mar 26 '20 at 20:34

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Your argument only works when $f$ is differentiable and has compact support. But you have to prove it for all $f\in L^1$. This works because differentiable functions with compact support are dense.

Also, I have no idea what you say is "clear", but the answer to your question is no. The Riemann-Lebesgue Lemma applies to any $f\in L^1$, and it states a property of the Fourier Transform. But there are lots of $f\in L^1$ such that $f(x)$ does not go to zero as $|x|\to\infty$. For instance take $$ f=\sum_n n\,1_{[n,n+\tfrac1{n^3}]}\in L^1(\mathbb R). $$

Martin Argerami
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  • So, the result only works if $f \in L^1$ have compact support, right? Then we can say $f(x) \rightarrow 0 \text{ as } |x| \rightarrow \infty$ – Atul Anurag Sharma Mar 27 '20 at 02:12
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    No, it works for any $f\in L^1$. What's limited is your argument, not the Riemann-Lebesgue Lemma. – Martin Argerami Mar 27 '20 at 02:14
  • What I figured out is that $f(x)=f(0)+\int_{0}^{x} f'(t) dt$ and from this as $f \in L^1$ will imply $f(x) \rightarrow 0.$ – Atul Anurag Sharma Mar 27 '20 at 02:17
  • Riemann Lebesgue lemma tells us that Fourier transform of any $f \in L^( \mathbb{R^n})$ will tend to zero but it does not give any information about $f(x)$ as $|x| \rightarrow \infty$. – Atul Anurag Sharma Mar 27 '20 at 02:20
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    Yes. Because, as I showed you in my answer, it is not true that a function in $L^1$ necessarily goes to zero at infinity. – Martin Argerami Mar 27 '20 at 02:21
  • So, what is the condition that guarantees that $f(x) \rightarrow 0$? – Atul Anurag Sharma Mar 27 '20 at 02:22
  • I don't know. What kind of condition would you expect? – Martin Argerami Mar 27 '20 at 02:24
  • So, I encountered this problem while finding the Fourier transform of $f'(x)$. $\hat{f'}(w)= \int_{-\infty}^{\infty}f'(x) e^{-2\pi iwx}dx =-\frac{f(x)e^{-2\pi iwx}}{2\pi iw}+2\pi iw \hat{f}(w)$, for the integral to exist we need $f(x) \rightarrow 0$ as $|x| \rightarrow \infty$ – Atul Anurag Sharma Mar 27 '20 at 02:26
  • https://math.stackexchange.com/questions/158652/every-absolutely-continuous-function-with-integrable-derivative-tends-to-zero-at . I think this what I was looking for. Thank you for your example. +1 – Atul Anurag Sharma Mar 27 '20 at 02:39
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    Yes, but that's a different point of view. You are assuming that $f$ is differentiable, and asking about the Fourier transform of its derivative. And then, like you said, the question you linked answers with the fact that for $f'$ to be integrable you need $f$ to go to zero at infinity. – Martin Argerami Mar 27 '20 at 02:47