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Let $f\in L^1(\mathbb R)$. Set $$\hat f(\alpha )=\int_{\mathbb R}f(x)e^{-2i\pi x\alpha }dx.$$

Prove $$\lim_{|\alpha |\to \infty }\int_{\mathbb R}f(x)e^{-2i\pi x\alpha }dx=0.$$

The thing is even if I put that limit in the integral, it doesn't work, so I don't know how to conclude this part (and it looks strange to me...).

MSE
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    This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^\infty_c(\mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(\mathbb R)$ functions by density. – User8128 Jul 20 '18 at 20:43
  • This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$. – Chappers Jul 20 '18 at 20:44

2 Answers2

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We prove the lemma first for $C^\infty_c(\mathbb R)$. Suppose that $f \in C^\infty_c(\mathbb R)$. Then we see using integration by parts $$\left \lvert \int_{\mathbb R} f(x) e^{i\xi x } dx \right \rvert = \left \lvert\frac{1}{i\xi} \int_{\mathbb R} f'(x) e^{i \xi x} dx \right \rvert \le \frac{1}{\lvert\xi \rvert} \int_{\mathbb R} \lvert f'(x)\rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $\lvert \xi \rvert \to \infty$, the above inequality shows that $$\int_\mathbb R f(x) e^{i \xi x} dx \to 0.$$ Thus the lemma holds for $f \in C^\infty_c(\mathbb R)$.

For general $f \in L^1(\mathbb R)$ and arbitrary $\epsilon > 0$, find $g \in C^\infty_c(\mathbb R)$ with $$ \int_{\mathbb R} \lvert f(x) - g(x) \rvert dx < \epsilon $$ (we can do this since $C_c^\infty(\mathbb R)$ is dense in $L^1(\mathbb R)$. By the above, there is $M > 0$ such that $\lvert \xi \rvert> M$ gives $$\left \lvert \int_\mathbb R g(x) e^{i\xi x} dx\right \rvert < \epsilon. $$ But then $$\left \lvert \int_{\mathbb R} f(x) e^{i\xi x} dx \right \rvert \le \left \lvert \int_\mathbb R g(x) e^{i\xi x} dx\right \rvert + \left \lvert \int_{\mathbb R} [f(x) - g(x)] e^{i\xi x}dx \right \rvert \le \epsilon + \int_{\mathbb R} \lvert f(x) - g(x) \rvert dx < 2\epsilon$$ for all $\xi \in \mathbb R$ with $\lvert \xi\rvert > M$. Thus $$\int_\mathbb R f(x) e^{i \xi x} dx \to 0$$ as $\lvert \xi\rvert \to\infty$ for any $f \in L^1(\mathbb R)$.

EDIT: I typed up this entire answer using $\xi$ rather than $\alpha$, and I forgot the $-2\pi$ in the exponent, but the substance of the proof is the same.

User8128
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  • It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(\mathbb R)\to S(\mathbb R)$). – Surb Jul 20 '18 at 21:03
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By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_n\to f$ in $L^1$. Set $\hat f_n$ Fourier transform of $f_n$. Then $$|\hat f(\alpha )|\leq |\hat f(\alpha) -\hat f_n(\alpha )|+|\hat f_n(\alpha )|\leq \|f_n-f\|_{L^1}+|\hat f_n(\alpha )|\underset{\substack{|\alpha |\to \infty \\ \hat f_n\ Schwartz}}{\longrightarrow }\|f_n-f\|_{L^1}\underset{n\to \infty }{\longrightarrow }0.$$

Surb
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  • Of course, this requires one to know that $\hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact. – User8128 Jul 20 '18 at 20:52
  • @User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-) – Surb Jul 20 '18 at 20:55