Prove Riemann-Lebesgue Theorem

Question

Prove the Riemann-Lebesgue Theorem: If $f$ is an integrable function on $(-\infty, \infty)$ then $$ \lim _{k \rightarrow \infty} \int_{-\infty}^{\infty} f(x) \cos k x \,dx=0 $$ So the Fourier transform $\hat{f}(k)$ of any $L^1$ function necessarily converges to zero as $|k| \rightarrow \infty$. Hint: First consider the case in which $f$ is a step function. Then consider the case that $f$ is bounded and vanishes outside a finite interval $[a, b] .$ Then consider $f \in L^1.$

I have proved the first two steps in the hint. But I don't see how to use them to prove the third step. Any help will be appreciated! Thank you!

Answer 1

For the third step:

Let $\epsilon>0$, then exists a simple function $\phi$ such that $\int|\phi-f| \leq \epsilon$

Thus $$|\int f(x)\cos{(kx)}dx| \leq \int |\phi -f|+|\int \phi(x)\cos{(kx)}dx|$$ $$\leq \epsilon+|\int \phi(x)\cos{(kx)}dx|$$

So $\limsup_{|k| \to +\infty}|\int f(x)\cos{(kx)}dx| \leq \epsilon+0=\epsilon$

Since $\epsilon>0$ is arbitrary,we have that $\limsup_{|k| \to +\infty}|\int f(x)\cos{(kx)}dx|=0$

Answer 2

I shall offer a slightly easier proof making use of the fact that for $f\in L^1$, $$\int_{-\infty}^\infty |f(x+a)-f(x)|dx \ \to 0 \ \text{ as } \ a\to 0.$$ which can be shown by e.g. the Dominated Convergence Theorem.

Thus, $$ \hat f(\xi) = \int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx = -\int_{-\infty}^{\infty}f(x+\frac{\pi}{\xi})e^{-ix\xi}dx $$ and so, $$|\hat{f}(\xi)| = \frac{1}{2}\Big|\int_{-\infty}^{\infty} (f(x) - f(x+\frac{\pi}{\xi}))e^{-ix\xi}dx| \leq \int_{-\infty}^{\infty} |f(x+\frac{\pi}{\xi})- f(x)|dx$$ which as above goes to $0$ as $|\xi|\to\infty$.