Let $f$ be Integrable over $(-\infty,\infty)$. Prove $\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}f(x)\cos(nx)\,dx=0$

Question

Let $f$ integrable in $(-\infty,\infty)$ prove $$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}f(x)\cos(nx)dx=0$$

My attempt

Let $g_n(x)=\cos(xn)f(x)$ note $g_n$ is measurable. Then $|g_n(x)|=|\cos(nx)f(x)|=|\cos(nx)||f(x)|\leq|f(x)|$ as $f$ is integrable then $|f(x)|$ is integrable.

Then, by Dominated Convergence Theorem we have:

$$\lim_{n\to\infty}\int_{-\infty}^\infty (\cos xn)f(x)dx = \int_{-\infty}^\infty \lim_{n \to \infty}(\cos xn)f(x)dx$$

Here I'm stuck. Can someone help me?

Answer 1

To apply the dominated convergence theorem directly, you need to identify a pointwise limit (which, as you noted, won’t work here). Instead, you should use the Riemann-Lebesgue lemma, or prove it, if you haven’t learned it (https://en.wikipedia.org/wiki/Riemann-Lebesgue_lemma).

Hints on proving this result: First, test on a characteristic function, which is easy to compute directly. Then, you automatically get the result for simple functions. Now, use simple function density in $L^1$. Alternatively, use the density of smooth functions with compact support (testing on this dense subspace is easy, since you can integrate by parts and not worry about boundary terms). Another option is, as @Jakobian noted), to use that translations are dense in $L^1$ (that is, if $L_tf(x)=f(x+t),$ then $\|f-L_tf\|_{L^1}\rightarrow 0$ as $t\rightarrow 0$).

So, there are a lot of potential ways to go about proving this!