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

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?

• You can approximate with simple functions. This question is asked many times on this site. Look for Riemann-Lebesgue lemma. – EpsilonDelta Jul 23 at 21:39

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 $$t\rightarrow 0$$).
• Agreed! $\ \ \$ – cmk Jul 23 at 21:49
• A more interesting exercise IMO is proving $uniform$ continuity of translations on compact sets. A nice application of Ascoli. – Matematleta Jul 23 at 22:16