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Let $h\in C_0([a,b])$ arbitrary, that is $h$ is continuous and vanishes on the boundary. I want to show that $\int\limits_a^b h(x)\sin(nx)dx \rightarrow 0$.

If $h\in C^1$, integration by parts immediately yields the claim, since $h'$ is continuous and thence bounded on the compact interval, using also the zero boundary condition.

However, I believe the statement is also true for all $h\in C_0([a,b])$. My idea is to approximate $h$ by functions $h_m \in C_0^1([a,b])$. Then for all $m$,

$$\begin{equation*} \lim_{n \to \infty} \int h_m(x) \sin(nx) dx = 0. \end{equation*}$$

$$\begin{align*} \Rightarrow ~~~ \lim_{n \to \infty} \int h(x)\sin(nx) dx &= \lim_{n \to \infty} \int \lim_{m \to \infty} h_m(x)\sin(nx) dx\\ &= \lim_{m \to \infty}(\lim_{n \to \infty} \int h_m(x)\sin(nx) dx)\\ &= \lim 0 = 0. \end{align*}$$

This is fine iff the second equality is. In fact, this is two different steps, as three limiting processes are involved. Hence the questions:

First, can I make sure that I can interchange the $m$-limit with the integral sign? (Can I assume that $h_m$ converges uniformly? Or use some sort of Dominated Convergence Theorem?)

And second, may I swap the $n$-limit for the $m$-limit? (The $n$-limit is in fact $C/n \to 0$)

I hope it's not too messy. Many thanks for any kind of help!

  • 2
    You might want to check [this](http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma)2012-05-27
  • 0
    Take a sequence $h_m$ that converges to $h$ uniformly on $[a,b]$ (this gives you $\sup_{t \in [a,b]} |h_n(t)-h(t)| \to 0$).2012-05-27
  • 0
    See [here](http://math.stackexchange.com/questions/105258/integral-involving-a-periodic-function/105275#105275) for some more approaches to a more general version of this problem.2012-05-28

2 Answers 2