Showing $\int\limits_a^b h(x)\sin(nx) dx \rightarrow 0$

Question

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!

Answer 1

I think this is from Apostol. It is an informal approach to the following Lemma, if I'm not recalling wrongly:

Let $f$ be integrable in $[a,b]$. Then

$$\lim \limits_{\lambda \to \infty } \int\limits_a^b f\left( x \right)\sin \lambda xdx = 0$$

$(1)$ Let $f$ be constant. Then

$$\lim \limits_{\lambda \to \infty } \int\limits_a^b k\sin \lambda xdx = \left.-k\frac{\cos \lambda x}{\lambda}\right]_a^b=0$$

$(2)$ Let $f$ be a step function over $[a,b]$, viz

$$f(x) = \begin{cases} k:a< x\leq a_1 \cr k_1: a_1<x \leq a_2 &\cr \cdots \cr k_n :a_n<x\leq b \end{cases}$$

Then by the last result,

$$\lim \limits_{\lambda \to \infty } \int\limits_a^b f\left( x \right)\sin \lambda xdx = 0$$

$(3)$ Since for an integrable $f$ there exist two step functions such that $$\int_a^b |f(x)-s(x)|dx<\epsilon$$

$$\int_a^b |s_1(x)-f(x)|dx<\epsilon$$

we can "conclude".

IMPORTANT: If anyone can make this more detailed, precise and formal, please, do so.

Answer 2

We can apply Stone-Weierstrass: polynomial are dense in $C_0([a,b])$ endowed with the supremum norm. We can also choose such a sequence vanishing at the boundary. Indeed, if $\{P_n\}$ is a sequence of polynomial converging uniformly to $h$, then $Q_n(x)=P_n(x)-P_n(a)-\frac{x-a}{b-a}(P_n(b)-P_n(a))$, we have $Q_n(a)=0=Q_n(b)$ and $$\sup_{a\leq x\leq b}|Q_n(x)-h(x)|\leq \sup_{a\leq x\leq b}|P_n(x)-h(x)|+2|P_n(a)|+|P_n(b)|. $$ Now, fix $\{P_m\}$ a sequence of polynomials such that $\sup_{a\leq x\leq b}|P_m(x)-h(x)|\leq \frac 1m$ and $P_m$ vanishes at the boundary. We have for a fixed $m$ that \begin{align}\left|\int_a^bh(x)\sin(nx)dx\right|&\leq \int_a^b|h(x)-h_m(x)||\sin(nx)|dx+ \left|\int_a^bh_m(x)\sin(nx)dx\right|\\ &\leq \frac{b-a}m+\left|\int_a^bh_m(x)\sin(nx)dx\right| \end{align} hence by the $C^1$ case, for each $m$ $$\limsup_{n\to +\infty}\left|\int_a^bh (x)\sin(nx)dx\right|\leq \frac{b-a}m$$ and we can conclude.