The problem statement is as follows: Given $f\in L^{1}(\mathbb{R})$, show that$$\lim_{n\to\infty}\int_{[-M,M]}f(x)\sin(nx)\,\mathrm dx=0$$for any fixed $M$.
My approach is as follows:
We have $\lim\limits_{n\to\infty}\int_{[-M,M]}\sin(nx)\,\mathrm dx=0$. If $f$ is a simple function taking values $a_{1},\cdots,a_{n}$, $$\lim_{n\to\infty}\int_{[-M,M]}f(x)\sin(nx)\,\mathrm dx = \lim_{n\to\infty}\sum_{i=1}^{n}a_{i}\int_{[-M,M]}\sin(nx)\,\mathrm dx\\
= \sum_{i=1}^{n}a_{i}\lim_{n\to\infty}\int_{[-M,M]}\sin(nx)\,\mathrm dx = 0.$$
My problem is when $f$ is a non-negative function. If $f\geq 0$, then there exists a sequence of simple functions $f_{k}\nearrow f$, then $f_{k}\sin(nx)\nearrow f\sin(nx)$ [n is fixed], then by Monotone convergence theorem, we have$$\lim_{n\to\infty}\int_{[-M,M]}f(x)\sin(nx)\,\mathrm dx = \lim_{n\to\infty}\int_{[-M,M]}\lim_{k\to\infty}f_{k}(x)\sin(nx)\,\mathrm dx\\
= \lim_{n\to\infty}\lim_{k\to\infty}\int_{[-M,M]}f_{k}(x)\sin(nx)\,\mathrm dx.$$
My question is:
- Can swap the two limits $\lim\limits_{n\to\infty}\lim\limits_{k\to\infty}\int_{[-M,M]}f_{k}(x)\sin(nx)\,\mathrm dx$ ?
- If I can't, then it seems like my approach doesn't work. What is wrong with my approach or are there other approaches ?
I've seen proofs u\sing Fourier Transforms, but I've not progressed that far yet, so I'm seeking proofs u\sing only basic integral theorems.
Any help is appreciated.
Best Answer
Your solution isn't right. If $\sin (nx)<0$, then the product $f_k(x)\sin(nx)$ no longer increases to $f(x)\sin(nx)$ as $k\to\infty$ (it decreases), so monotone convergence cannot be applied anymore.
Note that the clause "for any fixed $M>0$" is actually a red herring; you can completely ignore it. In other words, we can prove the following: for any $f\in L^1(\Bbb{R})$, \begin{align} \lim\limits_{n\to\infty}\int_{\Bbb{R}}f(x)\sin(nx)\,dx&=0. \end{align} The version you wrote about is a special case obtained by replacing $f$ with $f\cdot \chi_{[-M,M]}$. Actually, one can prove the stronger statement that for any $f\in L^1(\Bbb{R})$, we have $\lim\limits_{t\to\infty}\int_{\Bbb{R}}f(x)e^{itx}\,dx=0$ (by looking at the imaginary part of this integral and considering $t=n\in\Bbb{N}$, we recover the version you're asking about).
Since the more general statement doesn't require any extra effort, we shall prove that instead. Verify first that for $f=\chi_{[a,b]}$, we have $\lim\limits_{t\to\infty}\int_{\Bbb{R}}f(x)e^{itx}\,dx=0$. This is almost obvious since the integral can be explicitly evaluated, and once you do, it's clear that it decays like $\frac{1}{t}$. By linearity, the statement is also true for finite linear combinations of such functions.
Next is a slightly technical measure theory approximation result: the space of "step functions", i.e linear combinations of characteristic functions of compact intervals is dense in $L^1(\Bbb{R})$. In other words, the space \begin{align} \text{Step}(\Bbb{R}):= \left\{\sum_{i=1}^nc_i\chi_{[a_i,b_i]}\,\bigg| \text{ $n\in\Bbb{N}$, $a_i,b_i\in\Bbb{R}, c_i\in\Bbb{C}$ for all $1\leq i \leq n$}\right\} \end{align} is dense in $L^1(\Bbb{R})$. To prove this, note first of all that the space of simple functions (finite linear combinations of characteristic functions of Lebesgue-measurable sets) is dense in $L^1(\Bbb{R})$. So, to prove that step functions are also dense, it is enough to show that any Lebesgue measurable set $A\subset\Bbb{R}$ of finite measure, can be approximated (in $L^1$ norm) by step functions.
This follows pretty much by regularity of Lebesuge measure. Suppose $A\subset\Bbb{R}$ has finite measure, and let $\epsilon>0$. By inner-regularity of Lebesgue measure, there exists a compact $K\subset A$ such that $m(A)-m(K)<\epsilon$. Also, by definition (if you use the Caratheodory construction) of Lebesgue measure, there exist countably many open intervals $\{I_j\}_{j=1}^{\infty}$ which cover $A$ and such that $\sum_{j=1}^{\infty}m(I_j)<m(A)+\epsilon$. Since these intervals cover the compact set $K$ as well, finitely many of them, say $I_1,\dots, I_n$ (relabel indices if necessary) will cover $K$. Then, the symmetric difference $\left(\bigcup_{j=1}^nI_j\right)\triangle A\subset \left(\bigcup_{i=1}^{\infty}I_i\right)\setminus K$, and the RHS has measure $\leq 2\epsilon$. This shows $\|\sum_{j=1}^n\chi_{I_j}-\chi_A\|_1\leq 2\epsilon$, which proves the required density.
Can you now conclude that the theorem holds for all $f\in L^1(\Bbb{R})$? (you only need the triangle inequality; no need for monotone/dominated convergence).
There are also several other proofs for the Riemann-Lebesgue lemma. If you know that $C^{\infty}_c(\Bbb{R})$ is dense in $L^1(\Bbb{R})$, then you can first prove the theorem for $f\in C^{\infty}_c(\Bbb{R})$ using integration by parts (this gives you the $\frac{1}{t}$ decay factor; and the boundary terms vanish due to compact support). Now another density argument allows you to conclude for all $f\in L^1(\Bbb{R})$.
The other proof I know makes use of continuity of translation in $L^1(\Bbb{R})$ (which itself is often proved by using density of $C_c(\Bbb{R})$ in $L^1(\Bbb{R})$). TO use this method (which I believe is how things are done in Stein and Shakarchi's text), note that due to translation-invariance of Lebesgue measure, for each $\alpha\in\Bbb{R}$, we have \begin{align} \int_{\Bbb{R}}f(x)e^{itx}\,dx&=\int_{\Bbb{R}}f(x-\alpha) e^{it(x-\alpha)}\,dx \end{align} So, \begin{align} \left|\int_{\Bbb{R}}f(x)e^{itx}\,dx\right| &=\left|\frac{1}{2}\int_{\Bbb{R}}[f(x)+ f(x-\alpha)e^{-it\alpha}]e^{itx}\,dx\right|\\ &\leq \frac{1}{2}\int_{\Bbb{R}}\left|f(x)+e^{-it\alpha}f(x-\alpha)\right|\,dt \end{align} In particular this holds when $\alpha=\frac{\pi}{t}$, so we have \begin{align} \left|\int_{\Bbb{R}}f(x)e^{itx}\,dx\right| &\leq \frac{1}{2}\int_{\Bbb{R}}\left|f(x)-f\left(x-\frac{\pi}{t}\right)\right|\,dx, \end{align} and the RHS vanishes as $t\to\infty$, due to continuity of translation in $L^1$.