Showing $\lim\limits_{n\to\infty}\int_{[-M,M]}f(x)\sin(nx)\,\mathrm dx=0$ for any mixed $M$
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.
Solution 1:
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 $L^1$.