Limits of an Integral: $\lim_{n \to \infty}\int f(x)\sin(nx)dx=0$

Question

I have a problem with the following exercise. I don't really have an idea where to start. I'm glad about every help. So here is the exercise:

Suppose $f\colon \Bbb{R} \to \Bbb{R}$ is a function in $L^1(\Bbb{R})$ (i.e integrable function). Show that $$\lim_{n \to \infty}\int f(x)\sin(nx)dx=0.$$

Thanks already for any help!

The question does not make sense as written: $\int f(x)\sin(nx),dx$ is a family of functions; what does it mean for the family of functions to "go to zero"? — Arturo Magidin, Dec 14 '11 at 20:19
obviously $\int_a^b\sin(nx)dx\to0$. now approximate $f$ by simple functions (see http://planetmath.org/encyclopedia/RiemannLebesgueLemma.html) — yoyo, Dec 14 '11 at 20:30

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

I'll assume you want to take definite integrals over $\Bbb R$ (see Arturo's comment).

Given $\epsilon>0$, there is an $M$ such that $\int_{[-M,M]^c} |f(x)\sin(nx)|<\epsilon$ for all positive integers $n$. From this, it follows that you need only show that $\lim\limits_{n\rightarrow\infty}\int_{-M}^M f(x)\sin(nx) =0 $ for any fixed $M$.

To do this, you can appeal to the theorem cited by yoyo in the comments; or, you can use the hints in this very similar post

edited Apr 13 '17 at 12:21

Community

1

answered Dec 14 '11 at 20:40

David Mitra

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I hope you can still answer questions here. How was the $M$ above chosen? – Kurome May 17 '16 at 15:32
@Kurome $f$ is in $L_1$ and the absolute value of the $\sin$ terms are uniformly bounded by $1$. So, you can just choose $M$ so that $\int_{[-M,M]^C }|f|<\epsilon$. – David Mitra May 17 '16 at 15:46
That's exactly what I want to know though, how is it possible that such an $M$ exists where $\int_{[-M,M]^C }|f|<\epsilon$. I originally thought that it was possible by the property that if $f$ is integrable in $\mathbb{R}$, then for every $\epsilon>0$, there is a $\delta>0$ s.t. if $A\subseteq R$ is measurable and $m(A)<\delta$ then $\int_{A} |f|<\epsilon$. But I can't seem to get $A=[-M,M]^C$ from that. – Kurome May 17 '16 at 16:21
@Kurome You could apply the Dominated Convergence Theorem to the sequence $(g_n)$ with $g_n =f\cdot\chi_{[-n,n]}$. – David Mitra May 17 '16 at 16:27

score 1 · Answer 2 · edited Jul 02 '13 at 07:11

Here's an answer assuming that you meant to write $\sin \left( \frac{x}{n}\right)$:

You can use the Lebesgue dominated convergence theorem:

You have $|f(x) \sin(\frac{x}{n})| \leq |f(x)|$ which you know is integrable hence you can swap the limit and the integral:

$$ \lim_{n \to \infty} \int f(x) \sin (\frac{x}{n})\ dx = \int f(x) \lim_{n \to \infty} \sin (\frac{x}{n})\ dx = \int f(x) \cdot 0\ dx = 0.$$

Limits of an Integral: $\lim_{n \to \infty}\int f(x)\sin(nx)dx=0$

2 Answers2