# Show that $\lim_{k\to\infty} \int_0^1 f(x) \sin(kx) \, dx = 0$

I'm working through some old test problems in preparation for a Real Analysis exam. I got stuck on the following problem:

Show that if $f$ is (Lebesgue) integrable on $[0,1]$, then $$\lim_{k\to\infty} \int_0^1 f(x) \sin(kx) \, dx = 0$$

It seems to me that the idea is to move the limit inside the integral. I know that $f(x)\sin(kx)$ is bounded by $f(x)$ which is integrable, so I thought maybe the bounded or dominated convergence theorems might be useful. However, I don't have a sequence of integrable functions converging point-wise to $0$. How can I proceed, or am I approaching this incorrectly?

• Apply the Riemann-Lebesgue lemma. – José Carlos Santos Jan 5 '18 at 16:33
• Hint: use simple function to approximate $f(x)$. – xpaul Jan 5 '18 at 16:36
• @JoséCarlosSantos Oh yeah that seems quite applicable indeed! This is just the limit of the imaginary component of $\int_0^1 f(x) e^{ikx} dx$. So since the Riemann-Lebesgue Lemma gives $\lim_{k\to\infty} \int_0^1 f(x) e^{ikx} dx = 0$, the limit of the imaginary component must also go to zero. Is it really that simple? – M47145 Jan 5 '18 at 16:52
• @M47145 Yes, it is that simple. – José Carlos Santos Jan 5 '18 at 18:03

This is a case of the Riemann–Lebesgue lemma.

If $f$ is has a continuous derivative and $f(1)=0$ then integration by parts yields $$\int_0^1 f(x) \sin(kx) \,dx = f(1)(\sin k) - \int_0^1 f'(x) \frac 1 k \sin(kx) \, dx = \frac 1 k\int_0^1 f'(x) \sin(kx)\,dx$$ And taking absolute values on both sides and showing that the last expression goes to $0$ (since $|f'(x)\sin(kx)| \le \max\{ |f'(x)| : 0\le x\le 1\}$) will do it.

But what if $f(1)\ne0$ or $f$ does not have a continuous derivative? The linked Wikipedia article says one can approximate $f$ in $L^1$ with another function $g$ that has those propoerties.

Assume function $f(x)$ has maximum $M$ and minimum $m$ in interval $[0,1].$ Since the integral exists so both $m$ and $M$ are real and bounded otherwise we will have counterexamples such as $f(x)={1\over x^2}$. Now we want to show that:

$$\lim_{k\to\infty} \int_0^1{f(x)\sin kx} \, dx=0$$

We try to bound the above integral. Since $m\le f(x)\le M$:

$$m\int_0^1 \sin k x \, dx \le \int_0^1f(x)\sin kx \,dx \le M \int_0^1\sin k x \, dx$$ or $$m{1-\cos k \over k}\le \int_0^1 f(x)\sin k x \, dx \le M{1-\cos k \over k}$$ By tending $k\to\infty$ both bounds get zero and we obtain what we want.