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What is the limit of $$\int_{\mathbb{R}}f(x)|\sin( \lambda x)| dx$$ for all integrable functions $f\colon\mathbb{R} \rightarrow \mathbb{C}$ when $|\lambda| \rightarrow \infty$?

I tried to use the Lemma of Riemann-Lebesgue. I want to prove that the integral of $|\sin(\lambda x)|$ is zero over one period when $|\lambda| \rightarrow \infty$.

I tried this: $$\int_{\mathbb{R}}\mathbb{1}_{[0,\pi]}\sin(\lambda x) dx= \frac{\cos(\lambda 0)-\cos(\lambda \pi)}{\lambda}\xrightarrow{|\lambda|\to\infty} 0$$

But then I doubted if this was indeed correct because $\lambda$ can be negative.

Can anybody help me to prove this?

Jochen
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notaMathqueen
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1 Answers1

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Using the fact that $f\in L^1$ can be approximated in $L^1$ by a step function (a function which can be expressed as $\sum_{j=1}^n c_j1_{I_j}$, where $\{I_j\}_{j=1}^n$ is a finite collection of disjoint open intervals and $c_j\in \mathbb{C}$) makes the problem much easier.

Fix $\epsilon>0$. Write $f=g+h$, where $g=\sum_{j=1}^n c_j1_{I_j}$ is a step function and $\|h\|_{L^1}<\epsilon$.

You can probably observe

$$\forall j,\lim_{\lambda\rightarrow\infty}\int_{I_j}|\sin(\lambda x)|dx=\frac{2}{\pi}m(I_j)$$

(For instance, you can show that by ignoring the rightmost portion of $I_j$, whose contribution to the integral shrinks as $\lambda\rightarrow\infty$, to make the integral periodic.)

Adding up these, you get

$$\lim_{\lambda\rightarrow\infty}\int_{\mathbb{R}}g|\sin(\lambda x)|dx=\frac{2}{\pi}\int_{\mathbb{R}}gdx$$

Now, since

$$\lim_{\epsilon\rightarrow0}\sup_{\lambda\neq0}\left|\int_{\mathbb{R}}gdx-\int_{\mathbb{R}}fdx\right|=\lim_{\epsilon\rightarrow0}\sup_{\lambda\neq0}\left|\int_{\mathbb{R}}h|\sin(\lambda x)|dx\right|=0$$

, taking $\epsilon\rightarrow0$, the answer is

$$\frac{2}{\pi}\int_{\mathbb{R}} fdx$$

Moonshine
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