For every integrable function $f \colon \mathbb{R} \to \mathbb{C}$ I have to calculate the limit $$\lim_{|\lambda| \to +\infty} \int_{\mathbb{R}}f(x)|\sin(\lambda x)|dx $$

We know that this limit exists because it is bounded by the integral of $f(x)$. From the lemma of Riemann-Lebesgue, we know that $$\lim_{|\lambda| \to +\infty} \int_{\mathbb{R}}f(x)\sin(\lambda x)dx = 0$$ So I assume that the limit that I need to calculate is bigger than zero, and is somewhere close to $$\int_{\mathbb{R}}f(x)dx$$ but I have no idea how I can prove this.


Elaborating on the idea of the Fourier series. This idea I have presented it at my blog here.

The Fourier series of $|\sin x |$ is of the form:

$$|\sin x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos nx$$


$$ |\sin \lambda x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos n \lambda x \tag{1}$$


\begin{align*} \int \limits_{\mathbb{R}} f(x) \left | \sin \lambda x \right | \, \mathrm{d}x &= \int \limits_{\mathbb{R}} f(x)\left ( \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos n\lambda x \right ) \, \mathrm{d}x\\ &=\frac{2}{\pi} \int \limits_{\mathbb{R}} f(x) \, \mathrm{d}x + \int \limits_{\mathbb{R}} f(x) \sum_{n=1}^{\infty} a_n \cos n \lambda x \, \mathrm{d}x \end{align*}

Hence, $\require{cancel}$

\begin{align*} \lim_{\lambda \rightarrow +\infty}\int \limits_{\mathbb{R}} f(x) \left | \sin \lambda x \right | \, \mathrm{d}x &=\frac{2}{\pi} \int \limits_{\mathbb{R}} f(x) \, \mathrm{d}x + \cancelto{0}{\lim_{\lambda \rightarrow +\infty}\int \limits_{\mathbb{R}} f(x) \sum_{n=1}^{\infty} a_n \cos n \lambda x \, \mathrm{d}x} \\ & = \frac{2}{\pi} \int \limits_{\mathbb{R}} f(x) \, \mathrm{d}x \end{align*}

The last limit is $0$ due to the Riemann - Lebesgue Lemma. Conclude!

  • $\begingroup$ The Riemann-Lebesgue step needs justification: you've interchanged the limit and integral with the summation. Why is this permissible? $\endgroup$ – Bungo Aug 27 '18 at 4:39

Consider the family of functions $$\left\{u_{\lambda}=|\sin (\lambda x)|\right\}_{\lambda>0}\subset L^{\infty}(\mathbb{R}) $$ This is a bounded subset of $L^{\infty}(\mathbb{R})$, with $$\|u_{\lambda}\|_{\infty}\leq 1,\qquad \forall \lambda>0 $$ Our goal is to find a function $u\in L^{\infty}(\mathbb{R})$ so that $$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}fu_{\lambda}=\int_{\mathbb{R}}fu,\qquad \forall f\in L^1(\mathbb{R})\qquad (*) $$ (if you know some functional analysis, thanks to the Frechet-Riesz representation theorem $u$ represents the weak-star limit of the net $\left\{u_{\lambda}\right\}_{\lambda>0}$ as $\lambda\to +\infty$, which exists because $\left\{u_\lambda\right\}$ is bounded and hence weakly-star compact in $L^{\infty}(\mathbb{R})$ by the Banach-Alaoglu theorem).

Now, if the above limit holds for all $f$ in a dense subspace $D\subset L^1(\mathbb{R})$, then it holds for all $f\in L^1(\mathbb{R})$. Let me prove this. Let $f_k\to f$, with $f_k\in D$, be an approximating sequence. Then \begin{align*}\left|\int fu_{\lambda}-\int fu\right|&\leq \left|\int fu_{\lambda}-\int f_ku_{\lambda}\right|+\left|\int f_ku_{\lambda}-\int f_ku\right|+\left|\int f_ku-\int fu\right|\leq \\ &\leq \|f-f_k\|_1+\left|\int f_ku_{\lambda}-\int f_ku\right|+\|f_k-f\|_1\|u\|_{\infty} \end{align*} and all the three summands vanish by assumption.

The dense subspace $D$ we choose is the space of step functions, i.e. of the form $$f=\sum_{i=1}^{N}\alpha_i\chi_{[a_i,b_i]} $$ where $\alpha_i\in \mathbb{R}$ and $\left\{[a_i,b_i]\right\}_{i=1}^{N}$ is a family of pairwise disjoint bounded intervals. It is not hard to show, using the fact that $\int_{n\pi}^{m\pi}|\sin x|dx=2(m-n)$ for all $m,n\in \mathbb{Z}$ and proceeding by approximation, we have $$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}\chi_{[a_i,b_i]}u_{\lambda}= \lim_{\lambda\to+\infty}\int_{a_i}^{b_i}|\sin (\lambda x)|dx=\lim_{\lambda\to+\infty}\frac{1}{\lambda}\int_{\lambda a_i}^{\lambda b_i}|\sin x|dx= \frac{2}{\pi}(b_i-a_i) $$ and therefore

$$\lim_{\lambda\to +\infty}\int_{\mathbb{R}}\chi_{[a_i,b_i]}u_{\lambda}=\int_{\mathbb{R}}\chi_{[a_i,b_i]}u$$ holds true when we choose $$u=\frac{2}{\pi} $$ By linearity this extends to all step functions, and hence by the above density argument to all integrable functions. Therefore we may substitute $u=\frac{2}{\pi}$ in $(*)$ to obtain

$$\lim_{\lambda \to +\infty}\int_{\mathbb{R}}fu_{\lambda}=\frac{2}{\pi}\int_{\mathbb{R}}f,\qquad \forall f\in L^1(\mathbb{R})$$

EDIT: More in general, the above argument shows that if $g:\mathbb{R}\to \mathbb{R}$ is a bounded periodic function such that its integral mean over a period is $\alpha$, then $$\lim_{\lambda\to +\infty}\int_{\mathbb{R}}f(x)g(\lambda x)dx=\alpha \int_{\mathbb{R}}f(x)dx,\qquad \forall f\in L^1(\mathbb{R}) $$


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