Calculate $\lim_{|\lambda| \to +\infty} \int_{\mathbb{R}}f(x)|\sin(\lambda x)|dx $

Question

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.

Answer 1

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:

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

Hence,

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

Thus,

\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!

Answer 2

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 $$\lim_{\lambda\to +\infty}\int_{\mathbb{R}}f(x)g(\lambda x)dx=\alpha \int_{\mathbb{R}}f(x)dx,\qquad \forall f\in L^1(\mathbb{R}) $$

Answer 3

There's a major gap in the accepted answer; at the end we assume without proof that the limit of an infinite sum is the sum of the limits. If that didn't require proof analysis would be easier...

Say $(c_n)$ are the Fourier coefficients of the $2\pi$-periodic function $|\sin(t)|$:$$|\sin(t)|\sim\sum_{-\infty}^\infty c_ne^{int}.$$

Lemma. $\sum|c_n|<\infty$.

Waving of the Hands: the function is Lip${}_1$, so the derivative is in $L^\infty(\Bbb T)\subset L^2(\Bbb T)$. So $\sum n^2|c_n|^2<\infty$, which implies $\sum|c_n|<\infty$ by Cauchy-Schwarz.

Now say $f\in L^1(\Bbb R)$. Up to an irrelvant $2\pi$ somewhere we have $$\int f(t)|\sin(\lambda t)|=c_0\int f+\sum_{n\ne0}c_n\hat f(\lambda n).$$ And $\sum_{n\ne0}c_n\hat f(\lambda n)\to0$ as $\lambda\to\infty$ by Riemann-Lebesgue plus Dominated Convergence (in $\ell_1$).

Answer 4

This can be analyzed The approach we follow is based on a simple extension of Fejér's formula

Let $g$ be a bounded measurable $T$-periodic function, $\sigma_n\xrightarrow{n\rightarrow\infty}\infty$, and $\alpha_n$ any sequence in $\mathbb{R}$. For any $\phi\in L_1(\mathbb{R},\lambda)$, where $\lambda$ is Lebesgue's measure, $$\begin{align} \lim_{n\rightarrow\infty}\int_{\mathbb{R}} \phi(x)g(\sigma_nx+\alpha_n)\,dx=\Big(\frac{1}{T}\int^T_0 g(x)\,dx\Big)\Big(\int_{\mathbb{R}} \phi(x)\,dx\Big)\tag{1}\label{one} \end{align}$$

A proof of this result can me found here

In the case of the OP, $g(x)=|\sin x|$ which is $\pi$-periodic and $\int^\pi_0|\sin x|\,dx =2$. Thus, for any $f\in\mathcal{L}_1(\mathbb{R})$ $$ \lim_{\lambda\rightarrow\infty}\int_\mathbb{R} f(x)|\sin \lambda x|\,dx = \frac{2}{\pi}\int_{\mathbb{R}}f(x)\,dx$$