Problem. Let $\lambda_{1},\lambda_{2},\dots$ be real numbers. Argue that $$ f\left(x\right)=\sum_{n=1}^{\infty}\frac{e^{i\lambda_{n}x}}{n^{2}} $$ defines a continuous bounded function on $\mathbb{R}$ and then show that the limit $$ \lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T}^{T}f\left(x\right)dx $$ exists.

It is easy to see that the above problem trivially holds when $|\lambda_n|$ has nonzero lower bound. The question has no additional condition on $\lambda_n$.

Attempt to solve this problem. First, it is clear that $f$ is continuous on $\mathbb{R}$ and it is bounded by Weierstrass M-test. So \begin{align*} \int_{-T}^{T}f\left(x\right)dx & =\sum_{n=1}^{\infty}\frac{1}{n^{2}}\int_{-T}^{T}e^{i\lambda_{n}x}dx\\ & =\sum_{n=1}^{\infty}\frac{1}{n^{2}}\frac{e^{i\lambda_{n}T}-e^{-i\lambda_{n}T}}{i\lambda_{n}}\\ & =\sum_{n=1}^{\infty}\frac{2}{n^{2}}\times\frac{\sin\left(\lambda_{n}T\right)}{\lambda_{n}}. \end{align*} So $$ \frac{1}{T}\int_{-T}^{T}f\left(x\right)dx=2\sum_{n=1}^{\infty}\frac{1}{n^{2}}\frac{\sin\left(\lambda_{n}T\right)}{\lambda_{n}T}. $$ For convenience, we define $$ g_{n}\left(T\right)=\sum_{k=1}^{n}\frac{1}{k^{2}}\frac{\sin\left(\lambda_{k}T\right)}{\lambda_{k}T}. $$ Since $$ \left|\frac{\sin x}{x}\right|\le1 $$ for all $x\in\mathbb{R}$, $g_{n}$ converges uniformly on $\mathbb{R}$ by M-test. Let us denote $g\left(T\right)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}\frac{\sin\left(\lambda_{n}T\right)}{\lambda_{n}T}$.

Due to uniform convergence, we have (this is my problem point) \begin{align*} \lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T}^{T}f\left(x\right)dx & =\lim_{T\rightarrow\infty}2g\left(T\right)\\ & =2\lim_{T\rightarrow\infty}\lim_{n\rightarrow\infty}g_{n}\left(T\right)\\ & =2\lim_{n\rightarrow\infty}\lim_{T\rightarrow\infty}g_{n}\left(T\right). \end{align*} Now note that $$ \lim_{T\rightarrow\infty}g_{n}\left(T\right)=0. $$ So we arrive $$ \lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T}^{T}f\left(x\right)dx=0, $$ which completes the proof.

The problem is the possibility of interchanging limit. We know that if $f_n \rightarrow f$ converges uniformly on $E$ and $x$ is a limit point of $E$, then we can interchange limit. $\infty$ is not a limit point in usual real number system. Can we have similar result for this situation? Or is there a counterexample for this situation?

Thank you in advanced.

  • $\begingroup$ You can certainly not have $\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T}^{T}f\left(x\right)dx=0$ for all real sequences $\lambda_1, \lambda_2, \dots$. For example, if this sequence is the always vanishing sequence, $f$ is constant and the limit you're looking equals twice this constant. $\endgroup$ – mathcounterexamples.net Dec 14 '16 at 14:45
  • $\begingroup$ Possible duplicate post here: math.stackexchange.com/questions/1851684/… (searched by approach0.xyz/search/…) $\endgroup$ – Wei Zhong Dec 14 '16 at 15:55
  • $\begingroup$ @mathcounterexamples.net Thank you for your comments. I didn't consider the case the zero cases. $\endgroup$ – Will Kwon Dec 14 '16 at 18:26

One easy way to deal with it is to just adjoin the "missing" point to the domain. Define $S \colon [0,+\infty] \to \mathbb{R}$ by

$$S(t) = \begin{cases} 1 &, t = 0 \\ \dfrac{\sin t}{t} &, 0 < t < +\infty \\ 0 &, t = +\infty.\end{cases}$$

Then $S$ is a continuous function on $[0,+\infty]$ with $\lvert S(t)\rvert \leqslant 1$ for all $t$. Then define $h_{\lambda}(t) = S(\lvert\lambda\rvert t)$ for $\lambda \neq 0$ and $h_0(t) = 1$. Every $h_\lambda$ is a continuous function on $[0,+\infty]$ with $\lvert h_\lambda(t)\rvert \leqslant 1$, and so

$$g = \sum_{n = 1}^\infty \frac{2}{n^2} h_{\lambda_n}$$

is a continuous function on $[0,+\infty]$ by the Weierstraß $M$-test. Hence

$$\frac{1}{T}\int_{-T}^T f(t)\,dt = g(T) \xrightarrow{T\to+\infty} g(+\infty) = \sum_{n = 1}^\infty \frac{2}{n^2} h_{\lambda_n}(+\infty) = \sum_{\lambda_n = 0} \frac{2}{n^2}.$$

  • $\begingroup$ I may be missing something... you have $h_{\lambda_n}(+\infty) = 0$ not $h_{\lambda_n}(+\infty) = 1$ ? $\endgroup$ – mathcounterexamples.net Dec 14 '16 at 15:32
  • $\begingroup$ For $\lambda \neq 0$, we have $h_{\lambda}(+\infty) = 0$. But $h_0(+\infty) = 1$. So at the end, the terms with $\lambda_n = 0$ remain. $\endgroup$ – Daniel Fischer Dec 14 '16 at 15:34
  • $\begingroup$ Yes... I missed the set of indexes of your last sum. $\endgroup$ – mathcounterexamples.net Dec 14 '16 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.