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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.

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  • $\begingroup$ 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
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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

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

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  • $\begingroup$ 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

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