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How can I show that $$ f_n=\frac{1}{n}\sum\limits_{k=1}^{n^2} e^{ikt} $$ converges weakly to $0$ in $L^2[-π,π] $ and that the sequence
$$ \left\Vert \frac{1}{n}(f_1+f_2+...+f_n)\right\Vert_2 $$ does not converge to $0$ ?

Thank you in advance.

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2 Answers 2

To prove the first part we need some estimations. We claim that the sequence $\{f_n:n\in\mathbb{N}\}$ is norm bounded. Indeed $$ \Vert f_n\Vert^2=\langle f_n, f_n\rangle= \frac{1}{n^2}\sum\limits_{k=1}^{n^2}\sum\limits_{m=1}^{n^2}\langle e^{ikt}, e^{imt}\rangle= \frac{1}{n^2}\sum\limits_{k=1}^{n^2}\sum\limits_{m=1}^{n^2}\int\limits_{[-\pi,\pi]} e^{ikt} e^{-imt}d\mu(t)= $$ $$ \frac{1}{n^2}\sum\limits_{k=1}^{n^2}\sum\limits_{m=1}^{n^2}2\pi \delta_{k,m}= \frac{1}{n^2}2\pi n^2=2\pi $$ Now note that for all $r\in\mathbb{N}$ we have $$ \lim\limits_{n\to\infty}\langle f_n, e^{irt}\rangle= \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n\langle e^{ikt}, e^{irt}\rangle= \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n2\pi \delta_{n,r}= \lim\limits_{n\to\infty}\frac{2\pi}{n}=0 $$ Since $\mathrm{span}\{e^{irt}:r\in\mathbb{N}\}$ is dense in $L^2([-\pi,\pi])$ and sequence $\{f_n:n\in\mathbb{N}\}$ is norm bounded, then we have $\lim\limits_{n\to\infty}\langle f_n, f\rangle=0$ for all $f\in L^2([-\pi,\pi])$. The last statement means that $f_n$ weakly converges to $0$ in $L^2([-\pi,\pi])$.

Now we proceed to the second part. Denote $$ F_n=\frac{1}{n}\sum\limits_{k=1}^n f_k $$ For all $k,m\in\{1,\ldots,n\}$ we have $$ \langle f_k, f_m\rangle= \frac{1}{km}\sum\limits_{r=1}^{k^2}\sum\limits_{s=1}^{m^2}\langle e^{irt},e^{ist}\rangle= \frac{1}{km}\sum\limits_{r=1}^{k^2}\sum\limits_{s=1}^{m^2}\int\limits_{[-\pi,\pi]}e^{irt}e^{-ist}d\mu(t)= $$ $$ \frac{1}{km}\sum\limits_{r=1}^{k^2}\sum\limits_{s=1}^{m^2}2\pi \delta_{r,s}=\frac{2\pi}{km}\min(k^2,m^2) $$ Hence the desired norm is $$ \Vert F_n\Vert^2=\langle F_n,F_n\rangle= \frac{1}{n^2}\sum\limits_{k=1}^n\sum\limits_{m=1}^n\langle f_k, f_m\rangle= \frac{1}{n^2}\sum\limits_{k=1}^n\sum\limits_{m=1}^n\frac{2\pi}{km}\min(k^2,m^2) $$ $$ \frac{2\pi}{n^2}\left(\sum\limits_{k=1}^n\sum\limits_{m=1}^{k-1}\frac{m^2}{km}+\sum\limits_{k=1}^n\frac{k^2}{kk}+\sum\limits_{m=1}^n\sum\limits_{k=1}^{m-1}\frac{k^2}{km}\right)= \frac{2\pi}{n^2}\left(\sum\limits_{k=1}^n\frac{1}{k}\sum\limits_{m=1}^{k-1}m+n+\sum\limits_{m=1}^n\frac{1}{m}\sum\limits_{k=1}^{m-1}k\right)= $$ $$ \frac{2\pi}{n^2}\left(\sum\limits_{k=1}^n\frac{k-1}{2}+n+\sum\limits_{m=1}^n\frac{m-1}{2}\right)= \frac{2\pi}{n^2}\left(\frac{n^2-n}{4}+n+\frac{n^2-n}{4}\right)= \pi\left(1+\frac{1}{n}\right) $$ Which yeilds $$ \lim\limits_{n\to\infty}\Vert F_n\Vert=\lim\limits_{n\to\infty}\sqrt{\pi\left(1+\frac{1}{n}\right)}=\sqrt{\pi} $$

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+1. If I may make a suggestion: You could shorten your proof for the first part somewhat by considering $(e^{inx})_{n}$ instead of the characteristic functions $\chi_{[a,b]}$. –  Sam Jun 24 '12 at 23:50
    
@SamL. That's a nice idea! Thanks! –  userNaN Jun 25 '12 at 5:10

Another proof can be achieved by using the isomorphism $L^2[-\pi,\pi]\simeq\ell^2(\mathbb{N})$, where you map $e^{ikt}\mapsto \delta_k$ (there's a factor $\sqrt{2\pi}$ around, but it affect convergence or not to zero).

Then, for any $a\in\ell^2(\mathbb{N})$, fix $\varepsilon>0$ and choose $M$ such that $\sum_{k>M}|a_k|^2<\varepsilon^2$. Then, using Hölder, $$ |\langle f_n,a\rangle|=\frac1n\,\left|\sum_{k=1}^{n^2}a_k\right| \leq\frac1n\,\sum_{k=1}^{n^2}|a_k| =\frac1n\,\sum_{k=1}^{M}|a_k|+\frac1n\,\sum_{k=M+1}^{n^2}|a_k| \leq\frac1n\,\sum_{k=1}^{M}|a_k|\\ +\frac1n\,\left(\sum_{k=M+1}^{n^2}|a_k|^2\right)^{1/2}(n^2-M)^{1/2} \\ \leq\frac1n\,\sum_{k=1}^{M}|a_k|+\varepsilon.$$ Letting $n\to\infty$ we get $\limsup_n|\langle f_n,a\rangle|<\varepsilon$, and as $\varepsilon$ was arbitrary, we get $\lim_n\langle f_n,a\rangle=0$.

For the sequence $\frac1n(f_1+\cdots+f_n)$ we can estimate: $$ \|\frac1n(f_1+\cdots+f_n)\|_2^2=\left\|\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\|_2^2=\left\langle\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j,\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\rangle\\ =\frac1{n^2}\sum_{j=1}^n\sum_{k=1}^{j^2}\sum_{l=1}^n\sum_{h=1}^{l^2}\frac{\langle\delta_k,\delta_h\rangle}{lj} =\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{n}\frac{\min\{l^2,j^2\}}{lj}\\ =\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{\min\{l^2,j^2\}}{lj} +\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{\min\{l^2,j^2\}}{lj} =\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l^2}{lj} +\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j^2}{lj}\\ =\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l}{j} +\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j}{l} =\frac1{n^2}\sum_{j=1}^n\frac{j(j+1)}{2j} +\frac1{n^2}\sum_{l=1}^n\sum_{j=1}^{l-1}\frac{j}l\\ =\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2} +\frac1{n^2}\sum_{l=1}^n\frac{(l-1)l}{2l} =\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2} +\frac1{n^2}\sum_{l=1}^n\frac{l-1}2\\ =\frac1{n^2}\sum_{k=1}^nk=\frac{n(n+1)}{2n^2}=\frac12+\frac1{2n} $$

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Not that it matters but there's an off-by-one error in the number of summands in the Schwarz inequality (should be $n^2-M$). –  WimC Jun 24 '12 at 19:13
    
Thanks a lot! I also corrected a missing 2 in the second part. –  Martin Argerami Jun 24 '12 at 23:15
    
Question: How many $\sum$ symbols did Martin use to get this answer? (+1 for endurance) –  draks ... Jun 25 '12 at 17:54
    
Actually, there was a lot of copy and paste... :D –  Martin Argerami Jun 25 '12 at 18:13

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