How can I show that $f_n$= $\frac 1n\sum\limits_{i=1}^{n^2} e^{ikt}$ converges weakly to $0$ in $L^2[-π,π] $ 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.
 A: 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
$$
\lim\limits_{n\to\infty}\Vert F_n\Vert=\lim\limits_{n\to\infty}\sqrt{\pi\left(1+\frac{1}{n}\right)}=\sqrt{\pi}
$$
A: 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:
$$
\|\frac1n(f_1+\cdots+f_n)\|_2^2=\left\|\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\|_2^2=\left\langle\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j,\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\rangle\\
=\frac1{n^2}\sum_{j=1}^n\sum_{k=1}^{j^2}\sum_{l=1}^n\sum_{h=1}^{l^2}\frac{\langle\delta_k,\delta_h\rangle}{lj}
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{n}\frac{\min\{l^2,j^2\}}{lj}\\
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{\min\{l^2,j^2\}}{lj}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{\min\{l^2,j^2\}}{lj}
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l^2}{lj}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j^2}{lj}\\
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l}{j}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j}{l}
=\frac1{n^2}\sum_{j=1}^n\frac{j(j+1)}{2j}
+\frac1{n^2}\sum_{l=1}^n\sum_{j=1}^{l-1}\frac{j}l\\
=\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2}
+\frac1{n^2}\sum_{l=1}^n\frac{(l-1)l}{2l}
=\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2}
+\frac1{n^2}\sum_{l=1}^n\frac{l-1}2\\
=\frac1{n^2}\sum_{k=1}^nk=\frac{n(n+1)}{2n^2}=\frac12+\frac1{2n}
$$
