Why $\sin(nx)$ converges weakly in $L^2(-\pi,\pi)$? Can anybody tell me why $\sin(nx)$ converges weakly in $L^2(-\pi,\pi)$. I can't see how $\sin(nx)$ can converge?
Explanation with any other example will be nice as well. 
 A: $\newcommand{\norm}[1]{\lVert{#1}\rVert}$Up to a constant, the functions $\sin nx$ are elements of the standard orthonormal basis in the Hilbert space $L_2(-\pi,\pi)$. (The constant is the same for all $n$'s.)
If $\{e_n; n\in\mathbb N\}$ is an orthonormal basis in some Hilbert space $X$, then for any $f\in X$ we have
$$f=\sum_{n\in\mathbb N} \langle f,e_n \rangle e_n$$
and 
$$\norm{f}^2 = \sum_{n\in\mathbb N} |\langle f,e_n \rangle|^2.$$
This formula is called Parseval's identity.
The fact that the series $\sum\limits_{n\in\mathbb N} |\langle f,e_n \rangle|^2$ converges implies that $\langle f,e_n \rangle \to 0$. This is true for each $f$, which means that $e_n$ converges weakly to $0$.
A: Edit: My previous proof was incorrect, showing that $\sin(x/n)\to 0$ weakly rather than $\sin(nx)$. 
Note that $\sin(nx)\in L^2([a,b])$ for any $a,b\in\mathbb R$ but $\sin(nx)\notin L^2(\mathbb R)$, so I assume you are working over some finite interval $[a,b]$. To see that $\sin(nx)\to 0$ weakly, note that the linear functionals on $L^2([a,b])$ are of the form $\int_a^b f(x)\cdot dx$ where $f\in L^2([a,b])$. For any $f\in L^2([a,b])$ and $\epsilon>0$, we have some step function $f'\in L^2([a,b])$ such that $\|f-f'\|_2<\epsilon$. Since $f'$ is a step function, we have (by definition) constants $c_1,\ldots,c_n$ such that $a=c_1<\cdots<c_n=b$ and $f'$ is constant on the intervals $[c_i,c_{i+1})$. Note that we need only show that $\lim\limits_{n\to\infty}\int_{c_i}^{c_{i+1}} f'(x)\sin(nx)dx=0$ for each $i$, as we can then put these together to get that $\lim\limits_{n\to\infty}\int_{a}^{b} f'(x)\sin(nx)dx=0$ and since $\int_a^b \sin(nx)\cdot dx$ is a continuous functional and $\|f-f'\|$ can be made arbitrarily small this shows that $\lim\limits_{n\to\infty}\int_{a}^{b} f(x)\sin(nx)dx=0$. But we see that 
$$\begin{eqnarray}\lim\limits_{n\to\infty}\int_{c_i}^{c_{i+1}} f'(x)\sin(nx)dx &=& f(c_i)\lim\limits_{n\to\infty}\int_{c_i}^{c_{i+1}} \sin(nx)dx\\
&=& \lim\limits_{n\to\infty}f(c_i) \left(\frac{\cos(nc_{i})}{n}-\frac{\cos(nc_{i + 1})}{n}\right)\\
&=&\lim\limits_{n\to\infty}\frac{f(c_{i})}{n}\big(\cos(nc_{i})-\cos(nc_{i + 1})\big)=0
\end{eqnarray}$$
thus $\sin(nx)\to 0$ weakly.
This technique is fairly standard for proving uniform convergence of $L^2$ functions: you prove convergence for any linear functional defined by a step function and then use the density of step functions in $L^2$ to complete the proof.
A: This actually holds for all $L^{p}([-\pi,\pi])$, $1<p<\infty$, and is a consequence of Riemann Lebesgue Lemma.

Fix a $1<p<\infty$, and set $p':=p/(p-1)$.
It is clear that $\sin(kx)\in L^{p}([-\pi,\pi])$ for any $1< p<\infty.$  Note that since $[-\pi,\pi]$ has finite lebesgue measure,  it is a standard fact from measure theory that $L^{q}([-\pi,\pi])\subset L^{1}([-\pi,\pi])$ for any $1< q<\infty.$ In particular, since $0<p-1< p\implies p'=p/(p-1)>p/p=1,$ we have $L^{p'}([-\pi,\pi])\subset L^{1}([-\pi,\pi])$. 
Let $g\in L^{p'}([-\pi,\pi])$, then by above $g\in L^{1}([-\pi,\pi])$. Then, it follows from Riemann-Lebesgue Lemma that $$\int_{-\pi}^{\pi}\sin(kx)g(x)dx=\dfrac{1}{2 i}\int_{-\pi}^{\pi}g(x)e^{ikx}dx-\dfrac{1}{2 i}\int_{-\pi}^{\pi}g(x)e^{-ikx}dx\longrightarrow 0,\text{as}\ k\longrightarrow\infty.$$
A: (I assume you mean $L^2$ on a bounded interval since $\sin(nx)$ has to be an element of $L^2$.)  A sequence converges weakly in a Hilbert space if its image under any bounded linear functional converges.  What are the bounded linear functionals on $L^2$?  Then use the Riemann Lebesgue lemma. 
