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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.

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By $L_2$ do you mean $L_2(-\pi,\pi)$, i.e. the space of functions such that $\int_{-\pi}^\pi f(x) dx < \infty$? – Martin Sleziak May 28 '12 at 6:45
@Martin Sleziak Yes – Theorem May 28 '12 at 6:47
This is probably an overkill, but from Parseval's identity we know that for each $f\in L_2$ we have $\sum (c_k^2+d_k^2)<\infty$, where $c_k=\int_{-\pi}^\pi f(x) \sin kx$ and $d_k$ is the same thing with cosines. This implies that $c_k\to 0$. Since this is true for each $f$, this is the weak convergence of $\sin nx$. – Martin Sleziak May 28 '12 at 6:48
up vote 5 down vote accepted

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 $$\begin{eqnarray}\lim\limits_{n\to\infty}\int_{c_i}^{c_{i+1}} f'(x)\sin(nx)dx &=& f(c_i)\lim\limits_{n\to\infty}\int_{c_i}^{c_{i+1}} \sin(nx)dx\\ &=& \lim\limits_{n\to\infty}f(c_i) \left(\frac{\cos(nc_{i+1})}{n}-\frac{\cos(nc_i)}{n}\right)\\ &=&\lim\limits_{n\to\infty}\frac{f(c_i)}{n}(\cos(nc_{i+1})-\cos(nc_i))=0 \end{eqnarray}$$ 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.

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I might have misunderstood something, but the graph of $\sin nx$ is not stretched, but the waves are getting "more dense" with increasing $n$. – Martin Sleziak May 28 '12 at 6:42
@MartinSleziak Whoops. Totally reversed that. Editing to fix. – Alex Becker May 28 '12 at 6:45

$\newcommand{\norm}[1]{\lVert{#1}\rVert}$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 $0$.

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I like your argument. It is quite neat. However, it doesn't work if you replace $L^2$ by $L^p$ for a general $p \in [1,\infty)$ because then you don't have an inner product. Alex's works for $p \in [1,\infty)$. – chango Jun 26 '12 at 14:22

(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.

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Riemann-Lebesgue lemma is appliable to $L^1$ functions. – Norbert May 28 '12 at 6:35
I've added the link to Riemann Lebesgue lemma - it might be useful for the users not familiar with it. I hope you don't mind.\\ In case you had some different version of this lemma in mind, you can replace the link or add the formulation to the post. (I had same question/objection as Norbert.) – Martin Sleziak May 28 '12 at 6:36
@Norbert: I assumed $L^2$ on a bounded interval. Otherwise $\sin(nx)$ would not be an element in $L^2$. – WimC May 28 '12 at 6:37

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