Does $\sin (nx)$ converge in $L^2$?

Question

I was just introduced the concept that if $(f_n)$ converges in $L^2$ topology to $g(x)\in L^2([0,2\pi])$ then $\lim_{n\to\infty}\int^{2\pi}_0|f_n(x)-g(x)|^2dx=0$. I would appreciate any hint to how to start proving for example $sin(nx)$ diverges in $L^2$. I know that in the normal sense it doesn't even have a pointwise limit in real.

Also given was a definition of weak convergence: a function $g(x)\in L^2([0,2\pi])$ such that for any continuous $h$ on $[0,2\pi]$ we have $\lim_{n\to\infty}\int^{2\pi}_0(f_n(x)-g(x))h(x)dx=0$. Is $\sin (nx)$ convergent in this sense?

Answer 1

In fact, $\sin(nx) \to 0$ weakly. Let's consider $X= L^2([0,2 \pi])$, then the dual (of $X$) space is $X^* = (L^2([0,2 \pi]))^*=L^2([0,2 \pi])$, which is the space of all linear continuous functionals $F$ from $X$ to $\mathbb{F}$ (say $\mathbb F =\mathbb R$ here). That is, for each $g \in X$ we define (an inner product) $ X^* \ni F_g: X \to \mathbb R$

$$\langle f, g \rangle=: F_g(f):=\int_0^{2 \pi} f(x) g(x)dx, \,\,\, \forall f \in X.$$

Let $f_n(x)= \sin(nx)$. Fix $\epsilon >0$, by the density of simple function in $L^p$ spaces ($L^2$ here) we have a step function $s$ such that

$$\|f-s\|_{L^2[0,2\pi]} < \epsilon.$$

Since simple function is a finite linear combination of characteristic functions, then is suffices to show the weak convergence on one characteristic functions, say $c_i \chi_{[a_i,a_{i+1}]}(x)$. Then,

$$\begin{eqnarray} \lim\limits_{n\to\infty}\int_{a_i}^{a_{i+1}} s(x)\sin(nx)dx &=& f(c_i)\lim\limits_{n\to\infty}\int_{a_i}^{a_{i+1}} \sin(nx)dx\\ &=&f(c_i) \lim\limits_{n\to\infty} \left(\frac{\cos(na_{i})}{n}-\frac{\cos(na_{i + 1})}{n}\right)\\ &=&f(c_i)\lim\limits_{n\to\infty}\frac{\big(\cos(na_{i})-\cos(na_{i + 1})\big)}{n}=0 \end{eqnarray}$$

therefore, $f_n= \sin(nx)\to 0$ weakly (or equivalently, $\langle f_n,g \rangle \to \langle 0,g\rangle \,\,\,\forall g \in X$).

Answer 2

Letting $f_n(x)=\sin(nx)$, note that $\|f_n\|_2=\sqrt\pi$.

Let us first show that $f_n\to 0$ weakly, using @HyperbolicPDEfriend's comment. For any $h\in L^2([0,2\pi])$ and any $\epsilon>0$, choose some $g\in C^1([0,2\pi])$ such that $\|h-g\|_2<\frac\epsilon{\sqrt\pi}$. Let $M$ be an upper bound for both $|g|$ and $|g'|$. Integrating by parts, $\left|\int^{2\pi}_0f_n(x)g(x)dx\right|=\frac1n\left|g(2\pi)-g(0)+\int^{2\pi}_0\cos(nx)g'(x)dx\right|\le\frac{2M+2\pi M}n<\epsilon$ for $n$ large enough, and then $\left|\int^{2\pi}_0f_n(x)h(x)dx\right|\le\|f_n\|_2\|h-g\|_2+\left|\int^{2\pi}_0f_n(x)g(x)dx\right|<2\epsilon$.

We can now derive that $(f_n)$ diverges in $L^2([0,2\pi])$ because if it did converge, it would necessarily be towards its weak limit $0$, which is impossible since $\|f_n\|_2\not\to0$.