Prove $\sin(kx) \rightharpoonup 0$ as $k \to \infty$ in $L^2(0,1)$ 
I want to show that $u_k(x)= \sin(kx) \rightharpoonup 0$ as $k \to \infty$ in $L^2(0,1)$.

We know trivially that $0 \in L^2(0,1)$.
I need to show that $\langle u^*,\sin(kx) \rangle \to \langle u^*, 0 \rangle$ for each bounded linear functional $u^* \in L^2(U)$, where $L^2(U)$ is a dual space of itself (since $L^2$ is a Hilbert space).
I think I need to show that, as $k \to \infty$,
$$\int_0^1 u^* \sin(kx) \, dx \to 0.$$

Let $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \sin (2\pi i \,n x) + b_n \cos (2\pi i \,n x) \in L^2[0,1]$ then we have Bessel inequality:
$$ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) < \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 < \infty$$
Then the Fourier coefficients tend to zero $|a_n| =|\langle f ,\sin (2\pi i \,n x)\rangle | \to 0$

Any orthogonormal basis tends the intuition here is that $\{ \sin kx \}$ or $\{e^{2\pi i \, kx}\}$ oscillate so violently we may as well just substitute $0$ there.  Weak convergence formalizes this asymptotic behavior in many cases.
