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:

$$ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) &lt; \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 &lt; \infty$$

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.
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:

$$ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) &lt; \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 &lt; \infty$$

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.