Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 23H / a

If $x=0$, the claim is immediate. Otherwise weak convergence givesBy Cauchy-Schwarz,Taking the lower limit and dividing by $\Vert x\Vert$ proves the weak lower semicontinuity of the Hilbert norm:

Strong convergence immediately implies convergence of norms by the reverse triangle inequality.

Conversely, suppose $x_n\rightharpoonup x$ and $\Vert x_n\Vert \to \Vert x\Vert$. ThenThus $x_n\to x$ strongly. This is the Radon-Riesz theorem for Hilbert spaces.

No. In $H={\ell }^2$, let $x_n=e_n$ be the standard orthonormal sequence. For any $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$. However,for $n\ne m$. No subsequence is Cauchy, hence none converges strongly. This is a weakly null orthonormal sequence.