Theorem (The Radon-Riesz Theorem): Let $H$ be a Hilbert space. Then a sequence $(h_n)$ in $H$ norm converges to $h \in H$ if and only if $(h_n)$ weakly converges to $h$ and $\displaystyle{\lim_{n \to \infty} \| h_n \| = \| h \|}$.
• Proof: $\Rightarrow$ Suppose that $(h_n)$ norm converges to $h$. Since the norm convergence implies weak convergence we have that $(h_n)$ weakly converges to $h$. Furthermore, since $(h_n)$ norm converges to $h$ we have that $\displaystyle{\lim_{n \to \infty} \| h_n - h \| = 0}$. Since $|\| h_n \| - \| h \|| \leq \| h_n - h \|$ we conclude that $\displaystyle{\lim_{n \to \infty} \| h_n \| = \| h \|}$ as well.
• $\Leftarrow$ Suppose that $(h_n)$ weakly converges to $h$ and $\displaystyle{\lim_{n \to \infty} \| h_n \| = \| h \|}$. Observe that:
• Since $(h_n)$ weakly converges to $h$, for all $f \in H^*$ we have that $\displaystyle{\lim_{n \to \infty} f(h_n) = f(h)}$. But since $H$ is a Hilbert space, by The Riesz Representation for Hilbert Spaces, we have that $\displaystyle{\lim_{n \to \infty} \langle h_n, h \rangle = \langle h, h \rangle = \| h \|^2}$. Hence:
• Therefore $(h_n)$ norm converges to $h$. $\blacksquare$