What extra assumption turns weak convergence into strong convergence?

Conditions

• $H$ is a Hilbert space.
• $(x_n)$ is a sequence in $H$ converging weakly to $l$.
• $\|x_n\|$ converges to $\|l\|$.

Reasoning, step by step

1. Assume $(x_n)$ converges weakly to $l$.
2. Assume additionally that $\|x_n\| \to \|l\|$.
3. Use the polarization identity or expand $\|x_n - l\|^2 = \|x_n\|^2 - 2\langle x_n, l \rangle + \|l\|^2$.
4. Note that $\langle x_n, l \rangle \to \langle l, l \rangle = \|l\|^2$ by weak convergence (choosing $v=l$).
5. Conclude that $\|x_n - l\|^2 \to \|l\|^2 - 2\|l\|^2 + \|l\|^2 = 0$, so $\|x_n - l\| \to 0$, which is strong convergence.

Example

The proposition slide states: 'Let ($x_n$) be a weakly converging sequence to l. Further assume that ||$x_n$|| converges to ||l||. Then ($x_n$) converges strongly to l.'

Common misconceptions

• Thinking that any bounded sequence converging weakly also converges strongly.
• Assuming that the extra assumption is about the convergence of inner products, which is already part of the definition of weak convergence.

Watch the explanation

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