Why does strong convergence imply weak convergence in a Hilbert space?

Conditions

• $H$ is a Hilbert space.
• $(x_n)$ is a sequence in $H$.
• $(x_n)$ converges strongly to $l$ (i.e., $\|x_n - l\| \to 0$).
• $v$ is an arbitrary element of $H$.

Reasoning, step by step

1. Start with the absolute difference of the inner products for an arbitrary test vector $v$: $|\langle x_n, v \rangle - \langle l, v \rangle|$.
2. Use the linearity of the inner product to rewrite this as $|\langle x_n - l, v \rangle|$.
3. Apply the Cauchy-Schwarz inequality to bound this expression: $|\langle x_n - l, v \rangle| \le \|x_n - l\| \|v\|$.
4. Observe that $\|v\|$ is a fixed finite number since $v$ is fixed.
5. Use the assumption of strong convergence, which means $\|x_n - l\| \to 0$ as $n \to \infty$.
6. Conclude that the upper bound $\|x_n - l\| \|v\|$ goes to zero, so $|\langle x_n, v \rangle - \langle l, v \rangle| \to 0$.
7. This satisfies the definition of weak convergence for all $v \in H$.

Example

The proof slide displays the chain: $|\langle x_n, v \rangle - \langle l, v \rangle| = |\langle x_n-l, v \rangle| \le \|x_n-l\|\,\|v\|$. The lecturer explains that the right-hand side goes to zero because $x_n$ strongly converges to $l$.

Common misconceptions

• Assuming that weak convergence implies strong convergence, which is generally false without additional conditions like norm convergence.
• Thinking that the Cauchy-Schwarz inequality is only used for bounding norms, rather than inner products.

Watch the explanation

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