Where is Cauchy-Schwarz used in the strong-to-weak proof?

Conditions

• $H$ is a Hilbert space.
• $(x_n)$ is a sequence in $H$ converging strongly to $l$.
• $v$ is an arbitrary element of $H$.

Reasoning, step by step

1. Start with the expression $|\langle x_n, v \rangle - \langle l, v \rangle|$.
2. Rewrite it using linearity as $|\langle x_n - l, v \rangle|$.
3. Apply the Cauchy-Schwarz inequality to get $|\langle x_n - l, v \rangle| \le \|x_n - l\| \|v\|$.
4. Use this inequality to establish an upper bound for the inner product difference.

Example

The displayed chain in the proof includes the step $|\langle x_n-l, v \rangle| \le \|x_n-l\|\,\|v\|$. The lecturer reads the right-hand side as the norm of $x_n$ minus l times the norm of v.

Common misconceptions

• Thinking that Cauchy-Schwarz is used to prove strong convergence itself.
• Assuming that Cauchy-Schwarz directly gives the limit, rather than providing a bound that is then evaluated using the definition of strong convergence.

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Explore next

• Why does strong convergence imply weak convergence in a Hilbert space?
• What extra assumption turns weak convergence into strong convergence?