Need a help in the proof of an example of Riesz representation theorem.

Question

The example said:"A functional $f$ on $l_{2}$ is linear and bounded iff there exists a $y = (\beta_{1}, \beta_{2}, .......) \in l_{2}$ such that for all $x = (a_{1}, a_{2}, .......) \in l_{2}$, $$f(x) = \sum_{k=1}^{\infty} a_{k} \bar{\beta_{k}}." $$

For the direction $\Rightarrow$, the statement is true by Riesz representation theorem, for the direction $\Leftarrow$, $f(x)$ is clearly linear. Now the proof that $f(x)$ is bounded is as follows:

$|f(x)| = |\sum_{k=1}^{\infty} a_{k} \bar{\beta_{k}}| = |<x,y>| \leq ||x||||y||$ (using Cauchy Schwarz inequality) . hence $||f|| \leq ||y||$..... am I right?

But If so, I have a question, why is Cauchy-Schwarz inequality is applicable in infinite dimensional spaces?

Thanks!

Answer 1

Yes, you are correct: Cauchy-Schwarz is applicable and hence the functional is bounded. The inequality holds because the proof of Cauchy-Schwarz only uses the structure of inner product spaces, i.e. only uses the properties of inner products and vectors. So no reference to basis, hence dimension, is necessary.

Answer 2

1. Schwarz not Schwartz.

2. If $X$ is a inner product space with inner product $<\cdot,\cdot>$ then the Cauchy-Schwarz inequality

$$|<x,y>| \le ||x|| \cdot||y||$$ holds, where $||x||=<x,x>^{1/2}$

$ \dim X < \infty $ is not needed.