I am given an inner product space $X$, with a norm induced by the inner product. Given an element $b\in X$ we define $f(a)=\langle a,b\rangle$. Now I want to prove that $f$ is a bounded linear functional as well as finding its operator norm.
My take:
I can prove that $f$ is linear by calculating $f(\alpha x+\beta y)$ and showing that this is $\alpha f(x)+\beta f(y)$. My uncertainty is when I am to trying to show that $f$ is bounded. For $f$ to be bounded there has to exist a $C$ such that
$$
\langle a,b \rangle \leq C\| a \|
$$ for every $a\in X$. An upper bound on $f(a)$ can be given by Cauchy Schwarz,
\begin{equation}\label{123}
\langle a,b \rangle \leq \| a \|\| b \|.
\end{equation}
My guess is that $\| b \|$ is the operator norm and that we can show this
by calculating $f(b)$. As the norm is induced by the inner product we know that $\langle b,b \rangle=\| b \|$ but I'm not sure on how to continue/finish this or if I'm on the right path.
Best Answer
Cauchy Schwarz inequality indeed implies that the norm of $f$ is less or equal than $\Vert b \Vert$.
If $b = 0$, $f$ is the always vanishing operator and we're done.
If $b \neq 0$ then $c = \frac{b}{\Vert b \Vert}$ is such that $\Vert c \Vert = 1$ and $\vert f(c) \vert = \Vert b \Vert$. Hence the norm of $f$ is at least equal to $\Vert b \Vert$. Finally the norm of $f$ is equal to $\Vert b \Vert$.