show that operator is bounded and find the norm

Question

Suppose that $V$ is a complex Hilbert space, and that $x_0 \in V$ is fixed. Show that the linear transformation $T: V \rightarrow \mathbb{C}$, defined by $Tx = <x, x_0>$ for every $x \in V$, is bounded, and find the norm $\|T\|$.

Answer 1

This is the norm of the map $\langle{x_0},{}\cdot{}\rangle$ in Riesz representation theorem, which has norm $\|x_0\|$.

Indeed, $\|T\|\leqslant \|x_0\|$ follows by Cauchy-Schwarz, since for $\|x\|=1$, $$\|Tx\|=|\langle x,x_0\rangle|\leqslant \|x\|\|x_0\|=\|x_0\|,$$ and for the other way, you have $$\|x_0\|=\frac{\langle x_0,x_0\rangle}{\|x_0\|}=\frac{Tx_0}{\|x_0\|}=T\Big(\frac{x_0}{\|x_0\|}\Big)\leqslant \sup_{\|x\|=1}|Tx|=\|T\|.$$

Answer 2

$|Tx| = |\langle x, x_0 \rangle| \leq \|x\| \|x_0\|$ by Cauchy-Schwarz. Hence, $T$ is bounded and $\|T\| \leq \|x_0\|.$ To show equality, you have to invent a vector $y$ that attains this bound. There is really only one reasonable choice for $y$ here. Do you see it?