How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$?

Question

I need to show that for a bounded linear operator, $T$, on a Hilbert space: \begin{align*} \|T\|^2=\|T^*T\| \end{align*} All I have so far:

\begin{align*} \|T^*T\|&=\sup\{|\langle T^*Tf,g \rangle |:\|f\|\le1,\|g\|\le1\} \\ &= \sup\{|\langle Tf,Tg \rangle|:\|f\|\le1,\|g\|\le1\} \\ &\le \sup_{\|f\|\le1}\|Tf\|\sup_{\|g\|\le1}\|Tg\| \end{align*}

Not too sure what to do from here (or even if this is right so far)

Any help is appreciated.

Answer 1

$ \newcommand{\norm}[1]{\left\|{#1}\right\|} \newcommand{\ip}[1]{\left\langle{#1}\right\rangle} $You might find it easier to use an alternative but equivalent definition of the operator norm: if $S \in B(H)$, then $$ \norm{S} = \sup_{\norm{x}=1} \norm{Sx}. $$

Now:

1. On the one hand, $$ \norm{T} = \sup_{\norm{x}=1} \norm{Tx} = \sup_{\norm{x}=1} \sqrt{\ip{Tx,Tx}}. $$ Since $\ip{Tx,Tx} = \ip{x, T^\ast T x} \leq \norm{T^\ast T}\norm{x}$, what can you conclude?

2. On the other hand, $$ \norm{T^\ast T} = \sup_{\norm{x}=1} \norm{T^\ast Tx}. $$ Since $\norm{T^\ast T x} \leq \norm{T^\ast} \norm{Tx}$, where $\norm{T^\ast}=\norm{T}$ (why?), what can you conclude?

In terms of wider context, the identity $\norm{T^\ast T} = \norm{T}^2$ is called the $C^\ast$-identity, and is the key fact that makes the theory of $C^\ast$-algebras (e.g., $B(H)$ for $H$ a Hilbert space) so much easier (for lack of a better word) than the theory of more general Banach ($\ast$-)algebras, just as the theory of Hilbert spaces is so much easier than the theory of more general Banach spaces.

Answer 2

The first notice that $\|T\|=\|T^*\|$: From definition of operator norm \begin{align*} \|T^*\|&=\sup_{\|y\|=1}\|T^*y\|=\sup_{\|y\|=1}\sup_{\|x\|=1}|(x,T^*y)|= \sup_{\|y\|=1}\sup_{\|x\|=1}|(Tx,y)|\\ &=\sup_{\|x\|=1}\sup_{\|y\|=1}|(Tx,y)|=\sup_{\|x\|=1}\|Tx\|=\|T\| \end{align*} Therefore $T^*\in\mathcal{L}(H,H)$ and $\|T\|=\|T^*\|$.

To conclude, let $x\in H$ with $\|x\|=1$. It follows from \begin{align} \|Tx\|^2=(Tx,Tx)=(x,T^*Tx)\leq\|x\|\|T^*Tx\|\leq \|T^*T\|\leq\|T^*\|\|T\|=\|T\|^2 \end{align} that $\|T\|^2\leq\|T^*T\|=\|T\|^2$.