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Let $H$ be a Hilbert space and $T \in B(H)$, the set of all bounded operator on $H$, such that $T$ is self-adjoint. Let $\left\{P_k\right\}_{k=1}^n$ be a finite family of orthogonal projections in $B(H)$ such that $\displaystyle\sum_{k=1}^n P_k=I,$ $I$ being the identity element in $B(H)$ and $TP_k=P_kT,~\forall k\in \{1,2,\cdots,n\}.$ Then I want to show that $\|T\|=\max\{\|TP_k\|:k\in\{1,2,\cdots,n\}\}$,

First I start with only two projections, say $P$ and $I-P$. Since $T=T^*$, then by definition we have $$\|T\|:=\sup\{\langle Tx,x\rangle: x \in H, \|x\|=1\}.$$ So we have, $$ \begin{align}\|T\| &=\sup\{\langle T(Px+(I-P)x),Px+(I-P)x\rangle: x \in H, \|x\|=1\} \\ &=\sup\{\langle TPx,Px\rangle+\langle T(I-P)x,(I-P)x\rangle: x \in H, \|x\|=1\}\end{align}$$ Now I am unable to proceed from here.Please help me to solve this. Thank you for your time and help.

abcdmath
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1 Answers1

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You don't need $T$ to be self-adjoint.

First, $$M:=\max_k\|TP_k\|\le\|T\|$$ because $\|TP_k\|\le\|T\|\|P_k\|\le\|T\|$ (since $P_k$ is an orthogonal projection).

Conversely, for every $x\in H,$ let $x_k:=P_kx$ and $y_k:=Tx_k=P_kTx.$ Then, since the $P_k$'s are mutually orthogonal, $$\begin{align}\|Tx\|^2&=\sum\|y_k\|^2\\&=\sum\|TP_kx_k\|^2\\&\le\sum\|TP_k\|^2\|x_k\|^2\\&\le\sum M^2\|x_k\|^2\\&=M^2\|x\|^2 \end{align}$$ hence $\|T\|\le M.$

Anne Bauval
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