I recently got some difficulty with my homework question. The question is:

Let $T_1,\dots,T_N$ be a finite collection of bounded linear operators on a hilbert space $H$, each of operator norm $\le 1$.

Suppose that $T_kT_j^\ast = T_k^\ast T_j = 0$ whenever $j \neq k$.

Show that $\displaystyle \sum_{i=1}^N T_i$ satisfies $\|T\| \le 1$.

For the condition $T_k^\ast T_j= 0$, $j \neq k$, I can show $T_k$ and $T_j$ have orthogonal ranges: since $T_k^\ast T_j= 0$, $T_k^\ast T_j f= 0$ for any $f \in H$, so it follows $(T_k^\ast T_j f,g)= 0$ for any $g$. But $(T_k^\ast T_j f,g)=(T_jf,T_kg)=0$, so $T_j$ and $T_k$ should have orthogonal ranges.

However, I cant do anything for condition $T_kT_j^\ast= 0$. But the hint says for $T_kT_j^\ast = 0$, $j \neq k$,introduce the orthogonal projection $P_i$ onto the closure of the range $T_i^\ast$, and show that $T_i f$ = $T_i P_i f$.

I dont quite understand the hint and I need some help with this question.

Beside, the question has a part 1, which is to show if $P_1$ and $P_2$ are two orthogonal projections, with orthogonal ranges, then $P_1 + P_2$ is also an orthogonal projection. I've done this part 1, but I guess this conclusion is helpful for solving this latter part.

  • Proof of the hint: we have to show that $T_j(I-P_j)f=0$ for each $j$ and $f\in H$. Note that $(I-P_j)f=:g$ is in the orthogonal complement of the range of $T_j^*$. Hence $\langle g,T_j^*h\rangle =0$ for each $h$, and $\langle T_jg,h\rangle=0$. Taking $h=T_jg$, we have what we wanted.

  • We have, using the hint: \begin{align} \left\lVert\sum_{j=1}^NT_jx\right\rVert^2&=\sum_{j=1}^N\sum_{k=1}^N\langle T_jP_jx,T_kP_kx\rangle\\ &=\sum_{j=1}^N\sum_{k=1}^N\langle P_jx, T_j^*T_kP_kx\rangle\\ &=\sum_{j=1}^N\langle P_jx, \color{green}{T_j^*}T_jP_jx\rangle\\ &=\sum_{j=1}^N\langle \color{green}{T_j}P_jx, T_jP_jx\rangle\\ &=\sum_{j=1}^N\lVert T_jP_j x\rVert^2\\ &\leq \sum_{j=1}^N\color{red}{\underbrace{\lVert T_j\rVert}_{\leq 1}}^2\lVert P_jx\rVert^2\\ &\leq \sum_{j=1}^N\lVert P_jx\rVert^2\\ &=\left\lVert\left(\sum_{j=1}^NP_j\right)x\right\rVert^2, \end{align} where the last line follows from the orthogonality of the projections.

By the first part, $P_1+\dots+P_N$ is a projection, and we can conclude $\lVert Tx\rVert \le \lVert x \rVert \implies \lVert T \rVert \le 1$ since $\lVert (P_1 + \dots P_N)x \rVert \le \lVert x\rVert$.

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  • $\begingroup$ Can you actually show why Tif=TiPif ? Also , I dont quite understand the last two step and how they imply the conclusion . $\endgroup$ – lindamac Aug 10 '12 at 13:45

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