Operator norm of the sum of a finite collection of bounded linear operator 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.
 A: *

*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 $\lVert (P_1 + \dots P_N)x \rVert \le \lVert x\rVert$.
A: Let's consider the simple case $T_1,T_2$.
(1) $V_1\perp V_2$, with $V_i=\overline{T_iH}$
$T_1H\perp T_2 H$, then $T_1H \subseteq(T_2 H)^{\perp}$ and $\overline{T_2H}\subseteq (T_2 H)^{\perp\perp}$.
(2) $H=V_1\oplus V_2 \oplus V_3$, with $V_3=(V_1\oplus V_2)^\perp$ and $V_3\subseteq Ker T_1^*\cap Ker T_2^*$.
If $z\in V_3\subseteq T_iH^\perp$, then $(T_ix,z)=(x,T_i^*z)=0$ for all $x\in H$, that is $V_3\subseteq Ker T_i^*$.
(3) Now we consider $S=T_1^*+T_2^*$ ( the * does not matter).
Let $v=v_1+v_2+v_3$ with $v_i\in V_i$.
$$\Vert Sv\Vert^2=(Sv,Sv)=\sum_{i,j,k}(T_iT_i^*v_j,v_k)=(T_1T_1^*v_1,v_1)+(T_2T_2^*v_2,v_2)\leq \Vert v_1\vert^2+\Vert v_2\Vert^2\leq \Vert v\Vert^2 .$$
The general case is similar by docomposite the space and analysis the Cauchy sequence $S_n v$ .
