I was reading an article by Hayashi and Nagaoka titled "General formulas for capacity of classical-quantum channels", here. On Page 9, the Hayashi Nagaoka lemma is stated and proved. I have difficulty grasping the first line of the proof. Without going into unnecessary details, I'll phrase my doubt as a question.
Let $0 \leq S \leq I$ and $T\geq 0$ be operators in a Hilbert space. Let $P$ be the orthogonal projection onto the range of $S+T$. Then $P$ commutes both $S$ and $T$.
I am unable to show how this is true. I think by $P$ commuting $S$ and $T$, the authors mean $PS = SP$ and $PT = TP$. Correct me if I am wrong in this understanding. It is clear to me that $P(S+T) = S+T$. The authors use this projection approach in order to justify their result when $S+T$ is not invertible.
I tried working with powers of $S+T$ (e.g. $(S+T)^2$), trying to see if that would help. I didn't get anywhere with that. I think it is something simple but I am unable to see it.
Update: We say $A \geq 0$ if $A$ is non-negative definite. So $A\geq B$ is the same as $A-B\geq 0$.
Update2: Based on my reading of Conway, if $S$ is a self adjoint operator and $P$ is the projection onto its range space, then $PS=SP$. This doesn't answer my question but is probably a step in the right direction.
Proof: Given $x \in H$, for Hilbert space $H$, $$PSx = P(Sx) = Sx$$ Also $$SPx = Sx - S(I-P)x = Sx$$ The last step uses that $Ran(S) = Ran(P)$, $Ran(I-P) = Ker(P) = Ran(P)^\perp$ and $Ker(S) = Ran(S^*)^\perp = Ran(S)^\perp$. Hence, $Ker(S) = Ran(I-P)$.
You have $P(S+T)=S+T$, so $(I-P)(S+T)=0$. Then, since $S\leq S+T$, $$ 0\leq (I-P)S(I-P)\leq (I-P)(S+T)(I-P)=0, $$ and we get that $$(I-P)S(I-P)=0.$$ As $S\geq0$, there exists $B$ with $S=B^*B$. Then $$ 0=(I-P)S(I-P)=(I-P)B^*B(I-P)=[(B(I-P)]^*B(I-P); $$ thus $B(I-P)=0$. Then $$S(I-P)=B^*B(I-P)=0.$$ This gives us $$S=SP.$$ As $S$ is selfadjoint, taking adjoints we get $$S=SP=PS.$$
Repeating the argument with the roles of $S$ and $T$ switched, we also get $TP=PT=T$. The hypothesis $S\leq I$ is not needed.