Spectral family of self-adjoint linear operator

Question

Let $T\colon H\to H$ be a self-adjoint bounded linear operator where $H$ is a complex Hilbert space. We define $B=(T^2)^{1/2}$, $T^+=\frac{1}{2}(B+T)$ and $T^-=\frac{1}{2}(B-T)$.

We denote with $E\colon H\to\mathcal{N}(T^+)$ the projection of $H$ onto the null space of $T^+$.

Question If I just proved that $B$, $T^+$ and $T^-$ commute with every bounded linear operator that $T$ commutes with, can I conclude that $BT=TB$?

My answer. Since $T$ commute itself, then $BT=TB$, it's correct?

Thanks!

Answer 1

First of all, rethink about your statement of the question: where does $E$ come in?

What you are asking is true in a much more general setting: Let $T\in B(H)$ be a normal operator (trivially, every self-adjoint operator is normal). If we denote by $C^*(1,T)$ the smallest $C^*$-algebra that contains $1=id_H$ and $T$, then it can be proved that there is a $*$-isomorphism (i.e. a linear map that preserves multiplication and involution, is 1-1 and onto)

$$C^*(1,T)\longleftrightarrow C(\sigma(T))$$

satisfying $$1\leftrightarrow 1,\text{ the constant function} $$ $$T\leftrightarrow id_{\sigma(T)} $$ $$p(T,T^*)\leftrightarrow p(z,\overline{z})$$ for any polynomial $p(z,w)$ with two variables.

Now the function $f:\sigma(T)\to\mathbb{C}$ given by $f(z)=|z|$ is continuous and it is true that $B:=(T^*T)^{1/2}\leftrightarrow f$. This shows that $B\in C^*(1,T)$, thus $B$ commutes with $T$, since $C^*(1,T)$ is isomorphic to $C(\sigma(T))$ which is commutative (multiplication here is pointwise multiplication of functions).

Your case follows from the above for $T^*=T$.

Answer 2

Let me give a different angle on this problem. For a subset $A$ of $B(H)$ let $A'=\{S\in B(H)\mid \forall T\in A\colon TS=ST\}$; this is called the commutant of $A$. If $B$ commutes with all operators that commute with $T$, this means that $B$ is in the commutant of the commutant of $\{T\}$, which is called the bicommutant and denoted by $\{T\}^{\prime\prime}$.

By the bicommutant theorem, every element of $\{T\}^{\prime\prime}$ can be approximated in the strong operator topology by polynomials in $T$ (and $T^\ast$ if $T$ is only assumed to be normal). These polynomials clearly commute with $T$ and commutation passes to strong limits. Thus $B$ commutes with $T$.

Let me summarize the difference to JustDroppedIn's answer: He showed that for normal $T$ the absolute value $\lvert{T}\rvert$ always commutes with $T$ without any further assumptions. I showed that if $B$ commutes with every operator commuting with the normal operator $T$, then $B$ commutes with $T$ (without assuming $B=\lvert T\rvert$).