Fuglede-Putnam-Rosenblum theorem

Let $A$ be a $C^{\ast }$-algebra with unit $e$.

The Fuglede-Putnam-Rosenblum theorem makes the assertion that for a normal element $a\in A$ the kernel of the commutator mapping $[a,-]:A\to A$ is a $\ast$-closed set.

The general formulation of the result is as follows:

Theorem. Let $A$ be a $C^{\ast }$-algebra with unit $e$. Let two normal elements$a,b\in A$ be given and $c\in A$ with $ac=cb$.Then it follows that $a^{\ast }c=c{b}^{\ast }$.

Lemma. For any $x\in A$ we have that $\exp (x-x^{\ast })$ is a element of $A$.

Proof. We have for $x\in A$ that$\exp {(x-x^{\ast })}^{\ast }\exp (x-x^{\ast })=\exp (x^{\ast }-x+x-x^{\ast })=\exp (0)=e$.And similarly $\exp (x-x^{\ast })\exp {(x-x^{\ast })}^{\ast }=e$. ∎

With this we can now give a proof the Theorem.

Proof. The condition $ac=cb$ implies by induction that $a^{k}c=c{b}^k$ holds for each $k\in \mathbb{N}$.Expanding in power series on both sides yields $\exp (a)c=c\exp (b)$.This is equivalent to $c=\exp (-a)c\exp (b)$. Set $U_1:=\exp (a^{\ast }-a),U_2:=\exp (b-b^{\ast })$. From the Lemma we obtain that ${\parallel U_1\parallel }_A={\parallel U_2\parallel }_A=1$.Since $a$ commutes with $a^{\ast }$ und $b$ with $b^{\ast }$ we obtain that

$$\exp (a^{\ast })c\exp (-b^{\ast })=\exp (a^{\ast })\exp (-a)c\exp (b)\exp (b^{\ast })$$

which equals $\exp (a^{\ast }-a)c\exp (b-b^{\ast })=U_{1}c{U}_2$.

Hence

$$\parallel \exp (a^{\ast })c\exp (-b^{\ast })\parallel \le \parallel c\parallel .$$

Define $f:ℂ\to A$ by $f(\lambda ):=\exp (\lambda a^{\ast })c\exp (-\lambda b^{\ast })$. If we substitute $a\mapsto \lambda a,b\mapsto \lambda b$ in the last estimate we obtain

$$\parallel f(\lambda )\parallel \le \parallel c\parallel ,\lambda \in ℂ.$$

But $f$ is clearly an entire function and therefore Liouville’s theorem implies that $f(\lambda )=f(0)=c$ for each $\lambda$.

This yields the equality

$$c\exp (\lambda b^{\ast })=\exp (\lambda a^{\ast })c.$$

Comparing the terms of first order for $\lambda$ small finishes the proof. ∎