Extending the spectral theorem for bounded self adjoint operators to bounded normal operators

Question

I'm currently preparing for an exam in functional analysis, and I have a question about the extension of the spectral theorem for bounded self adjoint operators to bounded normal operators.

Starting point is the spectral theorem for bounded self adjoint operators: Let $T$ be a bounded self adjoint operator in an Hilbert space $X$, then there exists a unique spectral measure $E : \Sigma_\mathbb{R} \rightarrow B(X)$, which has compact support in $\mathbb{R}$ (Here $\Sigma_\mathbb{R}$ is the Borel-$\sigma$-algebra on $\mathbb{R}$ and $B(X)$ is the set of all bounded and linear operators in $X$) and $T = \int\limits_{\mathbb{R}}\lambda dE_\lambda$. Moreover the mapping $f \rightarrow f(T) := \int\limits_{\mathbb{R}} f(\lambda) dE_\lambda$, for bounded and measurable functions $f$, satisfies the conditions of the (unique) measurable functional calculus.

If a normal operator $T \in B(X)$ is given, one can define the Operators: $S_1 := \frac{1}{2} \left( T+T^{\ast} \right)$ and $S_2 := \frac{1}{2i} \left( T-T^{\ast} \right)$. Then we get that $T = S_1 + i S_2$ and that $S_1$ and $S_2$ are self adjoint. Then by the spectral theorem for self adjoint operators there exist two spectral measures $E^1$ and $E^2$. Since $T$ is normal, $S_1$ and $S_2$ commute, and therefore the spectral measures $E^1$ and $E^2$.

Then there exists a unique spectral measure $E : \Sigma_{\mathbb{R}^2} \rightarrow B(X)$ such that for all $A, B \in \Sigma_\mathbb{R}$ we have that $E(A \times B) = E^1(A)E^2(B)$. (See: Schmüdgen - Thm. 4.10)

By identifying $\mathbb{R}^2$ with $\mathbb{C}$ one gets a unique specral measure $E : \Sigma_\mathbb{C} \rightarrow B(X)$ and is able to define integrals with respect to this spectral measure in the natural way: First for step functions and then for bounded measurable functions by approximation.

Now I have to show that $E$ has the same properties as the spectral measure for self adjoint operators, i.e.: $T = \int\limits_{\mathbb{C}} z dE_z$ and the mapping $f \rightarrow f(T) := \int\limits_{\mathbb{C}} f(z) dE_z$, for bounded and measurable functions $f$, satisfies the conditions of the (unique) measurable functional calculus.

My question now is: is there any other way to show that, beside re-do the proof of the spectral theorem for self adjoint operators? It's not that much work, once one has the proof of the self adjoint case. I'm just curious if there's an more elegant way ...

Thanks in advance, GordonFreeman

Answer 1

The proof of the spectral theorem for normal operators doesn't rely on the proof of the spectral theorem for self-adjoint operators, instead the proofs are basically identical.

How do you construct the spectral measure in the self-adjoint case? One way to do it is to look at the $C^*$-algebra generated by the self-adjoint operator $T$ on the Hilbert space $X$, let's call it $C^*(T)$. Since $C^*(T)$ is commutative, by Gelfand theory it is isomorphic to the algebra of continuous functions on the spectrum of $T$, $C(\sigma(T))$. Given $x,y\in H$, the map $C^*(T)\to\mathbb C$ given by $S\mapsto \langle Sx,y\rangle$ is a bounded linear functional, hence defines a Borel measure $\mu_{x,y}$ on $\mathbb R$, supported in $\sigma(T)$. Using these measures, we can extend the isomorphism $C(\sigma(T))\to C^*(T)$ to a homomorphism of $B(\mathbb R)\to \mathcal B(X)$ from the algebra bounded Borel functions on $\mathbb R$ to bounded operators on $X$. The spectral measure is just the restriction of this homomorphism to characteristic functions of Borel sets.

If now $T$ is normal, $C^*(T)$ is still commutative, and (again by Gelfand theory) is isomorphic to $C(\sigma(T))$, where now $\sigma(T)\subset\mathbb C$. Given $x,y\in X$, the measure $\mu_{x,y}$ is now a Borel measure on $\mathbb C$ supported in $\sigma(T)$, and in this way we obtain a homomorphism $B(\mathbb C)\to\mathcal B(X)$ from the algebra of bounded Borel functions on $\mathbb C$ to $\mathcal B(X)$, and obtain the spectral measure.

The rest of the proof of the spectral theorem should be the same.

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EDIT

Hopefully this will help translate my response to language you are familiar with.

Firstly, yes, $C^*(T)$ is as you have defined it.

Secondly, basically the only difference between the two cases is that if $T$ is normal, we define the map $\Phi_0$ from polynomials in two variables $p=p(z,\overline z)$ to $B(X)$ by $\sum_{ij}a_{ij}z^i\overline z^j\mapsto \sum_{ij}a_{ij}T^i(T^*)^j$ and extend this by Stone-Weierstrass to a map $\Phi:C(\sigma(T))\to B(X)$. We need to consider bivariate polynomials in the normal case because if the set $X\subset\mathbb C$ is not a subset of $\mathbb R$, polynomials in one variable are not closed under conjugation, hence the Stone-Weierstrass theorem cannot be applied.

Thirdly, there are plenty of books out there that prove the spectral theorem for normal operators, leaving the case for self-adjoint operators as a corollary, but most of the one's I'm familiar with develop some basic $C^*$-algebra theory to make the proofs more transparent. See for instance Conway's or Rudin's functional analysis books, or Murphy's $C^*$-algebras and operator theory.

Answer 2

@Aweygan: Thanks for the quick reply! HL3 has to wait, since I lately have some troubles concerning math :)

Sadly, in my lecture, we didn't really touch the topics of $C^{\ast}$-algebras or Gelfand-theory and only proved the spectral theorem for self adjoint operators. We started out by defining a map $\Phi_0 : P(\sigma(T)) \rightarrow B(X)$ by $p(T) = \sum_{i = 0}^{n} a_i T^i$ and then extending this map to a map $ \Phi : C(\sigma(T)) \rightarrow B(X)$ by density of the polynomials $P(\sigma(T))$ in $C(\sigma(T))$ due to the Stone-Weierstrass theorem. Then, we did the rest of the construction the same way you described it by obtaining a complex measure by Riesz represenation theorem.

I'm assuming that the space $C^{\ast}(T)$ you are talking about is the set $\{ \Phi(f) \,|\, f \in C(\sigma(T)) \},$ right?

So, by setting $\Phi_0$ the same way we did for self adjoint operators, we would again get a complex measure by the Riesz represenation theorem. So, it's basically the same proof for normal operators?

I'm just asking this question because every book (Werner, Schmüdgen, etc.) that proves the spectral theorem by using the functional calculus (not the Stieltjes integrals) just proves it for self adjoint operators, and for normal ones, "it's left for the reader." I'm guessing the authors of the mentioned books don't want to do the whole proof again and are using the method I described in the first post.