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