Recently, I worked through the proof of the spectral theorem presented in the book "Introduction to Hilbert Space and the Theory of Spectral Multiplicity" by Halmos. The proof is notably concise compared to other proofs of the spectral theorem. My questions are:
- Is this brevity due to the use of the Riesz-Markov-Kakutani Theorem, or am I missing that only some special case is being addressed?
- Additionally, I noticed that this proof appears to be easier and more elementary compared to other proofs. However, I am only aware of this particular reference where the Spectral Theorem is proven with the assistance of the Riesz-Markov-Kakutani theorem. Other proofs that I am familiar with either involve the use of $C^{*}$-Algebras or cyclic vectors. Is there a reason for this that I am overlooking?
Note: The author refers to self-adjoint operators as "Hermitian".
Theorem
If $A$ is a Hermitian operator, then there exists a (necessarily real and necessarily unique) compact, complex spectral measure $E$, called the spectral measure of $A$, such that $A=\int \lambda d E (\lambda)$.
Proof
Let $x$ and $y$ be any two fixed vectors and write
$$
L(p)=(p(A) x, y)
$$
for every real polynomial $p$.
It follows from 34.3 that
$$
\vert L(p) \vert \leq \mathbf{N}_{A}(p) \| x \| \| y \|
$$
and hence that, with respect to the norm $\mathbf{N}_A$, $L$ is a bounded linear functional of its argument.
There exists consequently a unique complex measure $\mu$ in the compact set $\Lambda(A)$ such that $(p(A) x, y)=\int(p(\lambda) d \mu(\lambda)$ every Borel set $M$ (Riesz-Markov-Kakutani).
We shall find it convenient to indicate the dependence of $\mu$ on $x$ and $y$ by writing $\mu_M(x, y)$ instead of $\mu(M)$.
Using the uniqueness of $\mu$, we may proceed by straightforward computations to prove that $\mu_M$ is a symmetric, bilinear functional for each Borel set $M$.
The proof of the fact that $\mu_y$ is additive in its first argument runs, for instance, as follows:
\begin{align}
\int p(\lambda) d \mu_\lambda\left(x_1+x_2, y\right)
&=
\left(p(A)\left(x_1+x_2\right), y\right)\\
&=\left(p(A) x_1, y\right)+\left(p(A) x_2, y\right) \\
&=\int p(\lambda) d \mu_\lambda\left(x_1, y\right)+\int p(\lambda) d \mu_\lambda\left(x_2, y\right) .
\end{align}
Since, in virtue of the relation $\left.\mid \mu_M(x, y)\right\} \leq\|x\| \cdot\|y\|$, valid for all $M, x$, and $y$, the bilinear functionals $\mu_{M}$ are bounded, it follows that for each $M$ there exists a unique Hermitian operator $E(I)$ such that $\mu_M(x, y)=(E(M) x, y)$ for all $x$ and $y$.
Consideration of the polynomials $p_0$ and $p_1$, defined by $p_0(\lambda)=1$ and $p_1(\lambda)=\lambda$, implies that
$$
\int d(E(\lambda) x, y)=(E(X) x, y)=(x, y)
$$
and
$$
\int \lambda d(E(\lambda) x, y)=(A x, y)
$$
for all $x$ and $y$.
In view of 36.3, all that remains in order to complete the proof of the theorem is to establish that the function $E$ is projection-valued; we shall do this by proving that $E$ is multiplicative.
For any fixed pair of vectors $x$ and $y$ and for any real polynomial $q$, we introduce the auxiliary complex measure $v$ defined for every Borel set $M$ by $\nu(M)=\int_M q(\lambda) d(E(\lambda) x, y)$. If $p$ is any real polynomial, then \begin{align} \int p(\lambda) d p(\lambda) &= \int p(\lambda) q(\lambda) d(E(\lambda) x, y)\\ &= (p(A) q(A) x, y) \\ &= (p(A) x, q(A) y)\\ &=\int p(\lambda) d\left(E(\lambda) x, q(A) y\right) \end{align} and therefore \begin{align} \nu(M) &= \int q(\lambda) \chi_M(\lambda) d(E(\lambda) x, y)\\ &= (E(M) x, q(A) y) & \\ &= (q(A) E(M) x, y)\\ &= \int q(\lambda) d(E(\lambda) E(M) x, y) \end{align} for every Borel set $M$. Since $q$ is arbitrary, it follows that \begin{align} \left( E (M \cap N) x, y\right) &= \int_{M \cap M} d(E(\lambda) x, y) \\ &= \int_N \chi_M(\lambda) d(E(\lambda) x, y)\\ &= (E(N) E(M) x, y) \end{align} for every Borel set, $N$, and hence that $E(M \cap N)=E(M) E(N)$. The proof of the spectral theorem for Hermitian operators is thereby complete.