How can I prove, that my constructed spectral measure $E(\Delta)$ is an orthogonal projection?

Question

Definitions

Representation

Let $X \subset \mathbb{C}^N$ and $\mathcal{A}$ be an algebra in $\mathcal{C}(X)$. Also, we denote $L(H)$ as the set of all linear operators on HIlbert space $H$.

We call $\Phi : \mathcal{A} \ni u \rightarrow \Phi(u) \in L(H)$ a representation, if:

1. $\Phi$ is linear and multiplicative
2. $\Phi(e) = I_H$, where $e$ is the neutral multiplicative element of the algebra $\mathcal{A}$
3. There exists a $K > 0$ such, that $\| \Phi(u) \| \le K \|u \|$ for all $u \in \mathcal{A}$

Measure $\mu_{xy}$

A measure $\mu_{xy}$ is called the "elementary measure for $\Phi$", if for any $x,y \in H$:

$$\langle \Phi(u) x, y \rangle = \int_X u(z) d\mu_{xy}$$ for $\| \mu_{xy} \| \le K \| x \| \|y \|$.

Spectral measure

On a Borel set $\mathcal{B}$ in $\Omega$, we call $E : \mathcal{B} \ni \Delta \rightarrow E(\Delta) \in B(H)$ a spectral measure, if:

1. $E(\Omega) = I_H$, $E(\varnothing) = 0$
2. $E(\Delta)$ is an orthogonal projection in $H$ for all $\Delta$
3. $E(\bigcup_{n}^{\infty} \Delta_n) = \sum_{n}^{\infty} E(\Delta_n)$ for all disjunctive $\Delta_n$

Important: In this context, we will assume, that $\Phi$ is a natural representation ("functional calculus") of an $N$-tuple of operators $(T_1, \ldots, T_N)$, i.e.

$$\Phi(u) := u(T_1, \ldots, T_n)$$

---

How can I prove, that my spectral measure $E(\Delta)$ constructed in the following way:

$$\mu_{xy} (\Delta) = \langle E(\Delta) x, y \rangle$$

is an orthogonal projection (i.e. it satisfies $E(\Delta)^* = E(\Delta)$ and $E(\Delta)^2 = E(\Delta)$)?

I've heard from my professor (I'm a 1st year PhD student) that it is important that in this context $K = 1$, but why? I can't find anything on the internet about that.

Answer 1

The following is a proof that, suppose you have a collection of complex-valued measures $\{\mu_{x, y}\}_{x, y \in H}$ on $\Omega$ such that $E(\Delta)$ defined by $\mu_{x, y}(\Delta) = \langle E(\Delta)x, y\rangle$ forms a spectral measure, then there exists a $\ast$-homomorphism (hence of norm 1) $\Phi: C(\Omega) \rightarrow B(H)$ s.t. $\mu_{x, y}$ is the elementary measure for $\Phi$.

We observe that as $\mu_{x, y}(\Delta) = \langle E(\Delta)x, y\rangle$ defines a bounded linear operator for all $\Delta$, we already have $\mu_{x, y}$ is sesquilinear in $x$ and $y$. Now, we show that $||\mu_{x, y}|| \leq ||x||||y||$. Let $f = \sum_{i = 1}^n a_i 1_{\Delta_i}$ be a simple function and we may assume $\Delta_i$ are disjoint. Then,

$$\int f d\mu_{x, y} = \sum_{i = 1}^n a_i \mu_{x, y}(\Delta_i) = \sum_{i = 1}^n a_i \langle E(\Delta_i)x, y\rangle = \langle (\sum_{i = 1}^n a_i E(\Delta_i))x, y\rangle$$

Since $\Delta_i$ are disjoint, $E(\Delta_i)$ are pairwise orthogonal (you can easily get this from condition 3 in your definition of spectral measures). Hence, $\sum_{i = 1}^n a_i E(\Delta_i)$ has operator norm the maximum of $|a_i|$, the same as the sup norm of $f$. Thus, $|\int f d\mu_{x, y}| \leq ||f||_\infty||x||||y||$, so $||\mu_{x, y}|| \leq ||x||||y||$.

This already tells you $K = 1$, but let’s keep going to construct $\Phi$. This is simply given by $\int u \mu_{x, y} = \langle \Phi(u)x, y\rangle$. Since $||\mu_{x, y}|| \leq ||x||||y||$, $\Phi(u)$ is a well-defined bounded operator of norm at most $||u||_\infty$. This can also be defined for bounded Borel $u$, which is known as the Borel functional calculus. To see that $\Phi(u)$ is self-adjoint whenever $u$ is, we see that, as $E(\Delta)$ is self-adjoint, $\mu_{x, x} = \langle E(\cdot)x, x\rangle$ is a real-valued measure. So if $u$ is self-adjoint, that is, a real-valued function, then $\langle \Phi(u)x, x\rangle = \int u \mu_{x, x}$ is real, so $\Phi(u)$ is self-adjoint. Proving multiplicativity is more annoying - we consider $\Phi$ as extended to all bounded Borel functions, defined the same way. Observe that $E(\Delta_1)E(\Delta_2) = E(\Delta_1 \cap \Delta_2)$. (Again, this follows from condition 3.) Also observe that $\Phi(\sum_{i = 1}^n a_i 1_{\Delta_i}) = \sum_{i = 1}^n a_i E(\Delta_i)$, as we have seen before. Then clearly $\Phi$ is multiplicative on simple functions, i.e., for simple functions $u$ and $v$,

$$\int u (v d\mu_{x, y}) = \int (uv) d\mu_{x, y} = \langle \Phi(uv)x, y\rangle = \langle \Phi(u)\Phi(v)x, y\rangle = \int u d\mu_{\Phi(v)x, y}$$

This holds for all simple $u$, so $v\mu_{x, y} = \mu_{\Phi(v)x, y}$. But then we can repeat the above argument with now $u$ bounded Borel and $v$ simple,

$$\langle \Phi(u)\Phi(v)x, y\rangle = \int u d\mu_{\Phi(v)x, y} = \int u (v d\mu_{x, y}) = \int (uv) d\mu_{x, y} = \langle \Phi(uv)x, y\rangle$$

So $\Phi(u)\Phi(v) = \Phi(uv)$ whenever $u$ is bounded Borel and $v$ is simple. Repeat the argument again with $u$ bounded Borel and $v$ simple,

$$\int v (u d\mu_{x, y}) = \int (uv) d\mu_{x, y} = \langle \Phi(uv)x, y\rangle = \langle \Phi(u)\Phi(v)x, y\rangle = \langle \Phi(v)x, \Phi(u)^\ast y\rangle = \int v d\mu_{x, \Phi(u)^\ast y}$$

So $u \mu_{x, y} = \mu_{x, \Phi(u)^\ast y}$ for all bounded Borel $u$. Finally, for $u$, $v$ both bounded Borel,

$$\langle \Phi(u)\Phi(v)x, y\rangle = \langle \Phi(v)x, \Phi(u)^\ast y\rangle = \int v d\mu_{x, \Phi(u)^\ast y} = \int v (u d\mu_{x, y}) = \int (uv) d\mu_{x, y} = \langle \Phi(uv)x, y\rangle$$

That is, $\Phi(u)\Phi(v) = \Phi(uv)$ for all bounded Borel $u$ and $v$, in particular continuous ones. So $\Phi$ is a $\ast$-homomorphism. By construction, $\mu_{x, y}$ is the elementary measure for $\Phi$.