# Uniqueness of the Spectral Measure

I am studying the spectral theorem for unbounded s.a. operators. The existence of a spectral measure E sucht that the Operator $$T=\int_{\sigma(t)}\lambda dE(\lambda):=E(id)$$, is already shown. What i want to understand now is the uniqueness of the spectral measure.

The Idea in my script is to show that any other spectral measure $$F$$ with $$T=\int_{\sigma(t)}\lambda dF(\lambda)$$ has to satisfy:

$$\int_{\sigma(T)} \frac{1}{\lambda-z}dE(\lambda) = \int_{\sigma(T)} \frac{1}{\lambda-z}dF(\lambda)$$ for $$z \in \mathbb{C}/\mathbb{R}$$,

and then to use the Theorem of Stone-Weierstraß to approximate the continuous functions by functions of the form $$p_N(\lambda)=\sum_{i=1}^{N}\frac{1}{\lambda - z_k}$$ uniformly.

And Finally to approximate the bounded measurable functions by the continuous functions pointwise in order to show that $$E=F$$.

My problem is to understand these approximation procedures. If $$\phi = \lim_{n \rightarrow \infty} p_n$$ is a continuous function then in which sense can i take the limit $$E(\phi)=\lim_{n \rightarrow \infty} E(p_n)$$?

Finally: why is it enough to proof $$E(f)=F(f)$$ only for bounded measurable $$f$$?

To answer the question in which sense one would understand $$E(\phi) = \lim\limits_{n\rightarrow\infty} E(p_n)$$ you should look at the definition of a spectral measure. A spectral measure is a map (with some properties) $$E:\mathcal{A}\rightarrow L(H)$$ for some $$\sigma$$-algebra $$\mathcal{A}$$ and Hilbert space $$H$$, where $$L(H)$$ is the space of bounded linear operators $$A:H\rightarrow H$$. Hence, I would understand $$E(\phi) = \lim\limits_{n\rightarrow\infty} E(p_n)$$ as convergence in $$H$$.
For your second question, I guess it is enough to show the equality for bounded measurable functions, since we can approximate an unbounded measurable function $$f$$ with $$f_n = f\boldsymbol{1}_{A_n}$$, where $$A_n = \{x: \vert f(x) \vert < n \}$$, and then one usually uses some convergence theorem (e.g. monotone convergence) to obtain equality even for unbounded measurable functions.
Hope this makes all sense and helps, it's been a while since I've been in touch with the topic.

• Thank you! I added one answer that actually proves the pointwise convergence in H. Commented Jun 29, 2023 at 21:30

I know how to obtain convergence in the strong operator topology know:

Since uniform convergence implies $$L^2$$-convergence, we obtain by using the measure $$\nu_x(A):= \langle x, E(A)x \rangle$$ with $$x \in H$$, that

$$||E(p_n)x-E(\phi)x||=\int_X |p_n - \phi|^2d \nu_x \rightarrow 0$$

For the approximation of the bounded measurable functions, using the dominated convergence theorem, again convergence in the sense of the strong operator topology follows.

Then as @stange pointed out, one obtains uniqueness for all measurable funktions, since simple functions are dense in the measurable functions.