# Relation between functional calculus on a bounded normal operator and on its restriction to a closed reducing subspace

In Spectral subspace is nontrivial iff it has a non-trivial intersection with an invariant closed subspace, the original poster sets the scene as follows:

$$A$$ is a bounded self-adjoint operator on a Hilbert space $$\mathcal{H}$$ and $$W$$ is a closed invariant subspace.

In his answer (https://math.stackexchange.com/a/3628794/527829) Martin Argerami points out that $$W$$ is actually reducing for $$T$$.

The original poster then goes on to say that there is a result that $$\sigma(A|_W)\subseteq\sigma(A)$$ and that $$f(A|_W)=f(A)|_W$$. I assume this should be interpreted as $$(f|_{\sigma(A|_W)})(A|_W)=f(A)|_W$$ for bounded Borel functions $$f$$ on $$\sigma(A)$$.

My questions are:

1. Is it true for a bounded normal operator A with an arbitrary closed reducing subspace $$W$$?
2. Does the result have a name? Or a reference?
3. How is the functional calculus part proved? (I can prove the spectrum part myself.)
• Let $\mu_A$ be the projection valued measure on $\sigma(A)$. For $X\subseteq \sigma(A)$ measurable check that: $$\mu_{A\lvert_W}(X\cap \sigma(A\lvert_W) ) = \mu_A(X)\lvert_W.$$ This is the ingredient that proves question 1. You can check that the statement is true by starting with $$\int_{\sigma(A\lvert_W)} \lambda d\mu_{A\lvert_W}(\lambda) v = A\lvert_W \,v = Av = \int_{\sigma(A)} \lambda d\mu_A (\lambda)v = \int_{\sigma(A\lvert_W)} \lambda d\mu_A(\lambda)v + \int_{\sigma(A\lvert_W)^C}\lambda d\mu_A(\lambda)v$$ for $v\in W$. Jun 30, 2020 at 10:47
• @s.harp Sorry, but I don't see how to get from the second equation (which I agree with) to the first. Do I need to approximate $\chi_X$ and $\chi_{X\cap\sigma(A|_W)}$ somehow with polynomials? How can I do that when $\chi_X$ is not continuous, so Stone-Weierstrass is not available? Jul 1, 2020 at 4:30
• @s.harp Well, I was able to get this to work after all. I will write it in a little more detail as an answer, just for documentation. But if you would like to write your two sentence comment as an answer, I will accept it. Jul 2, 2020 at 4:16
• you should post an answer, I'll upvote it and tell you if it looks ok. Jul 2, 2020 at 18:37

Thanks to the suggestion by s.harp, I can give an answer to my questions. It is amazing how long the answer is when written out fully. In order to match my handwritten notes, I am going to change the notation. So here is the original question, written as a theorem in a different notation:

Theorem. Suppose $$T$$ is a bounded normal operator on the (not necessarily separable) Hilbert space $$H$$, and suppose $$H_\alpha$$ is a closed reducing subspace for $$T$$. That is, $$TH_\alpha\subseteq H_\alpha$$, and $$T^*H_\alpha\subseteq H_\alpha$$ (or equivalently, the second condition can be replaced with $$TH_\alpha^\perp\subseteq H_\alpha^\perp$$). Then

1. $$T_\alpha=T|_{H_\alpha}$$ is a bounded normal operator on $$H_\alpha$$,
2. $$\sigma(T_\alpha)\subseteq\sigma(T)$$
3. Let $$E$$ and $$E'$$ be the resolutions of the identity for $$T$$ and $$T_\alpha$$, respectively, and let $$B_T$$ and $$B_{T_\alpha}$$ be the Borel subsets of $$\sigma(T)$$ and $$\sigma(T_\alpha)$$, respectively. Then for all $$e\in B_T$$, $$\begin{equation*} e\cap\sigma(T_\alpha)\in B_{T_\alpha}\quad\text{and}\quad E'(e\cap\sigma(T_\alpha))=E(e)|_{H_\alpha}, \end{equation*}$$
4. If $$\mathscr{B}_T$$ and $$\mathscr{B}_{T_\alpha}$$ are the families of bounded complex Borel functions on $$\sigma(T)$$ and $$\sigma(T_\alpha)$$, respectively, and if $$f_\alpha=f|_{\sigma(T_\alpha)}$$ then for all $$f\in\mathscr{B}_T$$, $$\begin{equation*} f_\alpha\in\mathscr{B}_{T_\alpha}\quad\text{and}\quad f_\alpha(T_\alpha)=f(T)|_{H_\alpha} \end{equation*}$$

Proof. When I mention the spectral theorem for bounded normal operators, I am referring to Rudin, Functional Analysis, Second Edition, 12.21-12.24. Since $$H_\alpha$$ is reducing for $$T$$, we see that $$T_\alpha$$ and $$(T^*)_\alpha$$ are bounded operators on the Hilbert space $$H_\alpha$$. For $$x_\alpha,y_\alpha\in H_\alpha$$, $$\begin{equation*} (x_\alpha,(T_\alpha)^*y_\alpha)=(T_\alpha x_\alpha,y_\alpha)=(Tx_\alpha,y_\alpha) =(x_\alpha,T^*y_\alpha)=(x_\alpha,(T^*)_\alpha y_\alpha), \end{equation*}$$ so $$(T_\alpha)^*=(T^*)_\alpha$$ and we will write it as $$T_\alpha^*$$. For $$x_\alpha\in H_\alpha$$, $$\begin{equation*} T_\alpha T_\alpha^*x_\alpha=T_\alpha T^*x_\alpha=TT^*x_\alpha =T^*Tx_\alpha=T^*T_\alpha x_\alpha=T_\alpha^*T_\alpha x_\alpha, \end{equation*}$$ so $$T_\alpha$$ is normal. This proves #1.

Suppose $$\lambda\in\rho(T)$$. Let $$S=(\lambda I-T)^{-1}\in\mathscr{B}(H)$$, the bounded linear operators on $$H$$. Let $$x\in SH_\alpha$$. Let $$x=x_\alpha+x_\alpha^\perp$$, where $$x_\alpha\in H_\alpha$$ and $$x_\alpha^\perp\in H_\alpha^\perp$$. Then $$(\lambda I-T)x_\alpha\in H_\alpha$$, $$(\lambda I-T)x_\alpha^\perp\in H_\alpha^\perp$$, and $$(\lambda I-T)x_\alpha+(\lambda I-T)x_\alpha^\perp=(\lambda I-T)x\in H_\alpha,$$ since $$x\in (\lambda I-T)^{-1}H_\alpha$$. Hence $$(\lambda I- T)x_\alpha^\perp=0$$, so $$x_\alpha^\perp=0$$ since $$\lambda I-T$$ is one-to-one. Therefore $$x=x_\alpha\in H_\alpha$$, so that $$SH_\alpha\subseteq H_\alpha$$. Conversely, if $$x_\alpha\in H_\alpha$$, then $$y=(\lambda I-T)x_\alpha\in H_\alpha$$ and $$x_\alpha=Sy$$ so $$SH_\alpha=H_\alpha$$. Define $$S_\alpha=S|_{H_\alpha}$$. Then $$S_\alpha$$ is linear, continuous, one-to-one onto $$H_\alpha$$, so $$S_\alpha\in\mathscr{B}(H_\alpha)$$, and for all $$x_\alpha\in H_\alpha$$, $$\begin{equation*} S_\alpha(\lambda I_\alpha-T_\alpha)x_\alpha =S(\lambda I-T)x_\alpha =x_\alpha =(\lambda I-T)Sx_\alpha =(\lambda I_\alpha-T_\alpha)S_\alpha x_\alpha, \end{equation*}$$ so $$\lambda I_\alpha-T_\alpha$$ has an inverse $$S_\alpha\in\mathscr{B}(H_\alpha)$$, whence $$\lambda\in\rho(T_\alpha)$$. This proves #2.

Let $$P_\alpha$$ be the orthogonal projection on $$H_\alpha$$. Then for $$x\in H$$, write $$x=x_\alpha+x_\alpha^\perp$$, where $$x_\alpha\in H_\alpha$$ and $$x_\alpha^\perp\in H_\alpha^\perp$$. Then since $$TH_\alpha^\perp\subseteq H_\alpha^\perp$$ and $$TH_\alpha\subseteq H_\alpha$$, $$\begin{equation*} P_\alpha Tx=P_\alpha Tx_\alpha+P_\alpha Tx_\alpha^\perp=P_\alpha Tx_\alpha =Tx_\alpha=TP_\alpha x, \end{equation*}$$ that is, $$P_\alpha$$ commutes with $$T$$, so by the spectral theorem, $$P_\alpha$$ commutes with $$E(e)$$ for every $$e\in B_T$$. Therefore $$E(e)H_\alpha\subseteq H_\alpha$$. Let $$p(\sigma(T))$$ be the algebra of all complex continuous functions on $$\sigma(T)$$ expressible as $$p(\lambda,\bar{\lambda})$$, where $$p(\nu,\gamma)$$ is a polynomial in two complex variables with complex coefficients. Then by the Stone-Weierstrass Theorem, $$p(\sigma(T))$$ is dense in $$C(\sigma(T))$$. If $$f\in C(\sigma(T))$$ then $$f_\alpha=f|_{\sigma(T_\alpha)}\in C(\sigma(T_\alpha))$$ by #2. If $$f(\lambda)=\sum_{i=1}^kc_i\lambda^{p_i}\bar{\lambda}^{q_i}\in p(\sigma(T))$$, then by the spectral theorem, for $$x_\alpha,y_\alpha\in H_\alpha$$, $$f(T)=\sum_{i=1}^kc_iT^{p_i}T^{*q_i}$$ and $$f_\alpha(T_\alpha)=\sum_{i=1}^kc_iT_\alpha^{p_i}T_\alpha^{*q_i}$$ and $$\begin{equation*} \begin{split} \int_{\sigma(T_\alpha)}\!f_\alpha(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda) &=\Biggl(\biggl(\sum_{i=1}^kc_iT_\alpha^{p_i}T_\alpha^{*q_i}\biggr)x_\alpha, y_\alpha\Biggr) =\Biggl(\biggl(\sum_{i=1}^kc_iT^{p_i}T^{*q_i}\biggr)x_\alpha,y_\alpha\Biggr)\\ &=\int_{\sigma(T)}\!f(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda) \qquad\qquad\qquad\qquad\qquad(f\in p(\sigma(T))).\quad\text{(1)} \end{split} \end{equation*}$$ If $$f\in C(\sigma(T))$$, let $$\{f_n\}\subseteq p(\sigma(T))$$ be such that $$f_n(\lambda)\to f(\lambda)$$ uniformly on $$\sigma(T)$$ (that is, in the norm of $$C(\sigma(T))$$). Then $$f_{n,\alpha}(\lambda)\to f_\alpha(\lambda)$$ uniformly on $$\sigma(T_\alpha)$$. By the spectral theorem $$\begin{equation*} \lvert((f(T)-f_n(T))x_\alpha,y_\alpha)\rvert \leq\lvert\lvert f(T)-f_n(T)\rvert\rvert\,\lvert\lvert x_\alpha\rvert\rvert\, \lvert\lvert y_\alpha\rvert\rvert\to 0 \end{equation*}$$ as $$n\to\infty$$. Similarly, $$\lvert((f_\alpha(T_\alpha)-f_{n,\alpha}(T_\alpha))x_\alpha,y_\alpha)\rvert\to 0$$. Therefore, by (1) and the spectral theorem, $$\begin{equation*} \begin{split} \int_{\sigma(T_\alpha)}\!f_\alpha(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda) &=\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\! f_{n,\alpha}(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda) =\lim_{n\to\infty}\int_{\sigma(T)}\! f_n(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda)\\ &=\int_{\sigma(T)}\!f(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda) \qquad\qquad\qquad\qquad\qquad(f\in C(\sigma(T))).\quad\text{(2)} \end{split} \end{equation*}$$ Let $$d(\lambda,S)$$ be the distance from $$\lambda$$ to $$S\subseteq\mathbb{C}$$: $$d(\lambda,S)=\inf\,\{\lvert\lambda-s\rvert:s\in S\}$$. $$d$$ is continuous as a function of $$\lambda\in\mathbb{C}$$. Let $$e$$ be a closed subset of $$\sigma(T)$$. For $$n=1,2,\dots$$, let $$f_n(\lambda)=\max(0,1-nd(\lambda,e))$$ for $$\lambda\in\sigma(T)$$. Then $$\{f_n\}\subseteq C(\sigma(T))$$, $$0\leq f_n,f_{n,\alpha}\leq 1$$, $$f_n(\lambda)\to\chi_e(\lambda)$$ for all $$\lambda\in\sigma(T)$$, and $$f_{n,\alpha}(\lambda)\to\chi_{e\cap\sigma(T_\alpha)}(\lambda)$$ for all $$\lambda\in\sigma(T_\alpha)$$. Since $$\chi_{\sigma(T)}\in L^1(\lvert E_{x_\alpha,y_\alpha}\rvert)$$, $$\chi_{\sigma(T_\alpha)}\in L^1(\lvert E'_{x_\alpha,y_\alpha}\rvert)$$, and $$\lvert f_n\rvert\leq\chi_{\sigma(T)}$$ and $$\lvert f_{n,\alpha}\rvert\leq\chi_{\sigma(T_\alpha)}$$ for $$n=1,2,\dots$$, we have by (2) and the Dominated Convergence Theorem for complex measures and by the spectral theorem, that $$\begin{equation*} \begin{split} (E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha) &=\int_{\sigma(T_\alpha)}\!\chi_{e\cap\sigma(T_\alpha)}(\lambda)\, dE'_{x_\alpha,y_\alpha}(\lambda) =\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\!f_{n,\alpha}(\lambda)\, dE'_{x_\alpha,y_\alpha}(\lambda)\\ &=\lim_{n\to\infty}\int_{\sigma(T)}\!f_n(\lambda)\, dE_{x_\alpha,y_\alpha}(\lambda) =\int_{\sigma(T)}\!\chi_e(\lambda)\, dE_{x_\alpha,y_\alpha}(\lambda)\\ &=(E(e)x_\alpha,y_\alpha)\qquad\qquad\qquad\qquad\qquad \text{(e closed \subseteq\sigma(T)).}\quad(3) \end{split} \end{equation*}$$

Let $$\begin{equation*} \mathscr{M}=\{e\in B_T: (E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha) =(E(e)x_\alpha,y_\alpha)\text{ for all }x_\alpha,y_\alpha\in H_\alpha\}. \end{equation*}$$ Suppose $$e\in\mathscr{M}$$. Then $$\begin{equation*} \begin{split} E'(e\cap\sigma(T_\alpha))+E'(e^c\cap\sigma(T_\alpha)) =E'(\sigma(T_\alpha))&=I_\alpha\quad\text{and}\\ E(e)+E(e^c)=E(\sigma(T))&=I, \end{split} \end{equation*}$$ so $$\begin{equation*} \begin{split} (E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha) &+(E'(e^c\cap\sigma(T_\alpha))x_\alpha,y_\alpha) =(I_\alpha x_\alpha,y_\alpha)\\ &=(I x_\alpha,y_\alpha) =(E(e)x_\alpha,y_\alpha)+(E(e^c)x_\alpha,y_\alpha), \end{split} \end{equation*}$$ hence $$(E'(e^c\cap\sigma(T_\alpha))x_\alpha,y_\alpha)=(E(e^c)x_\alpha,y_\alpha),$$ so $$e^c\in\mathscr{M}$$.

If $$e,e'\in\mathscr{M}$$, then $$E'(e'\cap\sigma(T_\alpha))x_\alpha, E(e)y_\alpha\in H_\alpha$$ and $$\begin{equation*} \begin{split} (E'((e\cap e')\cap\sigma(T_\alpha))x_\alpha,y_\alpha) &=(E'(e\cap\sigma(T_\alpha))E'(e'\cap\sigma(T_\alpha))x_\alpha,y_\alpha)\\ &=(E(e)E'(e'\cap\sigma(T_\alpha))x_\alpha,y_\alpha) =(E'(e'\cap\sigma(T_\alpha))x_\alpha,E(e)y_\alpha)\\ &=(E(e')x_\alpha,E(e)y_\alpha) =(E(e)E(e')x_\alpha,y_\alpha) =(E(e\cap e')x_\alpha,y_\alpha), \end{split} \end{equation*}$$ so $$e\cap e'\in\mathscr{M}$$.

Suppose $$\{e_1,e_2,\dots\}\subseteq\mathscr{M}$$ are disjoint. Then $$\begin{equation*} (E'((\cup_n e_n)\cap\sigma(T_\alpha))x_\alpha,y_\alpha) =\sum_{n=1}^\infty(E'(e_n\cap\sigma(T_\alpha))x_\alpha,y_\alpha) =\sum_{n=1}^\infty(E(e_n)x_\alpha,y_\alpha) =(E(\cup_n e_n)x_\alpha,y_\alpha), \end{equation*}$$ so $$e=\bigcup_{n=1}^\infty e_n\in\mathscr{M}.$$

If $$\{e_1,e_2,\dots\}$$ are not disjoint, let $$e_n'=e_n\cap\biggl(\bigcup_{i=1}^{n-1}e_n'\biggr)^c\qquad(n=1,2,\dots).$$ By induction, $$\{e_1',e_2',\dots\}$$ are disjoint and by induction along with what has been shown so far, $$\{e_1',e_2',\dots\}\subseteq\mathscr{M}$$. Since $$\bigcup_{n=1}^\infty e_n=\bigcup_{n=1}^\infty e_n',$$ we have that $$e=\cup_n e_n\in\mathscr{M}$$, so $$\mathscr{M}$$ is a $$\sigma$$-algebra containing the closed sets of $$\sigma(T)$$, so $$\mathscr{M}=B_T$$. This proves that $$\begin{equation*} (E'(e\cap\sigma(T_\alpha))x_\alpha,y_\alpha)=(E(e)x_\alpha,y_\alpha) \qquad(x_\alpha,y_\alpha\in H_\alpha,\,e\in B_T).\quad(4) \end{equation*}$$ Since (4) holds for all $$y_\alpha\in H_\alpha$$, we have that $$\begin{equation*} E'(e\cap\sigma(T_\alpha))x_\alpha=E(e)x_\alpha\qquad(x_\alpha\in H_\alpha,\, e\in B_T) \end{equation*}$$ hence $$E'(e\cap\sigma(T_\alpha))=E(e)|_{H_\alpha}\quad(e\in B_T)$$, which proves #3.

Suppose $$f\in\mathscr{B}_T$$ and $$f\geq 0$$. Let $$\{s_n\}$$ be a sequence of simple Borel measurable maps, bounded by $$f$$, converging uniformly on $$\sigma(T)$$ to $$f$$. Then $$\{s_{n,\alpha}\}$$ converges uniformly on $$\sigma(T_\alpha)$$ to $$f_\alpha$$. Say $$s_n(\lambda)=\sum_{i=1}^{k_n} c_{n,i}\chi_{e_{n,i}}(\lambda).$$ By the spectral theorem $$\begin{equation*} \lvert((f(T)-s_n(T))x_\alpha,y_\alpha)\rvert \leq\lvert\lvert f(T)-s_n(T)\rvert\rvert\,\lvert\lvert x_\alpha\rvert\rvert\, \lvert\lvert y_\alpha\rvert\rvert\to 0 \end{equation*}$$ as $$n\to\infty$$. Similarly, $$\lvert((f_\alpha(T_\alpha)-s_{n,\alpha}(T_\alpha))x_\alpha,y_\alpha)\rvert\to 0$$. By the spectral theorem and by (4), $$\begin{equation*} \begin{split} (f_\alpha(T_\alpha)x_\alpha,y_\alpha) &=\lim_{n\to\infty}\,(s_{n,\alpha}(T_\alpha)x_\alpha,y_\alpha) =\lim_{n\to\infty}\int_{\sigma(T_\alpha)}\! s_{n,\alpha}(\lambda)\,dE'_{x_\alpha,y_\alpha}(\lambda)\\ &=\lim_{n\to\infty}\sum_{i=1}^{k_n} c_{n,i}E'_{x_\alpha,y_\alpha}(e_{n,i}\cap\sigma(T_\alpha)) =\lim_{n\to\infty}\sum_{i=1}^{k_n} c_{n,i}E_{x_\alpha,y_\alpha}(e_{n,i})\\ &=\lim_{n\to\infty} \int_{\sigma(T)}\!s_n(\lambda)\,dE_{x_\alpha,y_\alpha}(\lambda) =\lim_{n\to\infty}\,(s_n(T)x_\alpha,y_\alpha)\\ &=(f(T)x_\alpha,y_\alpha)\qquad\qquad\qquad(x_\alpha,y_\alpha\in H_\alpha). \end{split} \end{equation*}$$ Since $$P_\alpha$$ commutes with $$E(e)$$ for every $$e\in B_T$$, by the spectral theorem, $$P_\alpha$$ also commutes with $$f(T)$$ for every $$f\in\mathscr{B}_T$$. Therefore, $$f(T)H_\alpha\subseteq H_\alpha$$, so $$f_\alpha(T_\alpha)x_\alpha=f(T)x_\alpha$$ for all $$x_\alpha\in H_\alpha$$, whence $$f_\alpha(T_\alpha)=f(T)|_{H_\alpha}$$. If $$f$$ is not positive, then we can decompose it into its real and imaginary positive and negative parts, all of which are positive, and finally get $$f_\alpha(T_\alpha)=f(T)|_{H_\alpha}$$, which proves #4 and completes the proof of the theorem.