Showing that the operator $T=\sum_{n\in J}f(n)P_n$ is bounded and finding its spectrum in a Hilbert space $H=\bigoplus_{n\in J}H_n,P_n$ projection map I have three questions related to the boundedness, operator norm and spectrum of the following operator in a Hilbert space $H$. I have bolded my exact questions for clarity.
Let $H$ be a Hilbert space over the field of complex numbers $\mathbb{C}$, $J$ a collection and assume that $H = \bigoplus_{n\in J}H_n$ with $P_n:H\to H_n$ being the orthogonal projection onto the closed subspace $H_n$ of $H$ with span dense in $H$, that is $\overline{\mathrm{span}(\bigcup_{n\in J}H_n)} = H$. Suppose that $f$ is a function such that $\forall n \in J:|f(n)| < 1$ and define the operator $T \equiv \sum_{n\in J}f(n)P_n$. I am both trying to show that the operator $T$ is bounded and finding its spectrum $\sigma(T)$.
As for boundedness, my idea was to take any finite index subset $I\subset J$ and use the Pythagorean theorem as follows for $u \in H$:
$$||\sum_{n\in I}f(n)P_n(u)||^2 = \sum_{n\in I}|f(n)|^2||P_n(u)||^2\Longleftrightarrow$$
$$\sum_{n\in I}|f(n)|^2||P_n(u)||^2\leq \sum_{n\in I}||P_n(u)||^2 \leq ||u||^2$$
by Bessel's inequality, showing that the series over the arbitrary collection $J$ converges.
(Question 1:) Here's my first question: It seems a bit too strong (and potentially incorrect) to use the Bessel's inequality at this part of the proof. I am thinking that if $\sum_{n=1}^NP_n$ is a sum over some unique orthogonal projection mappings onto some pairwise orthogonal subspaces, then
$$||\sum_{n=1}^NP_n(u)||^2 = \sum_{n=1}^N||P_n(u)||^2 \leq ||u||^2$$
seems reasonable, but I am not entirely sure how to prove this. If I start with any $P_n$, then $u = P_n(u) + Q_n(u)$ with $Q_n(u) = I - P_n(u)$. But this gives you only the inequality $||P_n(u)||^2 \leq ||u||^2$ for any single $P_n$, and not for the sum.
(Question 2:) The next question is related to the operator norm of $T$, which is connected to my prior boundedness question. Namely, how could you argue rigorously that $||T|| = \sup_{u \in H:||u||\leq 1}||Tu|| = \sup_{n\in J}|f(n)| \leq 1$? At the moment my high-level argument is that since the sum of the individual norms of the projected parts of any $u \in H$ is at most $u$'s norm, then the operator norm is maximized at the maximal modulus value of $f(n)$. But I don't know how to massage this argument into a more mature take on $T$'s operator norm.
(Question 3:) My last question is related to determining the spectrum $\sigma(T)$ of $T$. Assume that we've argued that $T$ is bounded. One spots almost immediately that the elements of the individual spaces $H_n$ are eigenvectors for $T$. Namely, $\forall n \in J:\forall u \in H_n:Tu = f(n)u$. Moreover these have to be the only eigenvectors of $T$ for if $u \not\in H_n$ for all $n \in J$, then $u = u_1 + u_2$ with $u_1 \in H_{n'}$ for some $n' \in J$ and $0 \neq u_2 = u - u_1 \not\in H_{n'}$ so that $Tu = f(n')u_1 + T(u_2) \neq \lambda u$ for a fixed $\lambda \in \mathbb{C}$.
Since $\sigma(T)$ consists of all $\lambda \in \mathbb{C}$ such that $\lambda I - T$ is not bijective, I think that this covers the non-injectivity part. But how do I even start to develop the non-surjectivity part? We know from the theory of Banach spaces that i.) $\sigma(T)$ is closed and ii.) $\sigma(T) \subset \overline{B_{||T||}(0)}$. We do know that the span of the union of all $H_n$ is dense in $H$ so that might reveal some new values for $\sigma(T)$ but I honestly don't know how to use this fact to my advantage.
Thanks!
 A: *

*Because the $P_n$ are pairwise orthogonal,
$$
\Big(\sum_{n=1}^NP_n\Big)^2=\sum_{n,m}P_nP_m=\sum_{n=1}^NP_n.
$$
Begin also selfadjoint, $\sum_{n=1}^NP_n$ is an orthogonal projection. Then $\Big\|\sum_{n=1}^NP_n\Big\|=1$.

Alternatively, if you want to work at the level of vectors, you can use that $u=\sum_n u_n$, and $\|u\|^2=\sum_n\|u_n\|^2$. Since $u_n=P_nu$, you get the result.


*You can do it rather explicitly:
$$
\|Tu\|^2=\sum_n|f(n)|^2\,\|P_nu\|^2\leq\|f\|_\infty\,\sum_n\|P_nu\|^2=\|f\|_\infty\,\|u\|^2. 
$$
Choosing an $n$ where $|f(n)|>\|f\|_\infty-\varepsilon$ and a unit vector $u\in H_n$, you get $$\|Tu\|=|f(n)>\|f\|_\infty-\varepsilon,$$ and hence $\|T\|=\|f\|_\infty$.


*You have shown that $f(n)$ is an eigenvalue for every $n$. Hence $\overline{f(\mathbb N)}\subset\sigma(T)$. Now suppose that $\lambda\not\in\overline{f(\mathbb N)}$. As $\overline{f(\mathbb N)}$ is closed, there exists $\delta>0$ such that $|\lambda-f(n)|>\delta$ for all $n$. Let $S$ be the operator defined by $$S=\sum_n \frac1{\lambda-f(n)}\,P_n.$$ From the previous computations, $S$ is bounded and $\|S\|\leq\frac1\delta$. And, by construction, $S(\lambda-T)=(\lambda-T)S=I$. Hence $\lambda\not\in\sigma(T)$ .Thus $$\sigma(T)=\overline{f(\mathbb N)}.$$
