# Show that an operator is bounded (from Reed and Simon)

I am currently reading Reed and Simon's IV: Analysis of Operators, Volume 4 (Methods of Modern Mathematical Physics). I don't understand something they do in Theorem XIII.64.

The problem is: Let $A$ be a self-adjoint operator that is bounded from below, and let $P_{\Omega}$ be the family of spectral projections corresponding to $A$. Then $AP_{(-\infty, \lambda)}$ is a bounded operator.

What I have thought so far: $A$ can be thought of as the identity on its spectrum and since the spectrum is bounded from below, and since $P_{(-\infty, \lambda)}$ corresponds to the characteristic function of $(-\infty, \lambda)$, $AP_{(-\infty, \lambda)}$ corresponds to the identity on a bounded set, and is then a bounded operator. I have no idea on how to give a rigorous proof of this.

Edit: Here $P_{\Omega} = \chi_{\Omega}(A)$ for a Borel set $\Omega$, and where $\chi$ is the characteristic function. $AP_{\Omega}$ is composition of the operators $A$ and $P_{\Omega}$.

Edit: spectral theorem (multiplication form) : Let $A$ be a self-adjoint operator on a separable Hilbert space $H$. Then there is a finite measure space $(M, \mu)$, and a unitary operator $U: H \rightarrow L^2(M, \mu)$ and a real-valued function $f$ which is finite a. e. such that $UAU^{-1}\phi(m) = f(m)\phi(m)$.

spectral theorem (projection valued measure form) There is a one-to-one correspondence between self-adjoint operators $A$ and projection-valued measures $\{P_{\Omega}\}$, the correspondence is given by $$A = \int_{-\infty}^{\infty}\lambda dP_{\lambda},$$ where $P_{\lambda} = P_{(-\infty, \lambda)}$

Thank you for your help!

• Can you write explicitly what $P_\Omega$ and $A P_\Omega$ are? I just want to clarify if there is any difference between your terminologies and mine. For instance I would say $A$ is not quite the identity, but it is a multiplication operator on its spectrum. Commented Sep 1, 2013 at 14:04
• @timur I have tried to answer your questions in the post.
– N.U.
Commented Sep 1, 2013 at 14:32

## 1 Answer

Let me give an answer first in a simple but very paradigmatic case. Suppose that the underlying Hilbert space is $L^2(\mathbb{R})$ and that the operator $A$ is given by $(Af)(\lambda)=\chi_\Omega\lambda f(\lambda)$, with $\Omega\subset(b,\infty)$ for some $b$. Then we have $$\|AP_{(-\infty,\lambda)}f\|_{L^2}^2 = \int_{\Omega\cap(-\infty,\lambda)} |\mu|^2|f(\mu)|^2\,\mathrm{d}\mu \leq \max\{b^2,\lambda^2\}\|f\|_{L^2}^2,$$ which shows the boundedness of $AP_{(-\infty,\lambda)}:L^2(\mathbb{R})\to L^2(\mathbb{R})$.

Next, let us consider the general case where the underlying Hilbert space is $H$. We use the spectral theorem in its projection valued measure form. In particular, this form implies that for any bounded Borel function $g$ on $\mathbb{R}$ and for $\varphi\in H$, we have $$(\varphi,g(A)\varphi) = \int_{-\infty}^\infty g(\mu)\,\mathrm{d}(\varphi,P_\mu\varphi),$$ where $\mathrm{d}(\varphi,P_\mu\varphi)$ is the Borel measure on $\mathbb{R}$ defined by $$\int_{\Omega} \mathrm{d}(\varphi,P_\mu\varphi) = (\varphi,P_\Omega\varphi),$$ for any Borel set $\Omega$. Since $A$ is bounded from below, the support of the measure $\mathrm{d}(\varphi,P_\mu\varphi)$ is contained in $(b,\infty)$ for some $b$. Putting $$g(\mu)=\mu\,\chi_{(-\infty,\lambda)}(\mu),$$ we get $$|(\varphi,AP_{(-\infty,\lambda)}\varphi)| \leq \int_{b}^\lambda |\mu|\,\mathrm{d}(\varphi,P_\mu\varphi) \leq \max\{|b|,|\lambda|\}(\varphi,P_{(b,\lambda)}\varphi) \leq \max\{|b|,|\lambda|\}\|\varphi\|^2,$$ where without loss of generality we have assumed that $b<\lambda$, and in the last step we have used the fact that $P_\Omega$ are orthogonal projectors. Let us summarize what we have: We have a self adjoint operator $B=AP_{(-\infty,\lambda)}$ satisfying $$|(\varphi,B\varphi)| \leq c \|\varphi\|^2 \qquad\textrm{for all}\quad \varphi\in H,$$ with the constant $c=\max\{|b|,|\lambda|\}$. Now it should not be difficult to show that $B:H\to H$ is bounded.

• Thanks, but do you have any suggestion on how to go from $L^2(\mathbb{R})$ to a general Hilbert space?
– N.U.
Commented Sep 1, 2013 at 15:15
• @N.U.: That is the spectral theorem, right? :) It should be somewhere in Reed and Simon! Commented Sep 1, 2013 at 15:20
• @N.U.: If you are having problems with how to apply the spectral theorem, can you include in your question the statement of the spectral theorem from the book? Perhaps also point out what difficulty you are having? Commented Sep 1, 2013 at 15:25
• Yes I'm having problems applying it. I included two of the definitions in Reed and Simon. Which one should I use, and how?
– N.U.
Commented Sep 1, 2013 at 15:37
• Very nice! Thanks for your patience.
– N.U.
Commented Sep 2, 2013 at 7:23