# 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 $P_{\lambda} = P_{(-\infty, \lambda)}$

• 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. – timur Sep 1 '13 at 14:04
• @timur I have tried to answer your questions in the post. – N.U. Sep 1 '13 at 14:32

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. Sep 1 '13 at 15:15