Convergence of Operators

Question

Suppose $\{A_{n}\}$ is a sequence of bounded positive operators on a Hilbert space H . This sequence is a decreasing sequence in the sense that $A_{n+1} \leq A_{n} $ (i.e. $A_{n} - A_{n+1}$ is a positive operator) for all $ n \in \mathbb{N} $ . I have to show that for any $ h \in H$ the sequence $\{A_{n}(h)\}$ converges in H . Moreover the limit produces a positive operator on H .

Answer 1

Since $$ \langle A_nh,h\rangle\leq\langle A_1h,h\rangle\leq\|A_1\|\,\|h\|^2, $$ it follows that $\|A_n\|\leq\|A_1\|$ for all $n$. So the sequence is bounded in norm.

For each $h$, the sequence of numbers $\{\langle A_nh,h\rangle\}$ is nonnegative and nonincreasing, so it is convergent. Define $$ \alpha_n(h,k)=\langle A_nh,k\rangle. $$ By polarization, $$ \alpha_n(h,k)=\sum_{m=0}^3 i^m \langle A_n(h+i^mk),h+i^mk\rangle, $$ so $$ \alpha(h,k)=\lim_{n\to\infty} \alpha_n(h,k) $$ exists. Because each $\alpha_n$ is sesquilinear, so is $\alpha$. Note also that $$ |\alpha(h,k)|\leq\|A_1\|\,\|h\|\,\|k\|, $$ so $\alpha$ is bounded. Now fix $k$; the map $h\longmapsto \alpha(h,k)$ is a bounded linear functional. By the Riesz Representation Theorem there exists a unique $z_k$ such that $$ \alpha(h,k)=\langle h,z_k\rangle. $$ The uniqueness, together with the properties of $\alpha$, imply that the map $k\longmapsto z_k$ is linear and bounded. So it is a linear operator, that we denote by $A$. So we have $$ \lim_n\langle A_nh,k\rangle=\langle h,Ak\rangle. $$ In particular $$ \langle A^*h,h\rangle =\lim_n\langle A_nh,h\rangle\geq0 $$ for all $h$, so $A^*$ is selfadjoint and positive, and $$ \lim_n\langle A_nh,k\rangle=\langle Ah,k\rangle. $$ Since every positive operator has a square root, for any $T\geq0$ we have $T\leq\|T\|\,I$, and $$ T^2=T^{1/2}TT^{1/2}\leq\|T\|\,T. $$ Then \begin{align} \|(A_n-A)h\|^2&=\langle(A_n-A)^2h,h\rangle \leq \|A_n-A\|\,\langle (A_n-A)h,h\rangle\\[0.2cm] &\leq 2\|A_1\|\,\langle (A_n-A)h,h\rangle \xrightarrow[n\to\infty]{}0. \end{align} So $A_nh\to Ah$ for all $h\in H$.