# Math Help - Sequence of the positive elements in B(H)

1. ## Sequence of the positive elements in B(H)

Let $B(H)$ - the algebra of bounded linear operators on a Hilbert space $H$, and $x\in B(H)$ is positive. Then sequence $x (\frac{1}{n}+x)^{-1}$ is monotone increasing to the range projection of $x:\quad [x].$

2. Originally Posted by karkusha

Let $B(H)$ - the algebra of bounded linear operators on a Hilbert space $H$, and $x\in B(H)$ is positive. Then sequence $x (\frac{1}{n}+x)^{-1}$ is monotone increasing to the range projection of $x:\quad [x].$
This makes no sense. Could you rephrase? What sequence? What projection?

3. What projection?
For each operator $x\in B(H)$ we define the range projection of $x$ as the projection on the closure of $xH$.

What sequence?
The sequence of operators is $x_n=x(\frac{1}{n}+ x)^{-1}$.

4. Originally Posted by karkusha
Let $B(H)$ - the algebra of bounded linear operators on a Hilbert space $H$, and $x\in B(H)$ is positive. Then sequence $x (\frac{1}{n}+x)^{-1}$ is monotone increasing to the range projection of $x:\quad [x].$
Have you come across the functional calculus for selfadjoint operators? If so, you can translate this problem into a problem about functions on the spectrum $S$ of the operator $x$. In fact, you just need to show that the sequence of functions $f_n(t) = t\bigl(\frac1n+t\bigr)^{-1}$ is monotone increasing (on the set $S$) to the function $p(t)$, where $p(0)=0$ and $p(t) = 1$ for $t>0.$

To do this without the functional calculus would be harder. You could start by observing that the operators $x$, $\bigl(\frac1m+x\bigr)^{-1}$ and $\bigl(\frac1n+x\bigr)^{-1}$ all commute with each other. Let $x_n = x \bigl(\frac{1}{n}+x\bigr)^{-1}$. If n>m then $x_n-x_m = \bigl(\frac1m-\frac1n\bigr)x\bigl(\frac{1}{m}+x\bigr)^{-1}\bigl(\frac{1}{n}+x\bigr)^{-1}$. The right side of that is a product of three commuting positive operators and is therefore positive. That shows that the sequence $(x_n)$ is monotone increasing.

To complete the proof, you need to know that $\|x_n\|\leqslant1$. This again is easy if you know a bit of spectral theory. In fact, it follows from the fact that $|f_n(t)|\leqslant1$ on $S$.

First, notice that $1-x_n = \bigl((\frac1n+x)-x\bigr)\bigl(\frac{1}{n}+x\bigr)^{-1} = \frac1n\bigl(\frac{1}{n}+x\bigr)^{-1}$. It follows that if $x\xi$ is in the range of $x$ (where $\xi\in H$) then $(1-x_n)x\xi = \frac1nx_n\xi\to0$ as $n\to\infty$. Thus $x_n\to 1$ strongly on the range of $x$ and hence (since $\|x_n\|\leqslant1$ for all n) on the closure of the range of $x$.

The orthogonal complement of the range of $x$ is the null space of $x$. So if $\eta$ is in that orthogonal complement then $x_n\eta = \bigl(\frac{1}{n}+x\bigr)^{-1}x\eta = 0$. Thus $x_n\to0$ strongly on the orthogonal complement of the range of $x$.

Put that all together to see that $x_n$ converges strongly to the range projection of $x$.