Thread: Sequence of the positive elements in B(H)

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

Hello, could you please help me to solve this problem:

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

2. Originally Posted by karkusha Hello, could you please help me to solve this problem:

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

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

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

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

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

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

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

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

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

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