Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).
I want to show by using the spectral theorem that: for any $x\in F$ there exists a sequence $(x_n)_n$ with
$$M^{1/2}x=\lim_{n\to \infty} Mx_n\;.$$
Best Answer
You can use the binomial expansion. Suppose that $0 \le M \le I$, which you can do by scaling $M$ by a positive constant, if necessary. Then the unique positive square root of $M$ is given by the binomial expansion, $$ M^{1/2} = (I-(I-M))^{1/2}=I-\sum_{n=1}^{\infty}c_n(I-M)^n, $$ where $c_n$ are positive constants such that $\sum_{n=1}^{\infty}c_n=1$. You can use that to show that the range of $M$ is dense in the range of $M^{1/2}$.
Using the spectral theorem, and assuming $L$ is an upper bound for the spectrum of $M$, $$ M\int_{1/n}^{L}\lambda^{-1/2}dE(\lambda)x= \int_{1/n}^{L}\lambda^{1/2}dE(\lambda)x. $$
So $x_n=\int_{1/n}^{L}\lambda^{-1/2}dE(\lambda)x$ is such that $Mx_n\rightarrow M^{1/2}x$ as $n\rightarrow\infty$, which proves that the range of $M$ is dense in the range of $M^{1/2}$.