There are a lot of questions here about square root operators, but none of them addresses the basic question of existence, and I didn't find a very beefy section in Wikipedia talking about this, so I'll ask it here. Let $A$ be a bounded positive-semidefinite Hermitian operator on Hilbert space. The claim is that there exists a positive-semidefinite Hermitian operator $B$ such that $B\circ B=A$. I've got no idea how you would prove this. Any tips or references would be appreciated.
[Math] Square root of a Hermitian operator exists
functional-analysishilbert-spaceslinear algebraoperator-theory
Best Answer
If $0 \le A \le I$ (which you can scale to get for a bounded $A$), then there are classical proofs out there to show that the following algorithm converges in the strong operator topology to $B$ such that $0 \le B$ and $B^{2}=A$: $$ B_{0}=0,\;\; B_{n+1}=B_{n}+\frac{1}{2}(A-B_{n}^{2}). $$ There are some basic facts you need in the proof, and these are not straightforward. First if $A$ and $B$ are bounded, commuting, selfadjoint operators, then $0 \le A \le B$ implies $A^{2} \le B^{2}$--this inequality is important in order to make sure that $\{ B_{n}\}$ is an operator monotone sequence, a proof which reduces to showing that the following is positive: $$ B_{n+2}-B_{n+1}=\frac{1}{2}(I-B_{n})^{2}-\frac{1}{2}(I-B_{n+1})^{2}. $$ Then, you need to know that, because $0 \le B_{1} \le B_{2} \le B_{3}\le \cdots \le I$, the limit $\lim_{n}B_{n}x=x$ exists for all $x \in X$. You end up with a unique $0 \le B \le I$ such that $B^{2}=A$, and such that $B$ commutes with every bounded operator which commutes with $A$.
Convergence: I'm assuming the underlying Hilbert space is complex. The argument for convergence is not terribly long; so I'll give you that part. The only basic proof I know for the fact that $0 \le A \le B \implies A^{2} \le B^{2}$ for commuting $A$, $B$ is too long.
Once you have $0 \le B_{1} \le B_{3} \le B_{3} \le \cdots \le I$, then $\lim_{n}(B_{n}x,x)$ exists for all $x$ because $\{ (B_{n}x,x)\}_{n=1}^{\infty}$ is a bounded, non-decreasing sequence of real numbers for each fixed $x$. By the Polarization Identity, $b(x,y)=\lim_{n}(B_{n}x,y)$ also exists for all $x,y$. $b$ is sesquilinear and symmetric with $0 \le b(x,x) =\lim_{n}(B_{n}x,x) \le \|x\|^{2}$. So $|b(x,y)|\le \|x\|\|y\|$ also holds.
By the Lax-Milgram theorem there is a unique $0 \le B=B^{\star} \le I$ such that $b(x,y)=(Bx,y)$. And, $0 \le B_{n} \le B \le I$ for all $n$. Because $(x,y)_{n}=((B-B_{n})x,y)$ has the properties of an inner-product, except possibly for positive definiteness, then the Cauchy-Schwarz inequality holds: $$ |(x,y)_{n}|^{2} \le (x,x)_{n}(y,y)_{n} \le (x,x)_{n}(y,y) $$ Now, let $y=(B-B_{n})x$ to obtain $$ \|(B-B_{n})x\|^{2} \le ((B-B_{n})x,x). $$ Therefore $\lim_{n}\|(B-B_{n})x\|=0$ for all $x$ because $\lim_{n}((B-B_{x})x,x)=0$.