Square root of a Hermitian operator exists

Question

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.

Answer 1

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 $\lim_{n}((B-B_{x})x,x)=0$.

Answer 2

You want to use the general Spectral Theorem. You obtain

$$ A = \int \lambda \, dE(\lambda),$$ where $dE(\lambda)$ is the spectral measure. Since $A$ is positive semidefinite, you can show that its spectral measure is supported on the nonnegative real line. Then it is well-defined to choose

$$ B = \int \sqrt{\lambda} \, dE(\lambda) $$

Answer 3

The Spectral Theorem is one way to do it. But it is like too strong a result: the ST gives you a Borel functional calculus, when all you need here is continuous functional calculus. The latter is achieved via the Gelfand Transform, that shows the isomorphism between C$^*(A) $ and $C(\sigma(A)) $; this way, $f(A) $ makes sense for all continuous functions on $\sigma(A) $, in particular $f(t)=t^{1/2} $.

Answer 4

There is a classical way to do what you want if $0 < \delta I \le A \in \mathcal{L}(X)$ for some $\delta > 0$, and this may be more interesting to you. The approach uses the well-defined series $e^{-tA}=\sum_{n=0}^{\infty}\frac{1}{n!}(-tA)^{n}$. The Laplace transform of $t^{-1+\alpha}$ for $0 < \alpha < \infty$ is $$ \int_{0}^{\infty}e^{-ts}t^{-1+\alpha}\,dt = s^{-\alpha}\int_{0}e^{-(ts)}(ts)^{-1+\alpha}d(ts) =\frac{1}{s^{\alpha}}\int_{0}^{\infty}e^{-u}u^{-1+\alpha}\,du=\frac{\Gamma(\alpha)}{s^{\alpha}}. $$ It turns out $$ A^{-\alpha} = \frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}e^{-tA}t^{-1+\alpha}\,dt, \;\;\; 0 < \alpha < \infty. $$ and there's no problem defining the integral on the right because $\|e^{-tA}\|\le e^{-t\delta}$. Using classical methods of Laplace transforms, one can show that $$ A^{-\alpha}|_{\alpha=1}=A^{-1},\;\;\; A^{-\alpha}A^{-\beta}=A^{-\alpha-\beta}, \;\; \alpha,\beta > 0. $$ The very form of the integral shows you that $A^{-\alpha} \ge 0$, and that $A^{-\alpha}$ commutes with $A$. Then you can get $\sqrt{A}=AA^{-1/2}$.

By the way, this approach works for a broad class of operators on a Banach space, which goes beyond what you can do with the Spectral Theorem. This is part of $C^{0}$ semigroup theory, where one can define $T(t)$ for $t \ge 0$ such that $T(t)T(s)=T(t+s)$, $T(0)=I$ and $\frac{d}{dt}T(t)x=-Ax$ for all $x \in \mathcal{D}(A)$ under well-known conditions on a densely-defined closed linear operator $A$ on a Banach space.

Final note: If $A$ is not positive definite, but only positive, then you can take a limit $$ \sqrt{A}x=\lim_{\epsilon\downarrow 0}(\epsilon I + A)(\epsilon I+A)^{-1/2}x. $$ Using the above, you can show this limit will exist.