This question is about a step in the proof of this answer, which is not directly clear to me.

Consider the following scenario:

$H$ is a Hilbert space, $A\in GL(H)$ a positive self-adjoint operator. We surround the spectrum $\sigma(A)$ by a contour $\gamma$ which does not intersect $]-\infty, 0]$. This allows to use the holomorphic functional calculus to define $\sqrt{A}$. I would like to see that $\sqrt{\cdot}$ is continuous at $A$.

**My problem is the following:** If $B$ is sufficiently close to $A$ wrt $\|\cdot\|_{op}$, how do we see that $\sigma(B)$ is still enclosed by $\gamma$?

If we know this, we get the desired result simply by continuity of parameter integrals.

I can't quite put the argument together. Intuitively it should be right, as the spectrum is compact, $\gamma$ encloses an open set, $GL(H)\subseteq (\mathcal L(H), \|\cdot\|_{op})$ is open, so everything seems nice and fitting. However, I struggle with pinning down the exact argument.

Any suggestions?

## Best Answer

Indeed there is a very well developed spectral perturbation theory which tells you that two nearby operators have close spectra.

Nevertheless, for the problem in hand, there is a much simpler alternative as follows: choose a big enough real number $M$ such that all operators under consideration have norm at most $M$. Then find a sequence $\{p_n\}_n$ of real polynomials converging uniformly to the function $x\mapsto \sqrt x$ on the interval $[0,M]$.

Then, for every positive operator $T$ with $\|T\|\le M$, one has that the spectrum of $T$ is contained in $[0,M]$ and

$$ \|p_n(T)-\sqrt {T}\|\le \sup_{x\in [0,M]}|p_n(x)-\sqrt {x}|, $$ from where it follows that $p_n(T)\to \sqrt{T}$, in norm.

With this it is now easy to see that $\sqrt{T}$ is continuous in the variable $T$.

Edit: Here is a proof of the inequality above in a nutshell, using only: (1) the fact that the spectrum of a self-adjoint operator is contained in $\mathbb R$, (2) the spectral mapping theorem for polynomials, that is, $\sigma(p(T))=p(\sigma(T))$, and (3) the fact that the norm of a self-adjoint operator coincides with its spectral radius $$ \text{spr}(T) =\sup _{\lambda \in\sigma(T)}|\lambda|. $$

Given a self-adjoint operator $T$, consider the subspace $P\subseteq C(\sigma(T))$ formed by the polynomial functions.

For each $p$ in $P$, define $\phi(p)=p(T)$, so that $\phi $ is a linear map from $P$ to $B(H)$. It is also norm preserving because $$ \|\phi(p)\|=\|p(T)\|=\text{spr}(p(T))= $$$$= \sup _{\lambda \in\sigma(p(T))}|\lambda|= \sup _{\lambda \in\sigma(T)}|p(\lambda)| =\|p\|. $$

Since $P$ is dense in $C(\sigma(T))$, we may extend $\phi$ to $C(\sigma(T))$ by continuity. The extended map is then clearly an isometric homomorphism of Banach $^*$-algebras $$ \phi:C(\sigma(T))\to B(H). $$ It is often also called the "continuous functional calculus".

A word about notation: whenever $f$ is in $C(\sigma(T))$, most people write $f(T)$ for $\phi(f)$.

Observe that, since $\sqrt{\cdot}$ is a function whose square is the identity, then $\sqrt{T}$ is indeed the square root of $T$.

We are thus ready for the punch line $$ \|p_n(T)-\sqrt {T}\|= \|\phi(p_n-\sqrt{\cdot})\| = $$$$= \|p_n-\sqrt{\cdot}\| = \sup_{x\in \sigma(T)}|p_n(x)-\sqrt {x}|\le $$$$\le \sup_{x\in [0,M]}|p_n(x)-\sqrt {x}|. $$