# Continuity of polar decomposition

continuityfunctional-analysisfunctional-calculusspectral-theory

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?

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 $$\|p_n(T)-\sqrt {T}\|= \|\phi(p_n-\sqrt{\cdot})\| =$$$$= \|p_n-\sqrt{\cdot}\| = \sup_{x\in \sigma(T)}|p_n(x)-\sqrt {x}|\le$$$$\le \sup_{x\in [0,M]}|p_n(x)-\sqrt {x}|.$$