# Continuity of polar decomposition

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

• Thanks! Sadly I'm not very well informed about spectral theory... would you mind explaining why $\|p_n(T) -\sqrt{T}\| \leq \sup_{x\in [0, M]} |p_n(x)-\sqrt{x}|$? I guess if we took $p_n$ to be polynomials in $z$ and $\overline z$, we could use Stone-Weierstraß to approximate $\sqrt{\cdot}$ on the compact set $\gamma([0,1])$, if $\gamma$ is our contour, and get the estimate up to a constant [using standard integral estimates] (which would suffice), but I guess there is some direct argument using the spectrum (?) May 20 at 15:54
• Nevertheless, the main argument seems to be that for positive operators we have that the spectral radius equals the norm (thus a continuous map), which allows us to see that for sufficiently "close" operators the spectrum is contained in the set $[0, M]$ (which itself is enclosed by $\gamma$), right? (It was a long day, so I hope I don't make trivial mistakes) May 20 at 15:56
• Please see the edit for a proof of that inequality. When dealing with self adjoint operators it is best to stay in the real line and avoid the analytic functional calculus.
– Ruy
May 20 at 19:29
• Oh right. I was so fixated on the holomorphic functional calculus, I forgot about the continuous… Thanks again. May 20 at 19:56

The set $$U\subseteq \Bbb C$$ enclosed by $$\gamma$$ is open. We may use the following result:

Theorem: Let $$A$$ be a Banach algebra, $$x\in A$$ and $$U\subseteq \Bbb C$$ open with $$\sigma(x) \subseteq U$$. Then $$\exists \delta >0$$ such that $$\sigma(x+y)\subseteq U$$ for all $$y$$ with $$\|y\| < \delta$$.

Proof: I will write $$\rho(x) = \Bbb C\setminus \sigma(x)$$ for the resolvent set. Let $$R_x(\lambda) = (\lambda\Bbb{1}-x)^{-1}$$ for $$\lambda\in\rho(x)$$. Then for $$x\neq 0$$ $$R_x(\lambda) = (\lambda(\Bbb{1}-\tfrac{1}{\lambda}x))^{-1} = \tfrac{1}{\lambda} (\Bbb{1}-\tfrac{1}{\lambda}x)^{-1} \to 0\cdot(\Bbb{1}-0)^{-1}=0 \quad\text{for}\quad |\lambda|\to\infty,$$ as inversion is continuous. We thus find $$M>0$$ such that $$\sup_{\lambda\in\Bbb C\setminus U} \|R_x(\lambda)\| \leq M,$$ where $$\Bbb C\setminus U$$ is a closed set, so $$(\Bbb C\setminus U) \cap B^{\Bbb C}_r(0)$$ is a compact set and $$\lambda\mapsto \|R_x(\lambda)\|$$ is continuous.

Now choose $$\delta = \frac{1}{2M}$$ and let $$y\in A$$ with $$\|y\| < \delta$$. For $$\lambda\in \Bbb C\setminus U$$ we have $$\lambda\Bbb{1} - (x+y) = (\lambda\Bbb{1} - x)+y = (\lambda\Bbb{1} - x)(\Bbb{1} - (\lambda\Bbb{1}-x)^{-1}y)\in A^{\times}A^{\times} \subseteq A^{\times}.$$ Note that $$(\lambda\Bbb{1}-x)\in A^{\times}$$ as $$\sigma(x)\subseteq U$$ and furthermore $$\|(\lambda\Bbb{1}-x)^{-1} y\| \leq \|(\lambda\Bbb{1}-x)^{-1}\| \|y\| \leq M \delta < \tfrac{1}{2},$$ so $$(\Bbb{1} - (\lambda\Bbb{1}-x)^{-1}y) \in A^{\times}$$ by the Neumann series. Thus $$\lambda\in \rho(x+y)$$, so indeed $$\sigma(x+y)\subseteq U$$.