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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?

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2 Answers 2

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

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  • $\begingroup$ 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 (?) $\endgroup$ May 20 at 15:54
  • $\begingroup$ 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) $\endgroup$ May 20 at 15:56
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    $\begingroup$ 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. $\endgroup$
    – Ruy
    May 20 at 19:29
  • $\begingroup$ Oh right. I was so fixated on the holomorphic functional calculus, I forgot about the continuous… Thanks again. $\endgroup$ May 20 at 19:56
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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 $\sigma(x+y)\subseteq U$.

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