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?


2 Answers 2


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $\sigma(x+y)\subseteq U$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.