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?
 A: 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
$$
\|p_n(T)-\sqrt {T}\|=
\|\phi(p_n-\sqrt{\cdot})\| = $$$$=  
\|p_n-\sqrt{\cdot}\| = \sup_{x\in \sigma(T)}|p_n(x)-\sqrt {x}|\le $$$$\le
\sup_{x\in [0,M]}|p_n(x)-\sqrt {x}|.
$$
A: 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 $\sigma(x+y)\subseteq U$.
