Is the square root function norm continuous?

Question

Let $\{a_n\}$ be a sequence of positive operators in $B(H)$. What about the following implication, True or false?

$$||a_n-a||\to 0\Longrightarrow ||a_n^{\frac{1}{2}}-a^{\frac{1}{2}}||\to0$$

Answer 1

Yes. This is true for any continuous function, not just for the square root.

Your assumption implies in particular that the spectrum of $a$ and all $a_n$ is contained in $I:=[0,R]$ for some $R>0$. On this set, the scalar square root is continuous, so that there is a sequence of polynomials $p_n$ which converge uniformly on $I$ to the square root function.

Now, by elementary properties of the spectral calculus, we have $$ \| \sqrt(b) - p_n (b)\|\leq \|\sqrt{\cdot} - p_n\|_{\sup,I} $$ for every $b \in B(H)$ with spectrum in $I$. Here, $\|f\|_{\sup,I}=\sup_{x\in I}|f(x)|$.

Thus, we see $p_n (b) \to \sqrt{b}$ uniformly with respect to $b$. But since $b \mapsto p_n (b)$ is continuous, so is $b \mapsto\sqrt{b}$ on the set of all operators with spectrum in $I$.