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If $A_n$ is a sequence of positive bounded linear operators converging in norm to $A$ on a Hilbert Space, show $\sqrt{A_n}\to\sqrt{A}$ in norm. I can show that $A$ would be positive and thus have a square root, but then I'm mostly stuck.

If $A_n=B_n^2$ and $A=B^2$, I have also shown that since $B_n$ is positive it is by definition self adjoint and so $$\|B_n x\| = \sqrt{\langle B_n x, B_n x \rangle} = \sqrt{\langle A_n x, x\rangle} \to \sqrt{\langle Ax,x\rangle} = \|Bx\|2$$ for all $x$ and so therefore $\|B_n\|\to \|B\|$. However, I am completely stuck on the desired result.

If I knew $B_n$ and $B$ would commute, then I'd use $$\|A_n^2 - A\| = \|(B_n - B)(B_n + B) \|$$ and play with the inner product, but I don't know this a priori.

Thanks for your help.

EDIT: For those wondering this question comes from Mathematical Physics I: Functional Analysis by Reed and Simon in chapter 7 question 14.

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Do you know about the continuous functional calculus for normal operators? If so this result follows right away from the functional calculus and the fact that this is true when $A_n$ is a positive real number. – mck Dec 6 '12 at 2:57
I tried proving this with the implicit function theorem, but seem to need to bound the spectrum of $A$ away from $0$ :-(. – copper.hat Dec 6 '12 at 3:26
No I do not. That is mentioned after proving the spectral theorem in the book I am using (Reed and Simon) which is the chapter after I am in. – toypajme Dec 6 '12 at 3:28
Related question on MathOverflow:… (The result follows from the inequality there, but you may not have the tools used in the proofs mentioned there.) – Jonas Meyer Dec 6 '12 at 4:09
@mck: I wouldn't say it follows "right away"; you have to be careful because $A_n$ and $A$ need not commute. – Jonas Meyer Dec 6 '12 at 4:12
up vote 5 down vote accepted

There is nothing particular about the square root here, as the result holds for any continuous function.

From $A_n\to A$, we get in particular that $\|A_n\|\to\|A\|$. This guarantees that $\sigma(A)\cup\bigcup_n\sigma(A_n)$ is contained in some interval $[a,b]$.

Let $f:[a,b]\to\mathbb R$ be continuous. Using the Spectral Theorem, we get that $\|f(X)\|\leq\|f\|_\infty$ for any selfadjoint $X$ with $\sigma(X)\subset[a,b]$. Indeed, $$ \left|\langle f(X)x,x\rangle\right|=\left|\langle \int_{\sigma(X)}\,f(\lambda)\,dE_X(\lambda)\,x,x\rangle\right|=\left|\int_{\sigma(X)}\,f(\lambda)\,\langle dE_X(\lambda)x,x\rangle \right|\\ \leq\int_{\sigma(X)}\,|f(\lambda)|\,\langle dE_X(\lambda)x,x\rangle\leq\|f\|_\infty\int_{\sigma(X)}\,1\,\langle dE_X(\lambda)x,x\rangle=\|f\|_\infty\,\langle x,x\rangle. $$ Now fix $\varepsilon>0$. Let $p$ be a polynomial such that $\|f-p\|_\infty<\varepsilon/3$ (this polynomial exists by Weierstrass Approximation Theorem). As $A_n^k\to A^k$ for all $k$, there exists $n_0$ such that $\|p(A_n)-p(A)\|<\varepsilon/3$ if $n\geq n_0$.

So, for $n\geq n_0$, $$ \|f(A_n)-f(A)\|\leq\|f(A_n)-p(A_n)\|+\|p(A_n)-p(A)\|+\|p(A)-f(A)\|\\ \leq\|f-p\|_\infty+\frac\varepsilon3+\|p-f\|_\infty<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. $$

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Although I was hoping to avoid the spectral theorem (as I had not encountered it yet at this point in the book), this seems like the best method we have. Thank you for yoru answer. – toypajme Dec 18 '12 at 6:19
You are welcome. The spectral theorem is not really needed; it can be replaced for example by the continuity of the Gelfand transform. I used the spectral theorem because you mentioned it in the comments. – Martin Argerami Dec 18 '12 at 12:03

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