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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.2012-12-06
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    I tried proving this with the implicit function theorem, but seem to need to bound the spectrum of $A$ away from $0$ :-(.2012-12-06
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    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.2012-12-06
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    Related question on MathOverflow: http://mathoverflow.net/questions/24392/simple-inequality-in-c-algebras (The result follows from the inequality there, but you may not have the tools used in the proofs mentioned there.)2012-12-06
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    @mck: I wouldn't say it follows "right away"; you have to be careful because $A_n$ and $A$ need not commute.2012-12-06
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    Yes the commuting was causing several headaches. The inequality there seems to be assuming too much in my situation.2012-12-07
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    @toypajme: What do you mean by "The inequality there seems to be assuming too much"? The inequality holds in your situation. Do you mean that you cannot use the tools mentioned in the proofs, e.g. because of what chapter you are in in the book?2012-12-10
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    @JonasMeyer Yes, that is what I mean. Sorry for being unclear.2012-12-11

1 Answers 1

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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, $$ \left|\langle f(X)x,x\rangle\right|=\left|\langle \int_{\sigma(X)}\,f(\lambda)\,dE_X(\lambda)\,x,x\rangle\right|=\left|\int_{\sigma(X)}\,f(\lambda)\,\langle dE_X(\lambda)x,x\rangle \right|\\ \leq\int_{\sigma(X)}\,|f(\lambda)|\,\langle dE_X(\lambda)x,x\rangle\leq\|f\|_\infty\int_{\sigma(X)}\,1\,\langle dE_X(\lambda)x,x\rangle=\|f\|_\infty\,\langle x,x\rangle. $$ (the same estimate can be obtained via the Gelfand Transform if one doesn't want to use the Spectral Theorem).

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.2012-12-18
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    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.2012-12-18
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    I'm really interested in learning the reason for the downvote.2017-07-20