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Let $A$ be a $C^*$ algebra.
Show that if $0 \le a \le b$ then $\sqrt a \le \sqrt b$.
I've shown that this is true in case $b$ is invertible, here is my proof:

$$\|a^{1/2}b^{-1/2}\|^2 = \|(a^{1/2}b^{-1/2})^*(a^{1/2}b^{-1/2})\| = \|b^{-1/2}ab^{-1/2}\| \le ||b^{-1/2}bb^{-1/2}||=1.$$

Thus $\|a^{1/2}b^{-1/2}\| \le 1$.

Now, as $\{0\}\cup \sigma(xy) = \{0\}\cup \sigma(yx)$ define $x=a^{1/2}b^{-1/4}$ and $y=b^{-1/4}$, denote by $\rho (z)$ the spectral radius of $z$, then $\rho (yx)=\|yx\|$ as $yx$ is self adjoint and $\rho(xy) \le \|xy\| \le 1$, but $\rho (xy)=\rho (yx)$ , so $\|yx\|=\|b^{-1/4}a^{1/2}b^{-1/4}\| \le 1$ and it's a positive element, therefore $0 \le b^{-1/4}a^{1/2}b^{-1/4} \le 1$. Multiply by $b^{1/4}$, which is self adjoint, from both sides and get $a^{1/2} \le b^{1/2}$ as required.

Now, for the general case:

I've tried to look at $0 \le a+1/n \le b+1/n$. By the previous part I know that $0 \le \sqrt {a+1/n} \le \sqrt{b+1/n}$ and by identification $C(\sigma(a))$ with $C^*(1,a)$, and same for $b$, I can show that $\lim_{n\to\infty} \| \sqrt{a+1/n}-\sqrt a\| =0$, and same for $b$ , even monotone convergence.

However, I can't just take limits to get the desired conclusion (or maybe I can?)

Any hints will be greatly appreciated.

Michael Hardy
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Shirly Geffen
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2 Answers2

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You have shown that the sequence of positive operators $$(b+1/n)^{1/2}-(a+1/n)^{1/2} $$ converges to $b^{1/2}-a^{1/2} $. So now all you have to show is that a limit of positives is positive.

The two ways that come to mind are: by representing $A\subset B (H) $ (and then using wot convergence); or, if you want to stay abstract, by looking at states.

Edit: With the second method, the key fact is that $a\in A$ is positive if and only if $\phi(a)\geq0$ for every state $\phi$. By continuity, for any state $\phi$ we have $$ \phi(b^{1/2}-a^{1/2})=\lim_n\phi((b+1/n)^{1/2}-(a+1/2)^{1/2})\geq0, $$ as a limit of non-negative numbers is non-negative. So $b^{1/2}-a^{1/2}\geq0$.

Martin Argerami
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  • Thank you. In the proposition: $a \ge 0$ iff $ \phi (a) \ge 0$ for every state $\phi$, do we need to assume $a$ is self adjoint? somehow I found it with this assumption. However, we deal with self adjoint element. – Shirly Geffen Apr 09 '16 at 15:35
  • Positive implies (or, depending on definitions, is defined as) selfadjoint. – Martin Argerami Apr 09 '16 at 15:38
  • @ShirlyGeffen$:$ If $\phi (a) \geq 0$ then $a$ is automatically self-adjoint. For that you need to first show that if for all states $\rho$ on $A,$ $\rho (b) = 0$ for some $b \in A$ then $b = 0.$ Try to first prove it for self-adjoint elements and then extend it for arbitrary elements by decomposing them into real and imaginary parts. Now suppose that for all states $\phi,$ $\phi (a) \geq 0$ i.e. in particular $\phi (a) \in \mathbb R.$ Then $\phi (a^{}) = \overline {\phi (a)} = \phi (a).$ Hence $\phi (a - a^{}) = 0$ for all states $\phi.$ So by the above remark it follows that $a = a^{*}.$ – ACB Jul 01 '22 at 06:48
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Thank you. If I understood it right then a possible solution is: There is a $*$-isomorphism $\phi : A \to L$ , where $L$ is a C*-subalgebra of $B(H)$ for some Hilbert space $H$.

$*$-isomorphism preserves positive elements, so we just need to verify that for any sequence of positive operators $T_n : H \to H$ that converges to $T$ in $B(H)$, $T$ is positive. Involution is norm-continuous so $T=T^*$ and it is sufficient to show that for any $x$ in $H$ we have $\langle Tx,x\rangle \ge 0$.

By WOT convergence $\lim_{n\to\infty} \langle T_nx,x\rangle=\langle Tx,x\rangle$ , but this is a sequence of real positive numbers, hence $\langle Tx,x\rangle \ge 0$. Is it correct?

I'm not sure about the second way you mentioned- looking at states.

Martin Argerami
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Shirly Geffen
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