# Simple inequality in C*-algebras

Sorry the title is a bit vague. Let A be a C*-algebra, and let x and y be positive elements in A. Is it true that $$\|x-y\|^2 \leq \|x^2-y^2\|?$$ Well, yes. But the proof I have is a bit of a hack, so I wonder if anyone has a "nice" proof, or a reference?

Aside: if $A=C_0(X)$ then this reduces to the inequality $(a-b)^2 \leq |a^2-b^2|$ for non-negative real numbers a and b.

Update: Jonas points me to http://www.springerlink.com/content/j4756m418220644r/ where Kittaneh has a proof pretty similar to what I had in mind (unpack the proof of Theorem 1). I guess I was interested in whether this sort of thing was standard (if I looked in the right textbook) or if it was a bit of a curiosity. I think the latter seems more likely...

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I don't know how "nice" it is compared to your hack, but Corollary 3 of springerlink.com/content/v74h482277288723 implies this as a special case. –  Jonas Meyer May 12 '10 at 17:02
Or Corollary 2 of springerlink.com/content/j4756m418220644r with $p=\infty$. This is an earlier reference, cited by the other I gave. (The case you're interested in actually follows from Theorem 1, so the more general Theorem 2 isn't needed.) –  Jonas Meyer May 12 '10 at 19:51
Thanks. Actually, Kittaneh's proof is pretty similar to the proof I had in mind... –  Matthew Daws May 12 '10 at 20:45
You're welcome. That's why I only commented--I didn't know if this would be an appropriate "answer". –  Jonas Meyer May 12 '10 at 20:53
It just occurred to me that this gives another way to see that the distance between 2 projections is at most 1. I find that amusing, even though there are much simpler proofs. More generally, it is straightforward to show that $\|x-y\|\leq\max(\|x\|,\|y\|)$ if $x$ and $y$ are positive. –  Jonas Meyer Jun 6 '10 at 1:37

Theorem 1.5 of this 1987 paper by J. Phillips says that if $f:[0,\infty)\to [0,\infty)$ is a continuous operator monotone function and $a$ and $b$ are positive operators on a Hilbert space, then $\|f(a)-f(b)\|\leq f(\|a-b\|)-f(0)$. I think that the proof is nice. Corollary 1.6 says that $\|a^{1/n}-b^{1/n}\|\leq\|a-b\|^{1/n}$, $n\geq1.$ Of course your inequality follows from taking $a=x^2$, $b=y^2$, and $n=2$.

Apparently Kittaneh and Kosaki have a similar approach in "Inequalities for the Schatten p-norm. V." Publ. Res. Inst. Math. Sci. 23 (1987), no. 2, 433--443 (MR link). I haven't read any of this article.

Perhaps I should add the following for a more general audience. A continuous function $f:[0,\infty)\to [0,\infty)$ is operator monotone if whenever $x$ and $y$ are positive operators such that $y-x$ is positive, it follows that $f(y)-f(x)$ is positive. The functions $t\mapsto t^\alpha$ are operator monotone for $0<\alpha\leq1$ (but not for $\alpha>1$).

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Ah, I like this! (So accepted). This operator monotone stuff is nice: I used to know about it some years ago, but forgot. I guess, with hindsight, it's a good place to look. –  Matthew Daws May 17 '10 at 8:11
By the way, that's some excellent detective work Jonas, as MathSciNet doesn't record this ever being published...? –  Matthew Daws May 17 '10 at 8:15
I found it with Google's help, and what changed from the last time I searched was that I thought of searching for "generalized Powers Stormer inequality" and the like. Yeah, I don't know how that worked in terms of publication: I guess it appeared in a U of Victoria "technical report", but I don't know if such things are refereed. It's not like the author had a shortage of other publications, and I was speculating that the independent overlapping results of Kittaneh and Kosaki may have been a factor in the decision. –  Jonas Meyer May 17 '10 at 8:24

Okay, so actually, decoding Phillip's work gives what I think is a very nice proof.

Let x and y be positive. Let $\epsilon=\|x-y\|$ so as $x-y$ is self-adjoint, it follows that $x \leq y+\epsilon 1$. What seems to be a very standard inequality is that then $x^{1/2} \leq (y+\epsilon 1)^{1/2}$. We then claim that $(y+\epsilon 1)^{1/2} \leq y^{1/2} + \epsilon^{1/2} 1$. This follows by working in the commutative C*-algebra generated by y and 1, and using that $(s+t)^{1/2} \leq s^{1/2} + t^{1/2}$ for positive real numbers s and t. So $$x^{1/2} - y^{1/2} \leq \epsilon^{1/2} 1$$and by symmetry, also $y^{1/2}-x^{1/2}\leq\epsilon^{1/2}1$. Thus $\|x^{1/2}-y^{1/2}\|\leq\epsilon^{1/2}=\|x-y\|^{1/2}$ as required.

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I agree that operator monotonicity of the square root is standard. A good reference is Proposition 4.2.8 on page 250 of Kadison-Ringrose volume 1: books.google.com/… –  Jonas Meyer May 17 '10 at 20:27