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How to prove Halmos’s Inequality?

If $A$ and $B$ are bounded linear operators on a Hilbert space such that $A$, or $B$, commutes with $AB-BA$ then $$\|I-(AB- BA)\|\ge 1.$$

I found it from http://www.staff.vu.edu.au/rgmia/monographs/bullen/Dict-Ineq-Supp-Comb.pdf at page 18.

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up vote 19 down vote accepted

There are several proofs and the inequality is given as problem 233 in Halmos's A Hilbert Space Problem Book. The inequality was originally proved in P.R. Halmos, Commutators of Operators, II, American Journal of Mathematics Vol. 76, No. 1 (Jan., 1954), pp. 191–198.

The following very nice proof is given by Philip Maher in A short commutator proof, International Journal of Science and Mathematics Education 27 (6) (1996), 934–935. If you have access, you find it online here.

Suppose $\|I-(AB-BA)\| \lt 1$. Then $(AB-BA)$ is invertible and as (without loss of generality) $A$ commutes with $(AB-BA)$ by hypothesis, we have that $A$ commutes with $(AB-BA)^{-1}$. Therefore $$I = (AB-BA)(AB-BA)^{-1} = A[B(AB-BA)^{-1}] - [B(AB-BA)^{-1}]A$$ and thus $I$ is exhibited as a commutator, and this is well-known to be impossible by a theorem of Wielandt and Wintner; see e.g. Qiaochu's question here for a proof of that fact. Therefore $\|I-(AB-BA)\| \geq 1$.

Note that this proof has nothing to do with bounded operators on a Hilbert space. It works just as well in any Banach algebra.

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+1. Neat and nice. – Did Aug 14 '11 at 16:52
    
comment as above. – Sunni Aug 14 '11 at 17:00

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