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$A,B$ are bounded, finite matrices (i.e. matrices on a finite-dimensional Hilbert space). Define $C:= AB-BA$. According to Hausdorff, $$W := \{x^*Cx: \|x\| = 1\}$$ is a convex set.

The above result is mentioned as part of a bigger proof in On Commutators of Bounded Matrices by C. R. Putnam. Unfortunately, the reference mentioned (where Hausdorff made the above claim) is not in English - and so I did not have access to the proof.

My approach is very simple-minded, and I am stuck. Take $x_1,x_2$ with unit norm and consider $tx_1^*Cx_1 + (1-t)x_2^*Cx_2$ for $0\le t\le 1$. We must find some $x$ satisfying $\|x\| = 1$ and $$x^*Cx = tx_1^*Cx_1 + (1-t)x_2^*Cx_2 = t(x_1^*Cx_1 - x_2^*Cx_2) + x_2^*Cx_2$$ Any ideas?

Thank you!

stoic-santiago
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    This set is called the numerical range and the fact that it is convex is called Toeplitz-Hausdorff theorem. It has been discussed on MSE before: https://math.stackexchange.com/questions/171729/why-is-the-numerical-range-of-an-operator-convex, https://math.stackexchange.com/questions/2974633/proof-explanation-convexity-of-the-numerical-range-of-an-operator-toeplitz-hau?noredirect=1&lq=1. – MaoWao Jan 20 '22 at 16:53
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    What prevents you from looking at the German reference? You will probably understand more than you think about it, and if you don't you can use online translation tools. Use the advantages the internet has to offer! – J. De Ro Jan 20 '22 at 16:54
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    @QuantumSpace Mathematical writing has changed a lot since Hausdorff's times, so even as a native speaker I'd prefer to read a modern exposition. – MaoWao Jan 20 '22 at 16:57
  • @MaoWao I surely agree, but sometimes we have to use what we have :) – J. De Ro Jan 20 '22 at 17:11
  • Thanks a lot @MaoWao! – stoic-santiago Jan 20 '22 at 17:54
  • In this paper, Gustafson's proof (in english) is a one pager: https://www.ams.org/journals/proc/1970-025-01/S0002-9939-1970-0262849-9/S0002-9939-1970-0262849-9.pdf – Maksim Jan 20 '22 at 20:41

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