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I know that if there are enough Hermitian elements in a Banach algebra, then the Banach algebra is stellar. In particular, I'm interested in the two spaces $B(L^1(S^1,\Sigma,\mu))$ the space of bounded linear operators on Lebesgue integrable functions of the circle and $B(ba(\Sigma))$ the space of bounded linear operators on finite, finitely-additive Borel measures. I know about the results that having enough Hermitian elements is sufficient, but I'm not quite sure how to apply them.

The issue comes up because I am trying to bound the inverse of a Hermitian element in terms of its spectral radius. From my reading, we have an equality for $C^\star$ algebras and an inequality for Banach algebras.

edited Aug 26 at 1:48

Yemon Choi
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asked Jul 26 at 18:28

Daniel
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Please link to a reference to the results about "enough Hermitian elements". This seems pretty neat! – Jon Bannon Jul 26 at 19:42

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Cross-posted at MSE: math.stackexchange.com/questions/175550/… – Philip Brooker Jul 26 at 20:00

I stole that phrase from the paper "The Spectral Theorem in Banach Algebras" S. Plafker (link below). From reading the references therein, if $H\subset B$ is the set of Hermitian elements in your Banach algebra, B, H+iH=B implies B is isometrically isomorphic to a $C^\star$ algebra. journals.cambridge.org/… – Daniel Jul 26 at 20:50

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What I'm looking for is a bound of $\|x^{-1}\|$ in terms of the spectral radius. I know $\rho(x^{-1})$ and that $x$ is Hermitian. If $B(L^1)$ is $C^\star$, I have $\|x^{-1}\|=\rho(x^{-1})$. If it is not, I have $\|x^{-1}\|\leq\frac{\pi}{2}\rho(x^{-1})$. My understanding for the involution is that if $H+iH=B$, we can write any element $x\in B$ as $x=a+ib$ where $a,\,b\in H$. In that case, $x^\star=a-ib$. I know my $x$ is Hermitian since $\|\exp[i\alpha x]\|=1$ for real $\alpha$, so I don't need to know the structure of the involution, just that it exists. – Daniel Jul 27 at 16:44

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Hi Daniel, your link didn't work. I had no idea there were results like this out there at all. Cool question! – Jon Bannon Jul 27 at 18:59

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Yemon Choi · Accepted Answer · 2012-08-26 09:09:07Z UTC

This is not an answer to the question posed in your first paragraph (which I think is asking for more than you need, and more importantly, more than you can really hope for). However, for the specific purpose outlined in your second paragraph, the following paper might be helpful:

H. König, A functional calculus for Hermitian elements of complex Banach algebras. Arch. Math. (Basel) 28 (1977), no. 4, 422–430.

IIRC, one has a C^1-functional calculus for Hermitian elements in Banach algebras (this is proved by taking the Fourier transform of your C^1 function after introducing a smooth cutoff outside the support of the spectrum of your Hermitian element). So if the spectrum of your $x$ is contained in $[a,b]$ for $0 \lt a$ then $\Vert x^{-1}\Vert$ should hopefully be bounded above by some universal constant times $a^{-2} = \rho(x^{-1})^2$. CAVEAT: I have not checked this in detail!

Update: I've just remembered that there are theorems to the effect that if $E$ is a Banach space, $A(E)$ the algebra of approximable operators, and $X$ is a reflexive Banach space, then an injective algebra homomorphism $A(E)\to B(X)$ must arise from some embedding of $E$ into $X$ as a closed, complemented subspace. In particular, if there is an injective HM $B(E) \to B(H)$ for some Hilbert space then $E=H$. So the two Banach algebras mentioned at the start of your question can't possibly be $C^\ast$-algebras.

When is a Banach Algebra $C^\star$

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