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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.

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  • $\begingroup$ Please link to a reference to the results about "enough Hermitian elements". This seems pretty neat! $\endgroup$
    – Jon Bannon
    Commented Jul 26, 2012 at 19:42
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    $\begingroup$ Cross-posted at MSE: math.stackexchange.com/questions/175550/… $\endgroup$
    – Philip Brooker
    Commented Jul 26, 2012 at 20:00
  • $\begingroup$ 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/… $\endgroup$
    – Daniel
    Commented Jul 26, 2012 at 20:50
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    $\begingroup$ 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. $\endgroup$
    – Daniel
    Commented Jul 27, 2012 at 16:44
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    $\begingroup$ Hi Daniel, your link didn't work. I had no idea there were results like this out there at all. Cool question! $\endgroup$
    – Jon Bannon
    Commented Jul 27, 2012 at 18:59

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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.

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    $\begingroup$ Thank for your insight. I haven't looked into the theorems you reference in your update, but I believe you that that road is not very productive. I was never able to get a strong enough bound on that universal constant for the the bound to be of much use to me. However, I have since been able to observe that the operator I'm working with is a perturbation of a resolvent positive operator and get some good results from there. $\endgroup$
    – Daniel
    Commented Oct 5, 2012 at 2:30
  • $\begingroup$ @Yemon, do you have by chance any reference for this theorem about homomorphisms into $B(X)$? $\endgroup$
    – Slavoj Žižek
    Commented Nov 15, 2012 at 21:47
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    $\begingroup$ @Slavoj I found this as Proposition 3.3.14 in the PhD thesis of Matthew Daws ( www1.maths.leeds.ac.uk/~mdaws/notes.html ) but apparently it may be folklore. $\endgroup$
    – Yemon Choi
    Commented Nov 16, 2012 at 0:35

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