When is a Banach Algebra $C^\star$

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.

• Please link to a reference to the results about "enough Hermitian elements". This seems pretty neat! – Jon Bannon Jul 26 '12 at 19:42
• Cross-posted at MSE: math.stackexchange.com/questions/175550/… – Philip Brooker Jul 26 '12 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 '12 at 20:50
• 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 '12 at 16:44
• 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 '12 at 18:59

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.
• @Yemon, do you have by chance any reference for this theorem about homomorphisms into $B(X)$? – Slavoj Žižek Nov 15 '12 at 21:47