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.



When is a Banach Algebra $C^\star$
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.](http://dx.doi.org/10.1007/BF01223945)
> 
> 
> 


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.


