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Let $f : A_1 \to A_2$ be an injective homomorphism of unital Banach algebras. It's a standard fact that if $f$ is has closed range, i.e. $A_1$ is embedded as a closed subalgebra of $A_2$, then for every $x \in A_1$ we have $\rho(f(x)) = \rho(x)$, where $\rho$ denotes the spectral radius in the corresponding algebra.

Now, is the same true if $f$ does not have closed range - e.g. if its image is dense? I suspect that the answer is negative, but I failed to construct a counterexample...

A less precise but more intuitive formulation of this question would be: is the spectral radius - and, more generally, the polynomially convex hull of spectrum - an intrinsic property of operator, independent of the ambient algebra?

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  • $\begingroup$ Is it a homomorphism of unital algebras? Must it map 1 of $A_1$ to 1 of $A_2$? $\endgroup$ Commented Aug 13, 2014 at 20:48
  • $\begingroup$ If the element 1 of $A_1$ does exist, then it is not very important what $f({\mathbf 1})$ is, but do you permit a non-unital $A_1$? $\endgroup$ Commented Aug 13, 2014 at 20:54
  • $\begingroup$ @IncnisMrsi: Assume that they are unital, and the homomorphism preserves the unit. Does that change the answer? $\endgroup$ Commented Aug 14, 2014 at 1:42
  • $\begingroup$ Then I can propose a reformulation: what a homomorphism makes with the function $( I - \frac{x}{\lambda} )^{-1}$ that is holomorphic at $\lambda = \infty$. Its radius of convergence, as a function of $\lambda^{-1}$, equals to the inverse of the spectral radius. Namely: can it have a singularity that is erased by the homomorphism? $\endgroup$ Commented Aug 14, 2014 at 5:47
  • $\begingroup$ On another thought, the unit is actually not used anywhere since we have a power series $\mathbf{1} + \frac{x}{\lambda} + \frac{x^2}{\lambda^2} + \frac{x^3}{\lambda^3} +\cdots $ whose behavior is not affected by the constant term. $\endgroup$ Commented Aug 14, 2014 at 5:52

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Please, see Banach subalgebra with a different spectrum Can this help you? In other words, consider polynomials on the interval $I=[1/2,3/4]$ with the uniform norm and with $\ell^1$-norm. The first norm leads to the $C^*$-algebra of continuous functions on $I$, the second norm leads to the Banach algebra of analytic functions inside the unite ball and restricted to the interval $I$. Hence $\rho_C(x)=3/4$ but $\rho_{\ell^1}(x)=1$ for the monomial $x$.

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