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Suppose $A$ is a non-unital $C^*$ algebra,$B$ is another $C^*$ algebra.Suppose $\phi:A\rightarrow B$ is a non-zero $*$ homomorphism and $x_0$ is a normal elememt in $A$,by continuous functional calculus,we have $\phi(f(x_0))=f(\phi(x_0))$ for any $f\in C_0(\sigma_{A}(x_0))$ .My question is:can we choose a function $f\in C_0(\sigma_{A}(x_0))$ such that $|\phi(f(x_0))\|>1$?

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  • $\begingroup$ If you can find a function $f$ such that $\phi(f)\neq0$, can you see why this is true? $\endgroup$ – Aweygan Dec 3 '18 at 18:30
  • $\begingroup$ If $\phi(x_0)\neq0$,you mean that $f(z)=z,z\in C_0(\sigma_{A}(x_0))$ is suitable?But how to ensure that $|\phi(f(x_0))\|\geq1$? $\endgroup$ – math112358 Dec 4 '18 at 2:16
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By Urysohn's lemma, we can always choose a continuous function that vanishes at infinity with $f(z)=10000$ at a fixed point $z$. So, $\|\phi(f(x_0))\|=\|f(\phi(x_0))\|=\|f\|>9999$.

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  • $\begingroup$ Probably, the OP wants that $f(\phi(x_0)) \in B$ and so one should also require that $f(0) = 0$. This however is no problem, since the spectrum of $\phi(x_0)$ contains points other than zero. E.g. $f(z) = \lambda z$ for some large $\lambda > 0$. $\endgroup$ – user42761 Dec 4 '18 at 13:11

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