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Here is my argument, please let me know if it works or not.

By Stone-Weierstrass Theorem (Complex Version), functions in $C_0((0, 1))$ can be uniformly approximated by polynomials in z and $\bar{z}$ which vanishes at 0 and 1. But on (0, 1), $z=\bar{z}$, so it suffices to consider only polynomials in $z$.

Larger Question: I am trying to define $f(a)$ for $a \in A$, $a$ is an element in Banach algebra with spectrum $\sigma(a)\subset [0,1]$, and $f \in C_0(0,1)$. I know that $f(a)$ can be defined using Holomorphic Functional Calculus1 for $f$ holomoprhic on a neighborhood of [0,1], and if $f_n \to f$ on compact subsets of an open neighborhood of $\sigma(a)$, then $f_n(a)$ converges. If the above were true, it seems that I can define $f(a)$ for general $f \in C_0((0,1))$ by using approximation with $f_n(a)$ with $f_n$ polynomials vanishing at 0 and 1.

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There's good news and bad news. The good news is that you can approximate every function in $C_0((0,1))$ uniformly by polynomials vanishing in $0$ and $1$ on $[0,1]$ per the Weierstraß approximation theorem. The bad news is that if such an approximating sequence of polynomials converges locally uniformly in a neighbourhood of the spectrum, then its limit function is holomorphic in that neighbourhood. So you may not gain anything, unfortunately. If the function you want to approximate is the restriction of a holomorphic function to the spectrum, you can directly use the holomorphic function calculus, and if it isn't, the approximation cannot be uniform in a neighbourhood of the spectrum, and it is in general not possible to deduce the convergence of $f_n(a)$.

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  • $\begingroup$ Thanks Daniel! Just one more point to clarify - for the good news part, do you mean I can approximate uniformly with just polynomials in z which vanishes at the endpoints (i.e. holomorphic functions is dense in C_0(0, 1)?) I know this would not help with my "larger question", but I am just curious. Thanks again!! $\endgroup$ Commented Nov 5, 2013 at 3:22
  • $\begingroup$ Yes. You (can) approximate with polynomials in $t \in\mathbb{R}$ on the interval without even considering that the domain is a subset of $\mathbb{C}$. Then you extend the approximating polynomials by replacing $t$ with $z$ (or, some $t$s with $z$, others with $\overline{z}$, as you noted, the value on $\mathbb{R}$ is the same). $\endgroup$ Commented Nov 5, 2013 at 9:30

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