The disk algebra is the set of all functions on the unit disc $D$ which are analytic on the interior of the disc and continuous on the boundary. Addition and multiplication are defined obviously. Why is the disk algebra generated by 1 and the map defined above?
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1The algebra generated by $1$ and the identity map contains only entire functioins, so it is not equal to the disk algebra. If you to define a topology on the disk algebra you can ask if the closure of this algbebra coincides with the disk algebra. – Kavi Rama Murthy Feb 03 '20 at 07:25
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Let $f\in A$ so that $f\in C(D) \cap H(\operatorname{int}(D))$.
Since $f$ is continuous on the compact set $D$, it is also uniformly continuous here. Let $\varepsilon >0$. There is an $0<r<1$ such that $|f(z)-f(rz)| < \varepsilon$ for all $z\in D$.
Since $f(r\cdot)$ is holomorphic in a neighbourhood of $D$, we can find a sequence of polynomials $p_N$ converging uniformly to $f(r\cdot)$ on $D$.
Using this uniform convergence we see that for $N$ sufficiently large, we have for each $z\in D$, $$ |f(z)-p_N(z)| \leq |f(z)-f(rz)| + |f(rz) - p_N(z)| \leq 2\varepsilon. $$ Hence the algebra generated by the functions $z\mapsto 1, z\mapsto z$ is dense in $A$.
