Are polynomials dense in $H^\infty$

Question

Let $H^\infty(D)=\{f:D\longrightarrow \mathbb{C}: f \;\text{is bounded and analytic on}\; {D}\}$ where $D=\{z\in\mathbb{C}: |z|<1\}$. Are the polynomials dense in $H^\infty(D)$? I know that they are dense in other Hardy spaces $H^p(D)$ for $0<p<\infty$.

Answer 1

No, polynomials are not dense in $H^\infty(D)$. The uniform limit of a sequence of polynomials (or, more generally, uniformly continuous functions) is always uniformly continuous and thus extends continously to $\bar D$. The set of bounded analytic functions on $D$ with continuous extension to $\bar D$ is the disk algebra $A(D)$, and polynomials are indeed dense in $A(D)$ by Mergelyan's theorem.

However, $A(D)$ is a proper subspace of $H^\infty(D)$ as Jonas Meyer explains in this answer. For a more explicit example, take $f(z)=\exp(\frac{z+1}{z-1})$. This function is bounded and analytic on $D$, but for every $w\in D$ one can find a sequence $(z_n)$ in $D$ converging to $1$ such that $f(z_n)\to w$.