Let $D = \{ z \in \mathbb{C} : |z| < 1 \}$ be the open unit disk and $\mathbb{T} = \{ z \in \mathbb{C} : |z| = 1 \}$ its boundary. We will naturally write $\bar{D}$ for its closure $\{ z \in \mathbb{C} : |z| ≤ 1 \}$. Consider the subalgebra $A(\bar{D})$ of $C(\bar{D})$ of functions $g \in C(\bar{D})$ such that $g|_D$ is holomorphic. By Morera’s Theorem, $A(\bar{D})$ is a closed subalgebra of $C(\bar{D})$. Let $B$ be the image in $C(\mathbb{T})$ of $A(\bar{D})$ via the restriction map. Since the map $g \rightarrow g|_{\mathbb{T}}$ is isometric by the Maximum modulus principle, $B$ is complete and therefore closed in $C(\mathbb{T})$. Hence $B$ is a closed subalgebra of $C(\mathbb{T})$ and is a Banach algebra in its own right usually called the disk algebra. It consists of those functions in $C(\mathbb{T})$ that have holomorphic extensions to D. As described above, $A(\bar{D})$ and $B$ are isometrically isomorphic.
Now set $P(\mathbb{T})$ as the closure of $\mathbb{C}[z]$ in $C(\mathbb{T})$, Prove that $P(\mathbb{T})=B.$
Hint: to prove that $B$ contains $P(\mathbb{T})$, use the maximum modulus principle. More explicitly, suppose that a sequence $(p_n)$ of polynomials uniformly converges on $\mathbb{T}$; what can we say about the convergence of the same sequence on $D$? To prove that $B$ is contained in $P(\mathbb{T})$, it suffices to show that the polynomials are dense in $A(\bar{D})$.