Discussion
Let \(S^{1} = \{z \in \mathbb{C} : |z| = 1\}\). For which one of the following functions \(f\) does there exist a sequence of polynomials in \(z\) that uniformly converges to \(f\) on \(S^{1}\)?
Answer: C
The function is \(f(z) = e^{z}\).
\(e^{z}\) is entire, and its Taylor polynomials \(\sum_{k \le n} z^{k}/k!\) converge to it uniformly on the compact set \(S^{1}\).
For a uniform limit \(f\) of polynomials on \(S^{1}\), \(\oint_{S^{1}} f(z)\,z^{m}\,dz = 0\) for every \(m \ge 0\), since each polynomial has this property. On \(S^{1}\), \(\overline{z} = 1/z\), and \(\oint \frac{1}{z}\,dz = 2\pi i \ne 0\). \(\text{Re}(z) = \frac{1}{2}(z + 1/z)\) and \(|z + 1|^{2} = 2 + z + 1/z\) also contain the \(1/z\) term, so their integrals against \(z^{0}\) are non-zero. Hence none of them is a uniform limit of polynomials on \(S^{1}\).
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