Question

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 \( S^1 \)?

• A. \( f(z) = \overline{z} \)
• B. \( f(z) = {Re}(z) \)
• C. \( f(z) = e^z \)
• D. \( f(z) = |z + 1|^2 \)

Hint:

For uniform approximation on compact sets, verify that the function is continuous and invariant on the set.

Answer 1

The Correct Option is C

1. Understanding Uniform Convergence of Polynomials on \( S^1 \): The Stone-Weierstrass theorem ensures that any continuous function on a compact subset of \( \mathbb{C} \) (such as \( S^1 \)) can be uniformly approximated by polynomials in \( z \) if and only if the function is holomorphic.
2. Analyzing the Options: - (1) \( f(z) = \overline{z} \): This is not holomorphic since \( \overline{z} \) involves the complex conjugate of \( z \). Hence, no sequence of polynomials can uniformly approximate \( f(z) \) on \( S^1 \).
- (2) \( f(z) = {Re}(z) \): This is not holomorphic since it depends on both \( z \) and \( \overline{z} \) (as \( {Re}(z) = \frac{z + \overline{z}}{2} \)).
- (3) \( f(z) = e^z \): This is holomorphic everywhere on \( \mathbb{C} \), so a sequence of polynomials can uniformly approximate \( f(z) \) on \( S^1 \).
- (4) \( f(z) = |z + 1|^2 \): This depends on both \( z \) and \( \overline{z} \) (as \( |z + 1|^2 = (z + 1)(\overline{z} + 1) \)), and thus it is not holomorphic.
3. Conclusion: The only function among the options that can be uniformly approximated by polynomials in \( z \) on \( S^1 \) is \( f(z) = e^z \).
Final Answer: \( f(z) = e^z \).