Question

Show that if f is differentiable and bounded on the entire complex plane, then it must be constant.

Answer 1

If f is differentiable and bounded on the entire complex plane, then it must be constant.

Step-by-step explanation:

The given statement is a consequence of Liouville's Theorem, which states that every bounded entire function is constant. Let's break down the proof.

Firstly, since f is differentiable on the entire complex plane, it is analytic everywhere. By Cauchy-Riemann equations, we know that the partial derivatives of u and v with respect to x and y are related, where f(z) = u(x, y) + iv(x, y). Suppose f is bounded, meaning there exists a constant M such that
`\(|f(z)| \leq M\)` for all z.

Now, consider the function
`\(g(z) = e^(-if(z))\)`. By construction,
`\(|g(z)| = e^(-u(x, y))\)`, and
`\(|g(z)| \leq e^M\)`. This implies g(z) is bounded. Applying Liouville's Theorem to g(z), we conclude that g(z) is constant. Now, let's express g(z) in terms of f(z):

`\[e^(-if(z)) = \text{constant} \implies e^(if(z)) = \text{constant}.\]`

This means f(z) is also a constant, as the exponential function with a constant argument is never zero. Therefore, if f is differentiable and bounded on the entire complex plane, it must be constant.