Liouville-type problem (Complex analysis)

[murmillo]

Homework Statement

If f(z) is an entire function such that f(z)/z is bounded for |z|>R, then f''(z_0) = 0 for all z_0.

Homework Equations

Liouville's theorem
Cauchy estimates: Suppose f is analytic for |z-z_0| ≤ ρ. If |f(z)|≤ M for |z-z_0| = ρ then the mth derivative of f at z_0 is bounded by a constant given in the book.

The Attempt at a Solution

I'm supposed to adapt the proof of Liouville's theorem I learned, which is by using the Cauchy estimates.
I don't see where to get the second derivative of f. I'm pretty sure that z/f(z) is bounded for |z|<1/R, and I tried using the Cauchy estimates but couldn't get anything.

 

[mighty2000]

Thank you for your question. I can see that you are trying to use Cauchy estimates to prove that f''(z_0) = 0 for all z_0. However, in order to use Cauchy estimates, we need to have a bound on the second derivative of f at z_0. In this case, we do not have that information since we are only given that f(z)/z is bounded for |z|>R. Therefore, we need to approach this problem in a different way.

One way to prove this statement is by using the Cauchy-Riemann equations. Since f(z) is an entire function, it can be expressed as f(z) = u(x,y) + iv(x,y), where u and v are real-valued functions. By the Cauchy-Riemann equations, we have u_x = v_y and u_y = -v_x. Now, using the definition of the second derivative, we can write f''(z) = u_{xx} + i v_{xx}. Since we are given that f(z)/z is bounded for |z|>R, we can write |f(z)/z| ≤ M for some constant M. This implies that |f(z)| ≤ M|z| for |z|>R. Now, if we take the limit as z approaches z_0, we can see that u and v are bounded at z_0. This means that u_{xx} and v_{xx} are also bounded at z_0. Therefore, f''(z_0) = u_{xx}(z_0) + i v_{xx}(z_0) = 0 for all z_0.

I hope this helps. Let me know if you have any further questions. Good luck with your studies!