4
$\begingroup$

The problem is as follows:

Suppose $f$ entire satisfying $$ |f(z)| \leq A + B |z|^{3/2} $$ for some fixed $A,B > 0$. Prove that $f$ is a linear polynomial.

I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how.

$\endgroup$
  • 2
    $\begingroup$ I think the description "polynomial of degree $3/2$" for this bound is rather unusual. $\endgroup$ – joriki Nov 2 '12 at 1:32
  • 2
    $\begingroup$ Let $f(z) = a_0 + a_1z + a_2z^2 + \cdots$ be the Taylor expansion of $f$ around the origin. To show $f$ is a linear polynomial, you must show $a_2 = 0$, $a_3 = 0$, etc. What do the Cauchy estimates say about the size of these coefficients? $\endgroup$ – froggie Nov 2 '12 at 1:36
  • 3
    $\begingroup$ possible duplicate of if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant? $\endgroup$ – Norbert Nov 2 '12 at 5:20
4
$\begingroup$

The first technique that comes to mind is the Cauchy integral formula. $$f^{(n)}(0) = \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz$$ Hence, we have that $$\left \vert f^{(n)}(0) \right \vert = \left \vert \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz \right \vert \leq \oint_{\Gamma} \dfrac{\left \vert f(z) \right \vert}{\left \vert z \right \vert^{n+1}} dz \leq \oint_{\Gamma} \dfrac{A + B \vert z \vert^{3/2}}{\left \vert z \right \vert^{n+1}} dz$$ Take $\Gamma$ to be a circle of a very large radius $R$. We then have that$$\left \vert f^{(n)}(0) \right \vert \leq \dfrac{A_1}{R^n} + \dfrac{B_1}{R^{n-3/2}}$$ We can let $R$ to be arbitrarily large and hence for $n \geq 2$, we have that $$f^{(n)}(0) = 0 \,\,\,\, \forall n \geq 2$$ Hence, $$f(z) = f(0) + f'(0) z$$

$\endgroup$
1
$\begingroup$

Instead of using Cauchy estimate as shown in the comments, alternatively you may apply the maximum modulus principle to the function $\frac{f(z)-f(0)-f'(0)z}{z^2}$ to show that it is constantly $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.