Let f be an entire function. Suppose there exist constants A, B such that $| f(z)| ≤ A + B|z|^\frac{1}{7}$ for every z ∈ C. Prove that f is constant.

I am unsure whether I have proved this correctly so any feedback would be greatly appreciated. My proof is as follows:

Here $| f(z)| ≤ A + B|z|^\frac{1}{7}$ for every z ∈ C so $A + B|z|^\frac{1}{7}$ is a bound for f. Liouville's theorem states that if f is entire and bounded then f is a constant, which we will prove now.

Let $a ∈ C$ and let $r>0$. As $| f(z)| ≤ A + B|z|^\frac{1}{7}$ we know that $| f'(a)| ≤ \frac{A + B|z|^\frac{1}{7}}{r}$ for every $r>0$. Hence $f'(a)=0$ for every $a ∈ C$.

Now we will show that any function with zero derivative is constant to complete the proof.

Let $z ∈ C$, then by the fundamental theorem of calculus: $f(z) - f(0)=\int_0^z f'(w)dz=0$ so $f(z)=f(0)$ is constant.

  • $\begingroup$ Where is your bound on $|f'(a)|$ from? Perhaps more details would be helpful there. $\endgroup$ – Michael Burr Nov 9 '16 at 15:42
  • $\begingroup$ This is from the Cauchy estimates theorem which states for $r=|z-a|$ and $M$, an upper bound for $|f(z)|$, we have $|f'(a)|≤\frac{1!M}{r^1}$ for the first derivative of $f(a)$. If I add this into the proof will it be satisfactory? $\endgroup$ – DavinaGoodman Nov 9 '16 at 15:52
  • $\begingroup$ Then, should $z$ be $r$ in the numerator of the bound? $\endgroup$ – Michael Burr Nov 9 '16 at 15:56
  • $\begingroup$ This looks like it will work, but your bound on $|f'(a)|$ should be explained in more detail. $\endgroup$ – Michael Burr Nov 9 '16 at 15:59

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