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I am tried to prove this:

Let a entire function $f$ such that $\lim_{ z \to \infty} \frac{f(z)}{z^N} = 0. $ Show that $f$ is a polynomial of degree at most $N$.

I find this results

Entire function bounded by a polynomial is a polynomial

So, I start my proof asuming that if $f$ is entire, it is equal to a power series centered at 0 with radius of convergence $\infty$, and its Taylor series is

$$f(z) = \sum_{n=0}^{\infty} \frac{f^{(N)}(0)}{N!}z^N$$

and I have the Cauchy formula says:

$$\frac{f^{n}(z_0)}{n!} = \frac{1}{2\pi i}\int_{|z-z_0|=r} \frac{f(z)}{(z-z_0)^{n+1}}dz$$

so if I consider $z_0 = 0$ and $$\frac{f^{n}(z_0)}{n!} {2\pi i} = \int_{|z|=r} \frac{f(z)}{z^{n+1}}dz$$

I tried to following the results of that, but I don't shure if I am in right way.

Can you help me to do it, please?

user111
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Ops
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1 Answers1

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Hint following your idea: write $f^{(n)}(0)$ as line integral $$ f^{(n)}(0) = \frac{n!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^{n+1}}\,dz $$ and apply the bound: $$\left|\int_{|z|=r}\frac{f(z)}{z^{n+1}}\,dz\right|\le \left(\sup_{|z|=r}\frac{f(z)}{z^N}\right)\frac1{r^{n+1-N}} \,2\pi r= \cdots$$ What happens when $n$ is large (how large?) enough?

Martín-Blas Pérez Pinilla
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  • The integral is equals to 0 and $f^{(n)}(0) = 0 $ but why I say that $f$ is a polynomial of grande at most N? It's because in the $n-th $ derivative is 0 and that's the limit? and where I use the hipotesis of my $lim_{z \to \infty} \frac{f(z)}{z^N} = 0$ – Ops Apr 07 '19 at 17:10
  • @Ops, the integral is $= 0$ for $n$ large enough. What is a Taylor series with coefficients $= 0$ for $n$ large enough? – Martín-Blas Pérez Pinilla Apr 07 '19 at 17:12