Prove if $k^\text{th}$ derivative of an entire function $f$ is polynomial, then $f$ itself is polynomial. Where's my mistake?

Question

The exact wording of the question is as follows:

Let $f$ be an entire function. Suppose there exists a positive integer $k$ such that $k^\text{th}$ derivative $f^{(k)}$ is a polynomial. Prove that $f$ is a polynomial.

Following the Taylor series representation of an arbitrary function $f(z) = \sum_{k \geq 0} c_k z^k$ and taking its derivative format as written in the textbook as an assumption, I provided the following as the solution:

Let $f^{(k)}(z) = \sum_{n\geq k} n(n-1)\ldots (n-k+1) c_k z^{n-k}$ and let $d_n = n(n-1)\ldots (n-k+1) c_k $. If $f^{(k)}$ is polynomial then $d_{k_0}$ is constant for each $k_0 \geq k$. Then $\underbrace{\int \ldots \int}_k f^{(k)} dz \ldots dz = \underbrace{\int \ldots \int}_k \sum_{n\geq k} d_n z^{n-k} dz \ldots dz = \sum_{n\geq k} \underbrace{\int \ldots \int}_k d_n z^{n-k} dz \ldots dz$

Since sum of polynomials is polynomial and $f = \underbrace{\int \ldots \int}_k f^{(k)} dz \ldots dz$, then $f$ is polynomial.

I was told the proof is wrong because I needed to show the Taylor series is finite; in other words, say $d_{k_0} = 0$ as opposed to just being a constant.

But now that I'm thinking about it:

1. What I said about $d_{k_0}$ is not wrong but it is redundant, since the coefficients of $z$ are constant by definition of the Taylor series.
2. I don't really see why it's necessary to write that the Taylor series is finite - even though it necessarily is. I'm showing it takes finitely many antiderivatives (i.e. $k$ many) of $f^{(k)}$ (which is assumed to be polynomial). Since I'm just adding and integrating terms of the Taylor series of some polynomial function (i.e. $f^{(k)}$), whatever property needed by a Taylor series to represent a polynomial function (i.e. $f$) must automatically be satisfied.

I'd very much appreciate it if anyone could specify my mistake.

Answer 1

If $f^{(k)}$ is polynomial then $d_{k_0}$ is constant for each $k_0 > k$.

At this point you already started on the wrong path. Note that the series $$\sum_{k=2}^\infty 2 x^k$$ for example satisfies your condition, but is not a polynomial!

Since you try to deduce from here that your function is a polynomial, your argument must be wrong, and here is why:

\smallskip

Since sum of polynomials is polynomial and $f=f = \underbrace{\int \ldots \int}_k f^{(k)} dz \ldots dz$, then f is polynomial.

Here you make the second mistake. It is true that a finite sum of polynomials is a polynomial, BUT your sum is

$$\sum_{n\geq k} \underbrace{\int \ldots \int}_k d_n z^{n-k} dz \ldots dz$$

is an infinite sum, and an infinite sum of polynomials is not always a polynomial.

Answer 2

Suppose $f(z) = \sum_{k=0}^\infty a_n z^n$, then $a_n = {1 \over n!} f^{(n)}(0)$.

If $p$ is a polynomial, then $p^{(m)} = 0$ for some $m$. Since $f^{(k)}$ is a polynomial, there is some $m$ such that $f^{(k+m)} = 0$. Hence $a_n =0$ for $n \ge k+m$. Then $f(z) = \sum_{k=0}^{k+m-1} a_n z^n$, which is a polynomial.

Another approach:

If $p$ is a polynomial, then $q(z) = c+\int_0^z p(w)dw$ is a polynomial.

Suppose $f^{(k)}$ is a polynomial. Then the fundamental theorem of calculus gives us $f^{(k-1)}(z) = f^{(k-1)}(0) + \int_0^z f^{(k)}(w) dw$, and so $f^{(k-1)}$ is a polynomial. Continue by induction to get that $f$ si a polynomial.

Answer 3

I would prove it by induction on $k$.

Base case - $k=1$.

If $f'(x)$ is a polynomial then $f'(x) =\sum_{j=0}^d a_jx^j $ so

$\begin{array}\\ f(x) &=f(0)+\int_0^x f'(t) dt\\ &=f(0)+\int_0^x (\sum_{j=0}^d a_jt^j) dt\\ &=f(0)+\sum_{j=0}^d a_j\int_0^x t^j dt\\ &=f(0)+\sum_{j=0}^d a_j\dfrac{x^{j+1}}{j+1}\\ \end{array} $

is a polynomial.

If true for $k$ then if $f^{(k+1)}(x)$ is a polynomial then $f'(x)$ is a polynomial by the induction hypothesis for $k$ since $(f'(x))^{(k)} =f^{(k+1)}(x) $. Therefore $f(x)$ is a polynomial by the base case with $k=1$.