Entire functions that satisfy $f(x) = |x^k|$ for all $x$ real in $(-1, 1)$, for some odd integer $k$

Question

For an odd integer $K\geq1$, let $F$ be the set of all entire functions $f$ such that $f(x) =|x^k|$ for all $x\in(-1, 1)$.

What is the cardinality of $F$?

I think cardinality of $F$ is zero because we can not remove mode in the function that make it real valued. And entire function can not be real valued.

Am i right?

Answer 1

Suppose $f$ is a function satisfying the condition and consider the function $g(z)=f(z)-z^k$. If $z$ is in $(0,1)$, $g(z)$ vanishes. That implies that $g$, which is an entire function, vanises identically on the whole place, that is, that $f(z)=z^k$. In particular, $f(-1/2)$ is a negative number: this contradicts the hypothesis.

Answer 2

Given that $f(x)=|x^k|$ for all $x \in (-1,1)$.

So $f(x)=-x^k$ when $x \in (-1,0]$, and $f(x)=x^k$ when $ x \in [0,1)$.

let $F(z)=-z^k$ for all $z \in (-1,1)$; then $f(z)-F(z)=g(z)$ is zero on $\{-1/2n\}$ and $z=0$ belongs to $(-1,0]$.

So , By identity theorem we have $ F(z)=f(z)=-z^k$ for all $ z \in (-1,1)$.

Similarly, taking $f(z)=z^k$ we can have the sequence $\{1/2n\}$, which says $f(z)=z^k$ for all $z \in (-1,1)$, by identity theorem.

So, $f(z)=z^k=-z^k$ for all $z \in (-1,1)$, which is a contradiction. So the cardinality of the set is $0.$