If $\sum_{n=0}^{N}a_nx^n-C_N|x|^N\le f(x)\le \sum_{n=0}^{N}a_nx^n+C_N|x|^N$ If $\sup_{n\in \mathbb{N}} C_n<\infty$ then $f$ is analytic

Question

Let $f$ be a function s.t. there exist two sequences $a_n$ and $C_n$ s.t. $$\sum_{n=0}^N a_nx^n-C_N|x|^N\le f(x)\le \sum_{n=0}^N a_nx^n+C_N|x|^N$$ If $\sup_{n\in \mathbb{N}} C_n<\infty$, then $f$ is analytic on $[-r,r] \ \forall |r|<1$. Is it true or false?

What I think is that this statement is false. If $\sup C_n$ is finite it doesn't mean that $\lim_{N\to\infty}C_N|x|^N$ exists (we can consider $C_n=-n!$) and we have no condition on a sequence $a_n$. But, if $a_n$ was bounded and $C_n$ as well, I guess that the statement would become true. Is it correct?

P.S For instance I don't have a counterexample (if the statement is false), I just would like to check if what I think is correct. Thank you in advance

Answer 1

The inequality can be rewritten as $|f(x) - \sum_{n=0}^{N}a_nx^n| \leq C_N|x|^N$

Let $C:=\sup_{n \in N} C_n$. Then $\lim_{n\to\infty}C_n|x|^n \leq \lim_{n\to\infty}C|x|^n = 0$. (note $|x|<1$) As -$C_N \leq C_N$ $C_N$ is also non-negative. Therefore, $\lim_{n\to\infty}C_n|x|^n=0$.

Fix $\epsilon>0$. $\exists n_0 \in N$ s.t. $\forall N>n_0$, $|f(x) - \sum_{n=0}^{N}a_n x^n| \leq \epsilon$

Therefore, $\sum_{n=0}^{\infty}a_n x^n=f(x)$. Therefore, $f$ is real analytic.