I'm a mathematics-majoring student trying to prove the validity of a certain formula that is on the literature but has not been fully proven. My proof involves analytic continuation as its key step- I, as a mathematics student, suspect that my proof is OK but I'm just asking here just to be sure.
Let $f(x)$ be a complex analytic function for $|x|\le 1+\delta$ and $g(x)$ be the function defined on $[-1,1]$ to be $=f(x)$. Suppose all the derivatives, as a real analytic function, of $g(x)$ at $x=0$ are known. Then, any function that equals $g(x)$ on $[-1,1]$, subject to the mere constraint that it be complex analytic at all for $|x|\le 1+\delta$, is unique and ($=f(x)$ for $|x|\le 1$).
Attempted proof: let $h(x)$ be such a function. Then, for $|x|\le 1$, $h(x)=\sum\limits_{k=0}^\infty \frac{h^{(k)}(0)}{k!} x^k=\sum\limits_{k=0}^\infty \frac{g^{(k)}(0)}{k!} x^k$, the latter equality being valid since $h(x)$ is complex analytic (by assumption) and real-analytic derivatives must equal complex-analytic derivatives whenever the latter exists at all.
Since the latter series converges for $|x|\le 1$, $h(x)$ is complex analytic for $|x|\le 1-\delta$. Now, $h(x)-f(x)$ is everywhere zero for $[-1+2 \delta,1-2 \delta]$; since both are complex analytic, $h(x)=f(x)$ everywhere else (i.e. for $|x|\le 1$), by Liouville's theorem. Q.E.D.
Is my intution correct, or is there an explicit counterexample the type of which I would need to avoid to complete my proof?
P.S. I'm doing this because all the derivatives of $f(x)$ that I'm using in my actual proof is much more simple to work with when restricted to the real line (w.l.o.g; one can cut it to $[-1,1]$).