Does there necessarily exist such a holomorphic function? This is an old qual problem I'm working on: Let $f:[0,1]\rightarrow \mathbb{R}$ be a $C^{\infty}$ function. Does there necessarily exist a holomorphic function $g: \mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}$ such that $f(x)-g(x)$ vanishes to infinite order at $0$ as $x$ tends to $0$ in $[0,1]$?
To be honest, maybe I couldn't make much progress on this. I tried to write Laurent series for $0$, and using the given condition, tried to conclude that if such a $g$ exists, it must be analytically extended to $0$ too. Then, I tried to find such an $f$ that cannot be extended analytically to the whole plane, but failed. But, my reasoning here might be very false, I'm not sure. I would appreciate any kind of help. Thanks!
 A: The answer is yes, and we can relax the condition on $f.$ We need only assume $f:(0,1]\to \mathbb {C}$ is continuous. We then consider  $\tilde {f}(t) = f(1/t),$ which is continuous on $[1,\infty).$ Let's extend $\tilde {f}$ to a continuous function on all of $\mathbb {R}.$
Now I pull out a big gun, Carleman's theorem (one of them anyway): Suppose $\varphi :\mathbb {R}\to \mathbb {C}$ is continuous, and $\epsilon:\mathbb {R}\to (0,\infty)$ is continuous. Then there is an entire function $g$ such that
$$|g(t)-\varphi (t)|<\epsilon(t), t \in \mathbb {R}.$$
That's a beautiful result. Anyway, let $\epsilon(t) = e^{-t^2}.$ Then Carleman says there is an entire $g$ with
$$ |\tilde {f}(t)-g(t)|< e^{-t^2}, t \in \mathbb {R}.$$
Flip back to $(0,1]$ and we have
$$|f(t)-g(1/t)|< e^{-1/t^2}, t \in (0,1],$$
which is more than enough in our problem (note $g(1/z)$ is holomorphic on $\mathbb {C}\setminus \{0\}).$
Perhaps the $f\in C^\infty([0,1])$ assumption is a red herring, or perhaps there is a more elementary approach to the problem at hand that uses smoothness.
