# Let $\psi$ be an analytic function. How do we construct a bounded smooth function $\Phi$ that equals $\psi$ over some domain?

Let $\psi: \mathbb R \to \Bbb R$ be a given analytic function.

How can I construct a smooth (that is, infinitely differentiable) function $\Phi:\Bbb R\to \Bbb R$ such that $\psi|_D=\Phi|_D$ for some interval $D$, and $\Phi$ is bounded above and below?

E: I've been reminded in the comments that such an analytic $\Phi$ isn't possible, I'm changing my request to just infinitely differentiable.

Example:

Red is $\psi(x)=x^3$, blue is $y=1.5$, green is $y=-1.5$. I want to construct a $\Phi(x)$ that looks exactly like $\psi$ for $-1\leq x\leq 1$ and is bounded like the black part of the diagram.

• You can't, the new function can almost be smooth, if two analytic functions are the same on an open set, then it has to be the same everywhere. math.stackexchange.com/questions/739476/… – user99914 Oct 8 '15 at 23:33
• The best you can do is that $\Phi$ is infinitely differentiable. – Alex S Oct 8 '15 at 23:45
• It should be "at most" instead of "almost" – user99914 Oct 8 '15 at 23:53
• Oh, that's true, I completely overlooked that fact, I'll change the question a bit then. – YoTengoUnLCD Oct 8 '15 at 23:57
• I'm pretty sure $\psi(x) = x^3$ or some other odd power. – Alfred Yerger Oct 9 '15 at 0:04

It still may be impossible.For example if $$f(0)=0,$$ $$f(x)=e^{(-1/x^2)} , x\neq 0$$ then $f$ cannot be extended to an analytic function on any open domain $D\subset C$ for which $0\in D$.