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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.

enter image description here

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    $\begingroup$ 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/… $\endgroup$ – user99914 Oct 8 '15 at 23:33
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    $\begingroup$ The best you can do is that $\Phi$ is infinitely differentiable. $\endgroup$ – Alex S Oct 8 '15 at 23:45
  • $\begingroup$ It should be "at most" instead of "almost" $\endgroup$ – user99914 Oct 8 '15 at 23:53
  • $\begingroup$ Oh, that's true, I completely overlooked that fact, I'll change the question a bit then. $\endgroup$ – YoTengoUnLCD Oct 8 '15 at 23:57
  • $\begingroup$ I'm pretty sure $\psi(x) = x^3$ or some other odd power. $\endgroup$ – Alfred Yerger Oct 9 '15 at 0:04
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Hint: look up "bump function".

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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 $0\in D$.

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