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I have $f:\mathbb{R} \to \mathbb{R}$ and I know $f(n)\ \forall n \in \mathbb{Z}$.

What should I have to assume about the function $f$, to know exactly $f(x)\ \forall x \in \mathbb{R}$?

Should analytic continuation of $f$ exist and be entire function on $\mathbb{C}$?

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up vote 4 down vote accepted

Multiply the function $\sin(\pi x)$ by any everywhere-defined function $g$. The result is identically $0$ on the integers. We have no chance of reconstructing $g(x)\sin(\pi x)$ from its values on the integers.

There seems to be no simple way to rule out this example by, say, insisting that we are interested in nowhere $0$ functions, since $f(x)+ \sin(\pi x)g(x)$ always agrees with $f(x)$ on the integers.

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If $f$ is rational, then your information will allow you to recover $f$, up to holes. That's an incredibly intense restriction on types of function though. I think Andre's answer makes it clear that there won't be many classes of functions that are less restrictive than this and meet your conditions.

To be complete: if $f$ and $g$ are known to be rational and have the same values on $\mathbb{Z}$, then $f-g$ is rational and equal to $0$ on $\mathbb{Z}$. Therefore the reduced numerator of $f-g$ is zero on $\mathbb{Z}$. A nonzero polynomial cannot have infinitely many roots. So the reduced numerator of $f-g$ is the zero polynomial. That means the unreduced numerator of $f-g$ is the zero polynomial, potentially with some holes at noninteger $x$-values. So $f-g=0$ except at finitely many $x$ where $f-g$ is undefined. At these $x$-values, either $f$ or $g$ is undefined. So $f=g$ up to where their nonintegral holes are.

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