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For every function $f$ from $\mathbb{Z}$ to $\mathbb{R}$, can we find a function $g$ from $\mathbb{R}$ to $\mathbb{R}$ that is infinitely differentiable and agrees with $f$ on $\mathbb{Z}$? Furthermore, would there be a simple way to explicitly define a $g$ for every $f$?

Could we impose stronger conditions on $g$ than this; I feel making $g$ infinitely differentiable wouldn't be too difficult in some piecewise way (perhaps using variations of $e^{-1/x^2}$ that are stretched and stitched together so all derivatives are 0 at every integer, though this seems ugly), but perhaps we can also find a $g$ equal to its Taylor series around a point, or a $g$ satisfying other nice conditions? Thanks.

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4  
Yes. Please see the Wikipedia article on bump functions. – André Nicolas Jan 10 at 1:13
2  
This seems relevant: math.stackexchange.com/questions/626492/… – Dejan Govc Jan 10 at 1:23
    
Check e.g. the gamma function as an extension of $n!$, and discussion on other extensions. – vonbrand Jan 10 at 1:31
    
Found the page on alternative extensions of $n!$ to real values here. – vonbrand Jan 13 at 19:19
up vote 7 down vote accepted

Yes. We can even make $g$ equal to its Taylor series around every point, everywhere -- or, in other word, every function $\mathbb Z\to\mathbb C$ can be extended to an analytic function $\mathbb C\to\mathbb C$.

We can construct $g$ as the sum of a sequence of polynomials $g_1, g_2, g_3, \ldots$, such that

  • $g_n$ is zero on $[-(n-1),n-1]\cap\mathbb Z$
  • $g_n$ has the right values at $-n$ and $n$ to give the sum the right value at these two points.
  • $|g_n(z)|<2^{-n}$ for $|z|\le n-1$.

The last of these properties will guarantee that the sum exists everywhere and is analytic everywhere (since it converges uniformly on every bounded subset of $\mathbb C$).

No matter what the values of $g_n(-n)$ and $g_n(n)$ need to be we can always achieve all the required properties by choosing $$ g_n(z) = (b+cz)z^m\prod_{k=-(n-1)}^{n-1} (z-k) $$ for appropriate constants $m$, $b$ and $c$. (Choosing $m$ large enough can make the magnitude of the value of the function on the disk $B_{n-1}(0)$ small enough compared with the values at $-n$ and $n$ that the third property can be achieved).

Of course, this extension is not unique -- at the very least we can add any multiple of $\sin(\pi z)$ we want to the result without losing analyticity.


Or, if we only want $g$ to be analytic on $\mathbb R$, we can simply choose $$ g(x) = \sum_{k\in\mathbb Z} \operatorname{sinc}(x-k)^{m_k} f(k) $$ where $\operatorname{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$ and $m_k$ is chosen large enough that $\operatorname{sinc}(x)^{m_k}f(k) < 2^{-|k|}$ for all $|x|>1$.

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Very nice, thanks! I imagine the first method could generalize to functions from the Gaussian integers (and similar things) to the reals in an analogous way. – B Gunsolus Jan 10 at 5:26
    
I would remark that though the polynomial sum is intriguing, it's somewhat infeasible to actually construct this unless the integral function happens to have some very simple pattern. The $\operatorname{sinc}$ definition, OTOH, is tremendously useful in all kinds of practical applications. Keyword sampling theorem‌​. – leftaroundabout Jan 10 at 11:58
    
@BGunsolus: Yes, it generalizes to any discrete subset of $\mathbb C$. This result is shown (with different details but the same core idea) in this question which Dejan Govc found while I was typing. – Henning Makholm Jan 10 at 12:02
    
(Sorry, it needs do be closed and discrete, of course). – Henning Makholm Jan 10 at 12:14

A uniqueness result can be obtained from Carlson's theorem:

Given a set of numbers $a_n$, $n\in\mathbb{N}$. We want to find an analytical function $f$ with $$ f(n) = a_n \tag{1}$$ (for $n\in\mathbb{N}$) and which fulfils the following growth conditions: $$|f(z)| \leq C e^{\tau |z|} \tag{2}$$ for all $z \in \mathbb{C}$ and for some $C,\tau < \infty$. And there exist $c< \pi$ such that $$|f(iy)| \leq C e^{c |y|} \tag{3}$$ for all $y\in\mathbb{R}$. Then the uniqueness of $f$ follows from Carlson's theorem.

Proof:

Given (on the contrary) two functions $f$ and $g$ obeying (1), (2), (3). Then we apply Carleson's theorem on $h=f-g$ and show $h=0$.

Note that (3) is needed to forbidd adding $\sin(\pi z)$ as pointed out by Henning Makholm.

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