Does there exist smooth or $C^2$ function for some infinite given points $a(n)$?

Question

I know that there exist some smooth function (polynomial) for finite numbers of values. The question is if there exists function (not necessary unique) which is twice differentiable and $f(n)=a(n)$?

Answer 1

Yes there are such functions. You can take a pair of values $a(n)$ and $a(n+1)$ and find the polynomial of degree 5, which fulfills $p(n)=a(n)$, $p(n+1)=a(n+1)$, $p'(n)=p''(n)=p'(n+1)=p''(n+1)=0$. Putting all these polynomials together piecewise, the result will be $C^2$ and will interpolate all given values.

Note that you can achieve $C^k$ smoothness by requiring $p^{(\ell)}(n) = p^{(\ell)}(n+1)$ for $\ell=1,\dots,k$. In this case you'll need piecewise polynomials of degree $2k+1$.

Answer 2

Define $f(x) = e^{4+1/(x^2-1/4)}, |x|<1/2, f(x) = 0, |x|\ge 1/2.$ Then $f\in C^\infty(\mathbb {R}),$ the support of $f$ is $[-1/2,1/2],$ and $f(0)=1.$ The function

$$g(x) = \sum_{n\in \mathbb {Z}}a_nf(x-n)$$

then belongs to $C^\infty(\mathbb {R})$ and $g(n) = a_n$ for all $n.$