An integer-valued polynomial is a polynomial with real coefficients mapping integers to integers. It is well known that all such polynomials $h(x)$ are generated as an additive group by the binomial coefficients $\binom{x}{n}$. My question concerns the problem of approximating an arbitrary polynomial $f$ with real coefficients on a skillfully chosen interval $I$ of length 1 by means of a skillfully chosen non-zero integer-valued polynomial $h$. Specifically, for $f\in\mathbb{R}[x]$ let $$N(f)=\inf_{I,h}\,\,\,\, ||f-h||_{I},$$ where $||\cdot ||_I$ means sup norm, $h$ runs through all non-zero integer-valued polynomials, and $I$ runs through all intervals of length 1. I am looking for a way to compute $N(f)$, and I have a conjecture (a guess really) that makes this computation very simple.

$\textbf{Conjecture:}$ Let $D(x)$ be the distance from the real number $x$ to the nearest integer. Then $$N(f)=\inf_{x\in\mathbb{Z}} \,\,\, D(f(x)).$$

My question is: Can anyone provide a proof or counter-example or some helpful references? In particular, as a simple test-case, can anyone compute $N(\frac{1}{2}x)$? Maybe the problem is impossibly difficult or ridiculously easy for some reason I don't see, and someone can put me out of my misery.

$\textbf{Remarks:}$

It is clear that $N(f)\ge\inf_{x\in\mathbb{Z}}D(f(x))$, because the closure of every interval of length 1 contains an integer.

I suspect that the conjecture is false if one defines $N$ so that $h$ ranges over polynomials with INTEGER coefficients, but I don't have an example to prove this.

It is easy to check that $N(0)$=0, and in general $N(c)=D(c)$ for any constant $c$, using the fact that the polynomial $\binom{x}{n}$ tends to 0 uniformly on $[0,1]$ as $n$ tends to infinity.

I'm already stuck on the computation of $N(\frac{1}{2}x)$, which is 0 according to the conjecture. One would naturally consider intervals $I$ of the form $[2n-\frac{1}{2},2n+\frac{1}{2}]$, and look at polynomials of the form $h(x):=\sum_{k}c_{k}\binom{x}{k}$, such that all the $c_{k}$ are integers and $h(2n)=n$.