Convergence of finite differences to zero and polynomials Assume that $f:\mathbb R \rightarrow \mathbb R$ is continuous and $h\in \mathbb R$.  Let $\Delta_h^n f(x)$ be a finite difference of $f$ of order $n$, i.e
$$
\Delta_h^1 f(x)=f(x+h)-f(x),
$$
$$
\Delta_h^2f(x)=\Delta_h^1f(x+h)-\Delta_h^1 f(x)=f(x+2h)-2f(x+h)+f(x),
$$
$$
\Delta_h^3 f(x)=\Delta_h^2f(x+h)-\Delta_h^2f(x)=f(x+3h)-3f(x+2h)+3f(x+h)-f(x),
$$
etc.
There is an explicite formula for $n$-th difference: 
$$
\Delta_h^n f(x)=\sum_{k=0}^n (-1)^{n-k}\frac{n!}{k!(n-k)!} f(x+kh).
$$
Assume now that $n\in \mathbb N$ and  $f:\mathbb R \rightarrow \mathbb R$ are such that for each $x \in \mathbb R$:
$$
\frac{\Delta_h^n f(x)}{h^n} \rightarrow 0 \textrm{ as } h \rightarrow 0.
$$
Is it then $f$ a polynomial of degree $\leq n-1$?
It is clear if $n=1$, because then $f'(x)=0$ for $x\in \mathbb R$.
Edit. Without continuity assumption about $f$ it is not true, because for $n-1$-additive function $F$ which is not $n-1$-linear we have $\Delta_h^nf(x)=0$, where $f(x)=F(x,...,x)$.
 A: Let $f(x) = |x|$ then $\Delta_h^2(f)$ has support $[-2h, 0]$.  In particular $\lim_{h \to 0}\Delta_h^2(f)/h^2 = 0$ pointwise, but $f$ is not a polynomial.
Edit: If the convergence in $x$ is uniform on an interval $[a, b]$ then I think that $f$ is a polynomial on that interval.  This may follow from Fourier expansion, but I don't have time now to hammer out the fine points (if it can be done).
A: The result holds if we assume that $f$ is $n$-times differentiable, otherwise, as WimC shows, it's not necessarily the case.
Using this thread and translated function ($f(\cdot)=h(x+\cdot)$), we can see that 
$$\frac{\Delta_h^nf(x)}{h^n}=f^{(n)}(x),$$
so the hypothesis yields $f^{(n)}\equiv 0$, hence $f$ is a polynomial of degree at most $n-1$.
A: This is actually a comment too long to fit in the usual format.
WimC’s claim about the uniform convergence case is correct : suppose that $\Gamma(h,x)=\frac{\Delta_h^2f(x)}{h^2} \to 0$, uniformly in $x$ on an interval $[a,b]$. 
Let us put $\beta(h)={\sf sup}_{x\in[a,b]}(\big| \Gamma(h,x)\big|)$ for $h>0$. Then the hypothesis states that $\beta(h) \to 0$ when $h \to 0$.
Now, the identity
$$
\Delta_{2h}^{2}f(x)=\Delta_h^{2}f(x+2h)+2\Delta_h^{2}f(x+h)+\Delta_h^{2}f(x)
$$
yields
$$
\Gamma(2h,x)=\frac{\Gamma(h,x+2h)+2\Gamma(h,x+h)+\Gamma(x,h)}{4}
$$
Taking sups above, we see that $\beta(2h) \leq \beta(h)$. So if the bound $|\beta(h)| \leq \varepsilon $ holds for $h\in [0,\eta]$, it will also hold for $h \in [0,2\eta]$ ; it will even hold everywhere, by induction. Since this holds for every $\varepsilon >0$, we see that $\beta=0$, as wished.
