How prove this $f(x)$ is polynomial function

Question

we define

$$f^{[n]}(x)=\lim_{h\to 0}\sum_{k=0}^{n}\binom{n}{k}\dfrac{(-1)^kf[(x+(n-2k)h]}{(2h)^n}$$ if $f(x)$ is continuous on $[a,b]$,and such $f^{[n]}(x)=0$ for all $x$.

prove or disprove :$f(x)$ is polynomial function,

My try: I find sometimes,and I find this following therom:

(Schwarz,1870):If $F$ is continuous and $DF(x)=0$ for all $x$, then $F$ is a linear function where $$DF(x):=\lim_{h\to 0}\dfrac{F(x-h)-2F(x)+F(x+h)}{h^2}$$ and this solution can see: http://math.depaul.edu/mash/newharder.pdf

I think my problem is true.But I can't prove it,Thank you

Answer 1

The statement in its current form is false for $n\geq 3$.

For odd $n\geq 3$ odd, $f(x)=\left |x\right |$ can serve as counterexample. This function is locally linear for $n\not =0$ and thus the corresponding limit is zero. For $x=0$, we can make use of the symmetry of binomial coefficients $\binom{n}{k} = \binom{n}{n-k}$ and the fact that for odd $n$, we have $(-1)^{n-k} = -(-1)^k$, to rewrite the limit as follows:

$$\lim_{h\rightarrow 0} \frac{1}{(2h)^n}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n}{k}\left[f((n-2k)h) - f((2k-n)h)\right]$$

Since we also have $f(x)=f(-x)$ for all $x$, the expression in square brackets is identically equal to zero; so is the sum and thus also the limit.

For $n\geq 4$ even, a counterexample is provided by $f(x)=x\cdot \left|x\right|$. This function is locally quadratic for $x\not =0$, so again, $x=0$ is the only interesting point. In this case, the symmetry of binomial coefficients and identity $f(-x) = -f(x)$, in addition to $f(0)=0$, are sufficient to show the limit to be equal to zero too.

Both counterexamples are clearly continuous and the latter is actually differentiable. In fact, the same approach can be used to obtain $(n-3)$-times differentiable counterexamples by considering $$f(x)=\left| x\right|\cdot x^{n-3}$$