If $\lim_{h\to 0} \frac{ f_1(a+ h) - 2f_1(a) + f_1(a-h) }{h^2}$ is constant then is $f_1$ a quadratic function?

Question

Let $f_2(x)$ be a contstant function and $f_1(x)$ be a continous funtion $f:\mathbb{R}\to\mathbb{R}$. Let $$\lim_{h\to 0} \frac{ f_1(a+ h) - 2f_1(a) + f_1(a-h) }{h^2}=f_2(a).$$

Is it true that if the limit above exist for all $a$s, then $f_1(a)$ is polynomial function with degree at most 2?

Here is the reverse of this problem:reverse Please help, I am thankful for every solution!

EDIT: In my original problem, “continous” was missing

Answer 1

Consider the Taylor expansion around $h=0$ $$f_1(a+n\,h)=f_1(a)+h n f_1'(a)+\frac{1}{2} h^2 n^2 f_1''(a)+\frac{1}{6} h^3 n^3 f_1'''(a)++\frac{1}{24} h^4 n^4 f''''(a)+O\left(h^5\right)$$ Use it to get $$ f_1(a+ h) - 2f_1(a) + f_1(a-h)=h^2 f_1''(a)+\frac{1}{12} h^4 f''''(a)+O\left(h^5\right)$$ $$\frac{ f_1(a+ h) - 2f_1(a) + f_1(a-h) }{h^2}=f_1''(a)+\frac{1}{12} h^2 f''''(a)+O\left(h^3\right)$$

Now, ??? I am sure that you can take it from here.

Answer 2

The identity expresses that

$$f_1''(a)=f_2$$ for all $a$.

Then integrating twice,

$$f_1(x)=\frac{f_2}2x^2+cx+d.$$

The antiderivative of a constant is a linear function and the antiderivative of a linear function, a quadratic one.

Answer 3

No, this is not true: Take $f_1$ defined via $$f_1(x) = \begin{cases} +1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0\end{cases}$$ Then, you get $f_2 \equiv 0$.