$\lim_{h\to 0} \frac{ f(a+ h) - 2f(a) + f(a-h) }{h^2}$ and a polynomial function with degree at most 2

Question

Let $f_1(x)$ be a contionous polynomial funtion $f:\mathbb{R}\to\mathbb{R}$ with degree at most 2. Let $$\lim_{h\to 0} \frac{ f_1(a+ h) - 2f_1(a) + f_1(a-h) }{h^2}=f_2(a).$$

Prove that if the limit above exist for all $a$s, then $f_2(a)$ is constant!

This problem comes from a problem also posted in this site. But I can't prove it even with the solutions posted to this: a similar problem

Please help! I am very thankful for every solution!!

Answer 1

You can easily check this directly.

If $f_1$ is a degree 1 or 0 polynomial, you can show that $f_2(a) = 0$ for all $a$. Just set $f_1(x) = c$ or $f_1(x) = mx + b$ and plug into the equation.

Similarly, if $f_1 = mx^2 + nx + b$ is a degree 2 polynomial, plug into equation and take the limit, you get (for all $a$)

$$ \lim_{h \to 0} \frac{ 2mh^2}{h^2} = 2m$$

And as discussed in the comments above, this is exactly the second derivative of $f_1$. The moral is that this represents the "definition of the second derivative" function and so it makes sense polynomials of degree $\le 2$ have constant second derivatives. It is possible to generalize this formula to the "third derivative" and so on.