I know that this has been asked on MSE before, but none of the answers seemed to answer my query and hence, I am making a new post.
$\textbf{Question:}$ $f:[a,b]\to\mathbb{R}$ is a continuous function. If $$\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}~=~0$$ holds for all $x$ in $[a,b]$ , prove that $f$ is linear on $[a,b]$ , i.e. $f(x)=cx+d$ for some $c,d\in\mathbb{R}$.
$\textbf{Note:}$ nothing has been said about the differentiability of $f(x)$ hence, we can't use L'Hôpital's rule or the Taylor Series expansion.
$\textbf{Attempt:}$ Given,
$$\displaystyle\lim_{h\to 0}\dfrac{f(x+h)+f(x-h)-2f(x)}{h^2}=0$$
Multiplying, $-1$ to the numerator and denominator and splitting $h^2$ and then proceeding further:
$$\displaystyle\lim_{h\to 0}\left(\dfrac{\left(\dfrac{f(x)-f(x+h)-f(x-h)+f(x)}{-h}\right)}{h}\right)$$
Let, $c$ be a constant such that, $c\to -h$, this automatically implies that, when $h\to 0$ then, $c\to 0$ as well. Now, introducing $c$ in the above equation:
$$\displaystyle\lim_{h\to 0}\left(\dfrac{\left(\displaystyle\lim_{c\to 0}\dfrac{f(x+h+c)-f(x+h)}{c}\right)-\left(\displaystyle\lim_{c\to 0}\dfrac{f(x+c)-f(x)}{c}\right)}{h}\right)$$
From the definition of the derivative, the numerator reduces to:
$$\lim_{h\to 0}\left(\dfrac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\right)$$
This is reduces to:
$$\lim_{h\to 0}\left(\dfrac{f^{\prime}(x+h)-f^{\prime}(x)}{h}\right)=f^{\prime\prime}(x)$$
Now, as per given condition, $f^{\prime\prime}(x)=0$, hence we may conclude that $f(x)$ is a linear equation of the form, $cx+d$ for some $c, d\in\mathbb{R}$.
I have been told that the constant $c$ which I introduced and applied the limits to, is not a legal move. I also feel the same way, but I can't figure out why.
Also, if the above process is wrong, then what is the right one? Note: we DO NOT know about the differentiability of the function, we only know that it is continuous that's all.