Question: My question is related to [this](https://math.stackexchange.com/questions/592576/how-can-we-prove-this-function-must-be-linear) as I posted earlier. But this time, we drop certain conditions:

Suppose $f:[a,b]\to\mathbb{R}$ be continuous and there exists a sequence $(\alpha\_n)\_{n=1}^{\infty}$ with $\alpha\_n\to 0$ such that for all $x\in(a,b)$:

$$\lim\_{n\to\infty}\frac{f(x+\alpha\_n)+f(x-\alpha\_n)-2f(x)}{\alpha\_n^2}=0\quad (\*)$$

Does this imply $f$ is linear? i.e. $f(x)=ax+b$. 

I try to prove that actually $(\*)$ holds for any sequence which converges to zero, then we can use the argument in the problem I posted earlier. I am not quite sure if I am right. If this statement is false, I would be very curious to see a counter example.

Answer: 
It's like this:
$$\forall x\ \ \ \ \ \qquad\lim\_{\alpha\_n\to 0}\frac{1}{\alpha\_n}(f^\prime(x^+)+f^\prime(x^-))=0$$
or even for differentiable function :
$$\forall x\ \ \ \ \ \qquad\lim\_{\alpha\_n\to 0}\frac{f^\prime(x)}{\alpha\_n}=0$$
but for every sequence $\{\alpha\_n\}$ which converge to $0$. So $f^\prime(x)=0$ for all $x$ and it means $f$ is line.
Consider $f(x)=0$ for $x\in\mathbb Q$ and $f(x)=1$ for $x\in\mathbb R\backslash \mathbb Q$. For any $x$ you can choose $\alpha\_n=1/n$.
You may need any continuity on $f$ to conclude $f$ is (affine) linear.