Is this $f$ a linear function?

Question

My question is related to this 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.

I assume this holds for every $x$ (with the same sequence)?! — Igor Rivin, Dec 04 '13 at 20:19

score 2 · Answer 1 · answered Dec 05 '13 at 19:54

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.

score 0 · Answer 2 · answered Dec 04 '13 at 20:10

0

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.

answered Dec 04 '13 at 20:10

sranthrop

8,497

I have just edit the post. Here I actually mean $f$ is continuous. Sorry for the misleading. – Frank Lu Dec 04 '13 at 20:14
But this is a very nice example, any way – Frank Lu Dec 04 '13 at 20:15

Is this $f$ a linear function?

2 Answers2