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Let $f:[a,b]\to\mathbb{R}$ be continuous. Suppose for any sequence $(r_n)_{n=0}^{\infty}$ with $\lim_{n\to\infty}r_n=0$, and any $x\in(a,b)$:

$$\lim_{n\to\infty}\frac{f(x-r_n)+f(x+r_n)-2f(x)}{r_n^2}=0$$

Show that $f$ is a linear function.

Clearly the limit of the numerator is zero, I want to conclude that the numerator must be zero for all $n$, then from this to conclude $f$ is linear. I am not sure if this is the right approach. Are there any hints?

Frank Lu
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  • I have fixed (what I believe to have been) a typo. Please make sure I have not altered your intended meaning. – Cameron Buie Dec 04 '13 at 15:56
  • @EricAuld $r_n$ is an arbitrary sequence which converges to zero – Frank Lu Dec 04 '13 at 15:58
  • @Cameron Buie I think now it looks good – Frank Lu Dec 04 '13 at 15:59
  • @EricAuld: I fixed the typo. It was supposed to be $r_n,$ not $r+n.$ – Cameron Buie Dec 04 '13 at 16:01
  • The condition could just as well be written $\lim_{r\to 0}\frac{f(x-r)+f(x+r)-2f(x)}{r^2}=0$. – Hagen von Eitzen Dec 04 '13 at 16:08
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    The expression you have is a second order difference quotient. That's related to the second derivative; if you had simply that $f''=0$ it would be straight-forward to integrate twice and determine $f$ was linear, so as a hint, how do you go about "undoing" the difference quotient to get a similar result? – postmortes Dec 04 '13 at 16:10
  • @postmortes I'm very interested in how to do this. I hope someone figures it out. If they don't, I'd love to read your answer. – Eric Auld Dec 04 '13 at 16:45

1 Answers1

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I think that you will need more than just $f$ to be continuous on $(a,b)$, but if you assume that:

$\exists x_0 \in (a,b)$ such that $f''(x_0) \neq 0$

then the property $\lim_{n \rightarrow \infty}\frac{f(x-r_n)+f(x+r_n)-2f(x)}{r_n^2}=0$ does not hold for $x=x_0$.

AdriMarteau
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  • And I think that the limit can be true for any piecewise linear function on $(a,b)$ – AdriMarteau Dec 04 '13 at 16:15
  • If we think backwards， that is, if $f$ is linear then $f''\equiv 0$, so the problem you mentioned would not occur. – Frank Lu Dec 04 '13 at 16:16
  • I am actually using a proof by contradiction argument (if it is not linear then there exists an $x_0$ such that $f''(x_0)\neq 0 $ which leads to a contradiction with the limit). – AdriMarteau Dec 04 '13 at 16:20
  • well, that is a good point, let me try this – Frank Lu Dec 04 '13 at 16:22
  • but I think if we want to use proof by contradiction, we also need to consider the case that $f$ is not differentiable – Frank Lu Dec 04 '13 at 16:23
  • Indeed, and this the reason of my initial remark about $f$. And moreover, if, for example, $f:x \rightarrow |x-\frac{a+b}{2}|$, $f$ is continuous and the property still holds but $f$ is not linear. – AdriMarteau Dec 04 '13 at 16:27
  • @AdriMarteau But then the property would not hold at $(b-a)/2$, right? – Eric Auld Dec 04 '13 at 16:44
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    @EricAuld, I think you mean $(a+b)/2$ (as $(b-a)/2 \not\in (a,b)$) and in that case you are right, it does not work. – AdriMarteau Dec 04 '13 at 16:56