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$f: \mathbb{R}\to\mathbb{R}$ and $f$ is continuous,

$\lim_{h\to 0}\frac{f(2h)-f(h)}{h}$ is exist then $f$ is differentiable at $x=0$?

If correct, prove.

If not correct, please give me some counterexample.

(This problem is not homework, only my concern... I tried many methods but I don't know it is true or false...)

M.E.R
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    have you tried anything yet? you know you would be more likely to get help if you show you tried. – Pink Panther Jan 10 '18 at 20:45
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    You can write $\frac{f(2h)-f(h)}{n}=2\frac{f(2h)-f(0)}{2h}-\frac{f(h)-f(0)}{h}$. Call $g(x)=\frac{f(x)-f(0)}{x}$. – orole Jan 10 '18 at 20:48
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    @Ellusionist I tried many method but all would be not exact. It is very hard. It is not homework(not in any books) and it is just my concern. – M.E.R Jan 10 '18 at 20:50
  • This is a duplicate, see https://math.stackexchange.com/a/1864092/72031 – Paramanand Singh Jan 11 '18 at 00:20

1 Answers1

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Define $\varphi(h) = \frac{f(2h) - f(h)}{h}$ for $h \neq 0$ and let $\varphi(0) = \lim_{h\to 0}\varphi(h)$. Then notice that for $h \neq 0$ and for $N \geq 1$,

$$ \frac{f(h) - f(2^{-N}h)}{h} = \sum_{k=1}^{N-1} 2^{-k} \varphi(2^{-k}h) + 2^{-N}\varphi(2^{-N}h). \tag{*} $$

By the construction we know that $\varphi$ is continuous. Taking $N\to\infty$, this converges to

$$ \frac{f(h) - f(0)}{h} = \sum_{k=1}^{\infty} 2^{-k}\varphi(2^{-k}h) $$

Moreover, for any $\delta > 0$, Weierstrass $M$-test tells that this convergence is in fact uniform over the set $[-\delta, \delta]\setminus\{0\}$. Thus, as $h \to 0$ we can interchange the order of the infinite sum and the limit to find that

$$ \lim_{h\to0} \sum_{k=1}^{\infty} 2^{-k}\varphi(2^{-k}h) = \sum_{k=1}^{\infty} \lim_{h\to0} 2^{-k}\varphi(2^{-k}h) = \varphi(0) $$

Therefore $f$ is differentiable at $0$ with $f'(0) = \varphi(0)$ as desired.

Sangchul Lee
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  • Then the argument in my answer is wrong? – user Jan 10 '18 at 21:08
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    @gimusi, Your argument assumes the existence of $f'(0) = \lim_{h\to0} \frac{f(h) - f(0)}{h}$, but it is not a free gift from OP's setting. – Sangchul Lee Jan 10 '18 at 21:09
  • Could you also give a counter example of this? Thanks! – user Jan 10 '18 at 21:11
  • @gimusi, A slight tweak of OP's problem does have a counter-example: Notice for $f(x) = |x|$, the limit $$\lim_{h\to0} \frac{f(x+h) - f(x-h)}{2h}$$ exists for any $x \in \mathbb{R}$ while $f$ is not differentiable at $x = 0$. – Sangchul Lee Jan 10 '18 at 21:13
  • @Sangchul Lee Thank you and I'll read it carefully. – M.E.R Jan 10 '18 at 21:16
  • @SangchulLee Yes I lnow of course this classical example but I can'y see it as a counter example in this case.Anyway many tjhanks for you explanation. Bye! – user Jan 10 '18 at 21:18
  • Is @orole's comment right way for this question, or where is his fault? I just wonder. – Domates Jan 10 '18 at 21:26
  • @Domates, As I explained in one of my comments above, existence of $\lim_{h\to0} \frac{f(h) - f(0)}{h}$ is not assumed in OP's setting but rather it is what OP wants to prove. – Sangchul Lee Jan 10 '18 at 21:28
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    @Domates This problem cannot solve by only use some (Stewart) Calculus knowledge, I assure this problem is in connection with sequences {f_n} and some real-analysis knowledge. – M.E.R Jan 10 '18 at 21:31