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Assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous at $x=0$, and the limit \begin{equation} \lim_{x\to0}\frac{f(2x)-f(-x)}{x} \end{equation} exists. Is $f$ necessarily differentiable at 0?

I was able to deduce that if $f$ is continuous in a neighbordhood of $0$ and has an isolated zero at $x=0$, it must be differentiable (unless I made an error). I believe that this result holds if there exists some $\delta>0$ such that $f(x)$ does not change sign in the neiborhoods $-\delta<x<0$ and $0<x<\delta$.

But for the life of me I can't figure out a general proof (or a counterexample).

Any help would be greatly appreicated!

KReiser
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jsdifjew
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  • if $\begin{equation} \lim_{x\to0}\frac{f(2x)-f(x)}{x} \end{equation}$exists,the$f$Derivable at 0 – Hilbert1994 Jan 12 '21 at 00:55
  • Except that $(-x) $ does not cause any problem. If $(f(2x)-f(-x))/x\to L$ then $f(-2x)-f(x))/x\to - L$ and the question in duplicate target applies with $a=-2,|a|>1$. You have $f'(0)=L/3$. – Paramanand Singh Jan 12 '21 at 02:19

1 Answers1

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Let $ \varepsilon >0 $, there exists some $ \eta >0 $ such that $ \left(\forall x\in\left(-\eta,\eta\right)\right),\ \left|f\left(2x\right)-f\left(-x\right)-\ell x\right|\leq\varepsilon\left|x\right| $.

Let $ n\in\mathbb{N}^{*} $, we have, for all $ x\in\left(-\eta,\eta\right) $ : \begin{aligned}\small\left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3} x\right|&\small=\left|\sum_{k=0}^{n-1}{\left(f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right)}+f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\left|f\left(\frac{\left(-1\right)^{k}}{2^{k}}x\right)-f\left(\frac{\left(-1\right)^{k+1}}{2^{k+1}}x\right)-\ell\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\sum_{k=0}^{n-1}{\varepsilon\left|\frac{\left(-1\right)^{k}}{2^{k+1}}x\right|}+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\\ &\small\leq\varepsilon\left(1-\frac{1}{2^{n}}\right)\left|x\right|+\left|f\left(\frac{\left(-1\right)^{n}}{2^{n}}x\right)-f\left(0\right)-\ell\frac{\left(-1\right)^{n}}{3\times2^{n}}x\right|\end{aligned}

Giving first some clarifications about what we did, we introduced first a telescopic sum combined with a geometric sum, then we applyied our hypothesis for $ y=\frac{\left(-1\right)^{k}}{2^{k+1}}x $, because whatever $ k $ is, $ y $ will always be in $ \left(-\eta,\eta\right) $, after that we simplified the geometric sum that we obtained to end up with the result.

How we've assumed that $ f'\left(0\right) $, if exists, must be $ \frac{\ell}{3} $ ? Well to get it, we must take look at the following, for $ x\in\left(-\eta,\eta\right) $ : $$ \frac{f\left(2x\right)-f\left(-x\right)}{x}=2\times\frac{f\left(2x\right)-f\left(0\right)}{2x}+\frac{f\left(-x\right)-f\left(0\right)}{-x} $$

Taking $ n $ to $ +\infty $, we end with : $$ \left|f\left(x\right)-f\left(0\right)-\frac{\ell}{3}x\right|\leq\varepsilon\left|x\right| $$

Which means $ f $ is differentiable at $ 0 $.

CHAMSI
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