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For a continuous function $f(x)$ with $f(0)=0$, the following limit exists. $$\lim_{x \to 0} \frac{f(2x)-f(-3x)}{x}.$$ For such a function, is $f(x)$ differentiable at $x=0$?

Idea. I tried to find a counterexample as there is no reason for $f(x)$ to be differentiable.

Also, as a famous example of $f(x)=|x|$ where $\displaystyle{\lim_{x \to 0}} \frac{f(x)-f(-x)}{x}$ exists but not differentiable, I tried to mimic this example.

But, because of the factor $2$ and $3$, I failed to find a counterexample or prove it.

gt6989b
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five manifold
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    Instead of $y = |x|,$ which arises by joining at the origin the $y > 0$ portions of the graphs of $y = x$ and $y = -x,$ use the function that arises by joining at the origin the $y>0$ portions of the graphs of $y = 2x$ and $y = -3x.$ – Dave L. Renfro Oct 01 '20 at 16:17
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    That example is not we want. That kind of exmaple gives a example where only right derivative or left derivative exists. – five manifold Oct 01 '20 at 16:20
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    @DaveL.Renfro That wouldn't be a counterexample, since $\lim_{x \to 0} \frac{f(2x)-f(-3x)}{x}$ wouldn't exist. – Varun Vejalla Oct 01 '20 at 16:28
  • @Varun Vejalla (and OP): Oops, you're right. I was thinking of what is sometimes called the pseudo-symmetric derivative, which is different than what we have here where, before the function evaluations, the coefficient of $x$ varies. – Dave L. Renfro Oct 01 '20 at 16:37
  • Recall that if $f$ is differentiable at $x=a,$ then the symmetric derivative exists and equals $f'(a).$ (This is a common calculus/real-analysis exercise. See here for example.) Thus, if $f$ is differentiable at $x=0,$ then both $\lim_\limits{x \to 0} \frac{f(2x)-f(-2x)}{x}$ and $\lim_\limits{x \to 0} \frac{f(3x)-f(-3x)}{x}$ exist, and you might be able to do something with this. – Dave L. Renfro Oct 01 '20 at 16:44

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Such a function will be differentiable at zero ($f(0)=0$ is a red herring, it doesn't contribute to the result). If we set $$r(x) = \frac{f(2x)-f(-3x)}x$$ and call $\ell$ its limit, then up to considering $x\mapsto f(x)-\frac\ell 5x$ we can assume $\ell=0$.

Fix an $\varepsilon>0$, and choose $\delta>0$ such that $|r(x)|<\varepsilon$ for all $x\in(-\delta,\delta)$. Then writing $$f(x)-f\left(\frac{4x}9\right) = -\big(f(2y)-f(-3y)\big)-\big(f(2z)-f(-3z)\big)$$ with $y=-x/3$ and $z=2x/9$, we see that $|f(x)-f(4x/9)|$ is at most $5\varepsilon|x|/9$. Now $$f(x)-f(0) = \lim_{k\to\infty}\left(f(x)-f\left(\frac{4^{k+1}x}{9^{k+1}}\right)\right) = \sum_{k\geq 0}\left(f\left(\frac{4^kx}{9^k}\right)-f\left(\frac{4^{k+1}x}{9^{k+1}}\right)\right)$$ by continuity of $f$ at zero, so $$\left|\frac{f(x)-f(0)}x\right| \leq \frac{5\varepsilon}9\sum_{k\geq 0}\frac{4^k}{9^k} = \varepsilon $$ on a neighbourhood of zero.

Since $\varepsilon$ was arbitrary, this shows that $f'(0)=0$.

Pierre PC
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