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Problem:
If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.

This approach is definitely wrong:

\begin{align} b&=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &=af'(0)-f'(0)\\ &=(a-1)f'(0) \end{align}

I will show you a case why this approach is wrong:

\[f(x)= \begin{cases} 1,&x\neq0\\ 0,&x=0 \end{cases}\] $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$
but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$

Does anyone know how to prove it? Thanks in advance!

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  • $\begingroup$ Do you have to also assume that $f'(0)$ exists or does not follow from the other hypotheses? $\endgroup$ – Jonas Meyer Jul 19 '16 at 8:41
  • $\begingroup$ the proof of the existence of $f'(0)$ is actually the problem is asking for $\endgroup$ – Spaceship222 Jul 19 '16 at 8:47
  • $\begingroup$ Your result holds even if $|a| < 1$. See update to my answer. $\endgroup$ – Paramanand Singh Jul 19 '16 at 12:14
  • $\begingroup$ Your counter-example violates the continuity hypothesis of $f$ at $x=0$, doesn't it? $\endgroup$ – BusyAnt Jul 19 '16 at 12:35
  • $\begingroup$ @BusyAnt yes it does.I am just not convinced by the approach using the continuity at $x=0$ $\endgroup$ – Spaceship222 Jul 19 '16 at 12:43
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This is a tricky question and the solution is somewhat non-obvious. We know that $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) \to 0$ as $x \to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, \ldots, n$ we get $$f(x) - f(x/a^{n}) = bx\sum_{k = 1}^{n}\frac{1}{a^{k}} + x\sum_{k = 1}^{n}\frac{g(x/a^{k})}{a^{k}}$$ Letting $n \to \infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = \frac{bx}{a - 1} + x\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x \to 0$ we get $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \frac{b}{a - 1} + \lim_{x \to 0}\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$

The sum $$\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x \to 0$ because $g(x) \to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)\to 0$ as $x \to 0$, it follows that for any $\epsilon > 0$ there is a $\delta > 0$ such that $|g(x)| < \epsilon$ for all $x$ with $0 <|x| < \delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < \delta$ if $|x| < \delta$ and therefore $|g(x/a^{k})| < \epsilon$. Thus if $0 < |x| < \delta$ we have $$\left|\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}\right| < \sum_{k = 1}^{\infty}\frac{\epsilon}{|a|^{k}} = \frac{\epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x \to 0$.

Hence $f'(0) = b/(a - 1)$.


BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ implies that $$\lim_{t \to 0}\frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = \frac{-bc}{c - 1} = \frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| \neq 1$.

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  • $\begingroup$ How "adding such equation" do you get the left side you did? Do you mean when adding over $\;k\;$ from $\;1\;$ to $\;n\;$? Then, you left $\;n\to\infty\;$ , but why would $\;x\to0\implies\;$ the second summand in the right tends to zero? Even if the right sum converges for all $\;x>R\;$ , for some $\;R>0\;$, how it depends on $\;x\;$ could makea difference. $\endgroup$ – DonAntonio Jul 19 '16 at 9:24
  • $\begingroup$ Yes adding over $k = 1$ to $n$. The infinite sum on right tends to $0$ as $x \to 0$. I have kept it as exercise for reader. But It appears I need to prove it. Wait for my updated answer. $\endgroup$ – Paramanand Singh Jul 19 '16 at 9:27
  • $\begingroup$ @DonAntonio: see my updated answer. $\endgroup$ – Paramanand Singh Jul 19 '16 at 9:32
  • $\begingroup$ Thank you. Yet I think this is way too convoluted for an answer, and it may be this exercise is way before infinite series is studied... $\endgroup$ – DonAntonio Jul 19 '16 at 9:32
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    $\begingroup$ @DonAntonio: The other answer only shows that if $f'(0)$ exists then it must be $b/(a - 1)$. But it does not show why $f'(0)$ exists. One of the downvotes for that answer is mine. In mathematics, correctness is more important than anything else. $\endgroup$ – Paramanand Singh Jul 19 '16 at 9:38
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In this quickly closed question the case $a=2$ is considered, which allows the following simpler solution:

Define $g(x):=f(x)- bx-f(0)$. Then $g$ is continuous at $0$, $g(0)=0$, and $$\lim_{x\to0}{g(2x)-g(x)\over x}=0\ .$$ We have to prove that $g'(0)=\lim_{x\to0}{g(x)\over x}=0$.

Let an $\epsilon>0$ be given. Then there is a $\delta>0$ such that $|g(2t)-g(t)|\leq\epsilon |t|$ for $0<t\leq\delta$. Assume $|x|\leq\delta$. Then for each $N\in{\mathbb N}$ one has $$g(x)=\sum_{k=1}^N\bigl(g(x/2^{k-1})-g(x/2^k)\bigr)+g(x/2^N)\ ,$$ and therefore $$\bigl|g(x)\bigr|\leq\sum_{k=1}^N\epsilon\,{|x|\over 2^k} \ +g(x/2^N)\leq\epsilon|x|+g(x/2^N)\ .$$ Since $N\in{\mathbb N}$ is arbitrary we in fact have $\bigl|g(x)\bigr|\leq \epsilon|x|$, or $\left|{g(x)\over x}\right|\leq\epsilon$, and this for all $x\in\>]0,\delta]$.

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Hint: You're very close.

Write the expression as $$a\frac{f(ax)-f(0)}{ax}-\frac{f(x)-f(0)}{x}$$ Note that $x\to 0$ if and only if $ax\to 0$ (since $a\neq 0$).

Can you see it from this?

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    $\begingroup$ $\lim_{x\to0}\frac{f(0)-f(x)}{x}$ is not the same as $\lim_{x\to0}\frac{f(0)-f(ax)}{ax}$ if we don't know the existence of $f'(0)$,so we can't put it together times $(a-1)$ $\endgroup$ – Spaceship222 Jul 19 '16 at 9:02
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    $\begingroup$ @MPW I'm rewriting this since I think I understand your comment above better now though it still is pretty messy (the limits are minus the usual one). $\endgroup$ – DonAntonio Jul 19 '16 at 9:03
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    $\begingroup$ I think, after making some order, both in the above answer and, in particular, in my mind, that this answer is correct: we can write$$b=\lim_{x\to0}\frac{f(ax)-f(x)}x=\lim_{x\to0}\left[a\frac{f(ax)-f(0)}{ax}-\frac{f(x)-f(0)}x\right]=\lim_{t\to0}(a-1)\frac{f(t)-t(0)}t$$because when $\;x\to0\;$ both limits (without the constant $\;a\;$) within the parentheses are the same, whether it exists or not, because $\;f\;$ is given continuous at zero and thus it is the same to take $\;\lim f(x)\;$ or $\;\lim f(ax)\;$ when $x\to0$. The rightmost expression, compared to the left side, answers all +1 $\endgroup$ – DonAntonio Jul 19 '16 at 9:10
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    $\begingroup$ I agree here with @Spaceship222: You must prove existence of $f'(0)$ by other means. see my answer. Your answer as it stands is incorrect. $\endgroup$ – Paramanand Singh Jul 19 '16 at 9:20
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    $\begingroup$ @DonAntonio: Let $F(x) = 1/x$ and $a > 0$. Then both the limits $\lim_{x \to 0^{+}}F(x)$ and $\lim_{x \to 0^{+}}F(ax)$ don't exist and yet $$\lim_{x \to 0^{+}}aF(ax) - F(x) = 0$$ so one should be very careful about the conditions under which laws of algebra of limits work. $\endgroup$ – Paramanand Singh Jul 19 '16 at 10:02

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