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!

  • $\begingroup$ 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

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 $|a| \neq 1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $x\in\>]0,\delta]$.


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?

  • 2
    $\begingroup$ $\lim_{x\to0}\frac{f(0)-f(x)}{x}$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.