Vrftsjtryk

Problem:

If $f(x)$ is continous at $x=0$, and $limlimits_{xto 0} dfrac{f(ax)-f(x)}{x}=b$, $a, b$ 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_{xto 0} frac{f(ax)-f(x)}{x}\
&=lim_{xto 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,&xneq0\
0,&x=0
end{cases}]
$lim_{xto0}dfrac{f(3x)-f(x)}{x}=lim_{xto0} dfrac{1-1}{x}=0$

but $lim_{xto0}dfrac{f(3x)}{x}=infty$,$lim_{xto0}dfrac{f(x)}{x}=infty$

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

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 $$f(x) - f(x/a^{n}) = bxsum_{k = 1}^{n}frac{1}{a^{k}} + xsum_{k = 1}^{n}frac{g(x/a^{k})}{a^{k}}$$ 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 $$f(x) - f(0) = frac{bx}{a - 1} + xsum_{k = 1}^{infty}frac{g(x/a^{k})}{a^{k}}$$ 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 $|a| neq 1$.

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 $$lim_{xto0}{g(2x)-g(x)over x}=0 .$$
We have to prove that $g'(0)=lim_{xto0}{g(x)over x}=0$.

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

Hint: You're very close.

Write the expression as $$afrac{f(ax)-f(0)}{ax}-frac{f(x)-f(0)}{x}$$ Note that $xto 0$ if and only if $axto 0$ (since $aneq 0$).

Can you see it from this?