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Given that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$.

Proof: Let $\epsilon\gt 0$ be fixed. \begin{align*} \frac {f(x)}x&=\sum_{k=0}^n\color{blue}{\frac{f(x/2^k)-f(x/2^{k+1})}{x/2^{k+1}}}(1/2^{k+1})+\frac{\color{green}{f(x/2^{n+1})}}{x}\\ &=\sum_{k=0}^n\color{blue}{g_k(x)}(1/2^{k+1})+\frac{\color{green}{h_n(x)}}{x}\\ \text{There exists $N$ such that }\\ \left|\frac {f(x)}x\right|&\le \sum_{k=0}^N|g_k(x)|\frac 1{2^{k+1}}+\epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\frac 1{|x|}\epsilon|x|\\ \text{It follows that } \\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\epsilon\\ \text{Taking $N\to \infty$,it follows that}\\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\tag 1 \end{align*}

Since $\epsilon\gt 0$ is arbitrary, it follows by $(1)$ that $\liminf \left|\frac {f(x)}x\right|= \limsup\left|\frac {f(x)}x\right|=0=\lim_{x\to 0} \frac{f(x)}x. \;\;\; \blacksquare$

Is my proof correct? Thanks.

Koro
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3 Answers3

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We could simplify all this. Let

\begin{equation}{\rho} \left(x\right) = \sup_{\left|y\right| \leqslant \left|x\right|} \left|\frac{f \left(2 y\right)-f \left(y\right)}{y}\right|\end{equation}

We have for $n \geqslant 0$

\begin{equation}\renewcommand{\arraystretch}{1.5} \begin{array}{rcl}\displaystyle \left|f \left(x\right)-f \left(\frac{x}{{2}^{n+1}}\right)\right|&=&\displaystyle \left|\sum _{k = 0}^{n} \left(f \left(\frac{x}{{2}^{k}}\right)-f \left(\frac{x}{{2}^{k+1}}\right)\right)\right|\\ & \leqslant &\displaystyle \sum _{k = 0}^{n} {\rho} \left(x\right) \frac{\left|x\right|}{{2}^{k+1}}\\ & \leqslant &\displaystyle \left|x\right| {\rho} \left(x\right) \left(1-\frac{1}{{2}^{n+1}}\right) \end{array}\end{equation}

By taking the limit of this inequality when $n \rightarrow \infty $, it follows that

\begin{equation}\left|\frac{f \left(x\right)}{x}\right| \leqslant {\rho} \left(x\right) \mathop{\longrightarrow}\limits_{x \rightarrow 0} 0\end{equation}

Gribouillis
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  • Hi, thanks a lot for the answer :). Can you please also let me know what is wrong with my working? Thanks. – Koro Jan 23 '22 at 18:13
  • @Koro Again I don't understand the $N$ in your proof. Does $N$ depend on $x$? Please be more specific about how $N$ is chosen. Another thing that worries me is that there are terms without $|\cdot|$ on the RHS. – Gribouillis Jan 23 '22 at 20:44
  • Missing $|.|$ was a typo which I have fixed now. I chose $N$ like this: For a fixed x, and a fixed $\epsilon\gt 0$, since $\lim_{n\to \infty}g_n(x)=0$, there is $N_1, N_2$ such that $n\ge N_1\implies |g_n(x)|<\epsilon$ and since $\lim_{n\to \infty} h_n(x)=0$, there exists $N_2$ such that $n\ge N_2\implies h_n(x)<\epsilon |x|$. Choosing $N:=\min (N_1,N_2)$, we get the $N$ that was used in OP and yes it depends on $x$. – Koro Jan 23 '22 at 21:31
  • I see. The line involving limsup, liminf was with the assumption that N was constant, which clearly does not seem to be the case here as $N$ depends upon $x$. Is my understanding correct? Thanks. – Koro Jan 23 '22 at 21:34
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    Yes, if $N$ depends on $x$, one cannot take the limit when $x\to 0$. – Gribouillis Jan 23 '22 at 22:03
  • Thanks a lot for clearing my confusion. :) – Koro Jan 23 '22 at 22:07
3

Nowhere in the question is there an estimate of how small $x$ needs to be to make $\left|\frac{f(x)}x\right|$ small.

Since $\lim\limits_{x\to0}\frac{f(2x)-f(x)}x=0$, for any $\epsilon\gt0$, there is an $x_\epsilon$ so that for $|x|\le x_\epsilon$, we have $\left|\frac{f(2x)-f(x)}{x}\right|\le\epsilon$.

For any $|x|\le x_\epsilon$, $$ \begin{align} \left|\frac{f(x)}x\right| &=\left|\sum_{k=0}^\infty\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}\,2^{-k-1}\right|\tag1\\ &\le\sum_{k=0}^\infty\epsilon\,2^{-k-1}\tag2\\[6pt] &=\epsilon\tag3 \end{align} $$ Explanation:
$(1)$: $\sum\limits_{k=0}^\infty\left(f(2^{-k}x)-f(2^{-k-1}x)\right)=f(x)-\lim\limits_{u\to0}f(u)$
$(2)$: $\left|\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}\right|\le\epsilon$
$(3)$: $\sum\limits_{k=0}^\infty2^{-k-1}=1$

robjohn
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0

Let $\epsilon > 0$ be arbitrary, and let $x_\epsilon$ be small enough so that $ \Bigg| \dfrac{f \big( x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg| \le \epsilon $ for all $|x| \le x_\epsilon$.

Then,

$$ \Bigg| \dfrac{f \big(x \big)}{x} \Bigg| = $$

$$ \dfrac{1}{2} \Bigg| \dfrac{f \big(x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} + \dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg| \le $$

$$ \dfrac{1}{2} \Bigg[ \epsilon + \Bigg| \dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg| \Bigg] \le $$

$$ \dfrac{1}{2} \Bigg[ \epsilon + \dfrac{1}{2} \Bigg[ \epsilon + \Bigg| \dfrac{f \big(\frac {x} {4} \big)}{\frac {x} {4}} \Bigg| \Bigg] \Bigg] \le \ldots \le $$

$$ \dfrac{1}{2} \Bigg[ \epsilon + \dfrac{1}{2} \Bigg[ \epsilon + \ldots + \dfrac{1}{2} \Bigg[ \epsilon + \Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{\frac {x} {2^{n}}} \Bigg| \Bigg] \Bigg] \Bigg] = $$

$$ \sum_{k=1}^n \dfrac{\epsilon}{2^k} + \Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \le $$

$$ \sum_{k=1}^\infty \dfrac{\epsilon}{2^k} + \Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| = $$

$$ \epsilon + \Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \xrightarrow{n \rightarrow \infty} $$

$$ \epsilon + 0 = $$

$$ \epsilon $$

Joe Shmo
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