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The question is this:

Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable at $x=0$ and suppose that there is a number $L$ such that $$\lim_{x\rightarrow0}\frac{f(x)-f(x/2)}{x/2}=L.$$ Prove that $f'(0)=L$.

Here's my answer with all theorems referenced being from Rudin:

Let $a_n$ be a positive sequence converging to zero and $$\varphi_n(x)=\frac{a_nf'(0)+2\big(f(x)-f(x/2)\big)}{x+a_n}.$$ Then $$\lim_{n\rightarrow\infty}\lim_{x\rightarrow0}\varphi_n(x)=f'(0)$$ while $$\lim_{x\rightarrow0}\lim_{n\rightarrow\infty}\varphi_n(x)=L.$$ By theorem 7.11 then, if $\varphi_n(x)$ converges uniformly to $\varphi(x)=\frac{f(x)-f(x/2)}{x/2}$ over a set $E$ and $0$ is a limit point of $E$, then $L=f'(0)$. Let $E=[0,1]$. Then for $x\in E$, $$\big|\varphi_n(x)-\varphi(x)\big|=a_n\bigg|\frac{xf'(0)-2\big(f(x)-f(x/2)\big)}{x+a_n}\bigg|=a_n\big|f'(0)-\varphi_n(x)\big|\leq a_n\big(|f'(0)|+|\varphi_n(x)|\big)\leq a_n\bigg(|f'(0)|+\bigg|\frac{a_nf'(0)}{x+a_n}\bigg|+\bigg|\frac{2\big(f(x)-f(x/2)\big)}{x+a_n}\bigg|\bigg)< a_n\big(|2f'(0)|+|L|\big)\rightarrow0.$$ So by theorem 7.9, $\varphi_n(x)$ converges uniformly to $\varphi(x)$ over $E$ and therefore $f'(0)=L$.

What I don't understand is that couldn't I have put basically anything, say $\pi$, in place of $f'(0)$ in $\varphi_n(x)$ and shown that in fact $L=\pi$? Not sure where I went wrong. Any help is greatly appreciated.

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    $\begingroup$ Your first equality for $| \varphi_n(x) - \varphi(x) | $ looks wrong, as does the subsequent equality. You will probably find an error if you try to work out a detailed derivation of that step. $\endgroup$
    – Erick Wong
    Commented Dec 12, 2015 at 18:49
  • $\begingroup$ You're right. My bad. Thanks. $\endgroup$
    – Andrew V
    Commented Dec 12, 2015 at 18:54
  • $\begingroup$ You should let $E=(0,1]$, not $[0,1]$ as $\psi (0)$ does not exist. $\endgroup$ Commented Dec 12, 2015 at 19:36

2 Answers 2

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Alternative proof:

\begin{align} \lim_{x\rightarrow0}\frac{f(x)-f(x/2)}{x/2}&=L\\ &=\lim_{h\rightarrow0}\frac{f(2h)-f(h)}{h}\\ &=\lim_{h\rightarrow0}\frac{f(2h)-f(0)+f(0)-f(h)}{h}\\ &=\lim_{h\rightarrow0}2\frac{f(2h)-f(0)}{2h}-\frac{f(h)-f(0)}{h}\\ &=2f'(0)-f'(0)\\ &=f'(0) \end{align} Therefore $f'(0)=L$.

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    $\begingroup$ Well done! ... +1 $\endgroup$
    – Mark Viola
    Commented Dec 12, 2015 at 18:52
  • $\begingroup$ Wow. Sorry, given my professor's track record I would never have imagined the answer might be this easy. Thanks. $\endgroup$
    – Andrew V
    Commented Dec 12, 2015 at 18:54
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    $\begingroup$ In fact this shows that we may dispense with the assumption that the limit $L$ already exists. $\endgroup$
    – Erick Wong
    Commented Dec 12, 2015 at 19:42
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    $\begingroup$ @AndrewV Does this actually answer the question? $\endgroup$
    – user541686
    Commented Dec 12, 2015 at 19:52
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I think you forgot an 1/x in the difference of the two functions when you start proving uniform convergence

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