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Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function, $A \subset \mathbf{R}$ be a closed set and $x_0, y \in \mathbf{R}$. Assume that for each $\varepsilon >0$ there exists $\delta >0$ such that

  1. if $x \in A$ , $|x-x_0|< \delta$ then $\left|\frac{f(x)-f(x_0)}{x-x_0}-y\right|< \varepsilon$,

  2. if $x \notin A$, $|x-x_0|<\delta$ then $f'(x)$ exists and $|f'(x)-y|<\varepsilon$.

How to prove that there exists $f'(x_0)$ and $f'(x_0)=y$.

Thanks.

P.S. My question concerns Lemma 1 on page 66 from the paper:

H. Whitney, Analytic extensions of differentiable functions, Trans. Amer. Math. Soc. 36 (1934), 63–89.

(In this paper there is no proof of Lemma 1.)

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Just to start with: Can you prove this for the special cases $A=\mathbf{R}$ and $A= \emptyset$ ? –  leonbloy Oct 21 '11 at 18:27
    
In case $A=\emptyset$ we have $f′(x_0)=lim_{x→x0} \frac{f(x)−f(x_0} {x−x_0}=lim_{x→ x_0} f'(x)=y$ by continuity of $f$ and de L'Hospital rule. In case $A=\mathbf{R}$ we have $f'(x_0)=y$ by definition of derivative. –  Richard Oct 21 '11 at 20:30

1 Answer 1

up vote 5 down vote accepted

Pick $\epsilon>0$ and choose $\delta>0$ that satisfies 1 and 2 above.

Suppose $x_0\not\in A$. Then, since $A$ is closed, there is a positive $\delta'\le\delta$ so that if $|x-x_0|<\delta'$, $x\not\in A$. Then, the Mean Value Theorem says $$ \left|\frac{f(x)-f(x_0)}{x-x_0}-y\right|=\left|f'(\xi)-y\right|<\epsilon\tag{1} $$ for some $\xi$ between $x$ and $x_0$, and therefore, $\xi\not\in A$.

Suppose $x_0\in A$ and $|x-x_0|<\delta$. If $x\in A$, then $$ \left|\frac{f(x)-f(x_0)}{x-x_0}-y\right|<\epsilon\tag{2} $$ If $x\not\in A$, let $a$ be the point in $A$ closest to $x$ so that $|x-a|+|a-x_0|=|x-x_0|$; that is, either $a=x_0$ or $a$ is between $x$ and $x_0$.

If $a=x_0$, then no point between $x$ and $x_0$ is in $A$ and the Mean Value Theorem then says $$ \left|\frac{f(x)-f(x_0)}{x-x_0}-y\right|=\left|f'(\xi)-y\right|<\epsilon\tag{3} $$ for some $\xi$ between $x$ and $x_0$, and therefore, $\xi\not\in A$.

If $a$ is between $x$ and $x_0$, then because no point between $x$ and $a$ is in $A$, $$ \left|\frac{f(x)-f(a)}{x-a}-y\right|=\left|f'(\xi)-y\right|<\epsilon\tag{4} $$ for some $\xi$ between $x$ and $a$. Furthermore, since $a\in A$, $$ \left|\frac{f(a)-f(x_0)}{a-x_0}-y\right|<\epsilon\tag{5} $$ Since $|x-a|+|a-x_0|=|x-x_0|$, $$ \begin{align} &\left|(x-x_0)\left(\frac{f(x)-f(x_0)}{x-x_0}-y\right)\right|\\ &=\left|(x-a)\left(\frac{f(x)-f(a)}{x-a}-y\right)+(a-x_0)\left(\frac{f(a)-f(x_0)}{a-x_0}-y\right)\right|\\ &<\epsilon|x-a|+\epsilon|a-x_0|\vphantom{\frac{f(x)-f(x_0)}{x-x_0}}\\ &=\epsilon|x-x_0|\vphantom{\frac{f(x)-f(x_0)}{x-x_0}} \end{align} $$ Therefore, $$ \left|\frac{f(x)-f(x_0)}{x-x_0}-y\right|<\epsilon\tag{6} $$ In conclusion, $(1)$, $(2)$, $(3)$, and $(6)$ cover all cases, and since $\epsilon>0$ was arbitrary, we get that $f'(x_0)=y$.

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Very thanks for answer. –  Richard Oct 22 '11 at 12:08

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