# Mean values theorem and countable sets

The mean values theorem says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u)$$ Assume that $f$ is a non-zero analytic function in the whole real line. My interest is about the real $c∈(u,v)$. I believe that the set of those $c$ is countable, otherwise we conclud that $f′(c)$ is constant (in an open set containing $c$) since $v$ is a constant and hence $f$ is identically zero. Also, I think that the set of those $c$ is finite in number or as equivalent classes (for analytic functions) but I am not able to prove that.

I can define the equivalence relation $ℜ$ by $$cℜd⇔f^{(1)}(c)=((1-t)/(1-u)).f^{(1)}(d)$$

where $u∈ℝ$ such that $g(u)=0$, $c∈(u,v)$ and $t∈ℝ$ such that $g(t)=0$, $d∈(t,v)$

• Check $f = \operatorname{id}$. – Lord_Farin Apr 26 '13 at 9:47

Zeroes of a non zero analytic function are isolated, and because $\mathbb{R}$ is separable, a discrete set is at most countable.
Therefore, if $\{c \in (u,v) \mid f(v)=f'(c)(v-u) \}$ is uncountable, $f'$ is identically equal to $f(v)/(v-u)$. In particular, $f$ is linear.
• @ Seirios: How can prove that the set of those $c$ is finite in number or as equivalent classes. – Germany Apr 26 '13 at 10:02
• @ Seirios: This is the question: Define that equivalence relation. I guess that its expression is related to this equation $f(v)=f(u)+f′(c)(v-u)$. – Germany Apr 26 '13 at 11:04
• Without additional conditions, you can define $x \mathcal{R} y$ for all $x,y$; then there is only one equivalent class. – Seirios Apr 26 '13 at 11:07
• @ Seirios: The relation can be defined by: $c∼d$⇔ $g′(c)=(((v-t)/(v-u)))g′(d)$ where $c∈(u,v)$ and $d∈(t,v)$ – Germany Apr 26 '13 at 11:09