4
$\begingroup$

Excuse me for the bad title, here's the question

Given a differentiable function defined on R. For a given number $a$, $\forall x\in \mathbb R, x\neq a$, by mean value theorem, there exists a $\xi$ between $a$ and $x$ such that $f'(\xi)=\frac {f(x)-f(a)}{x-a}$. Suppose it's unique for all $x\neq a$. Therefore $\forall x\neq a$ we define $f_a$ as $f_a(x)=\xi$. Then is $f_a$ continious?

$\endgroup$
9
  • 3
    $\begingroup$ How can there be a unique number between two unequal real numbers? $\endgroup$ Nov 21 '14 at 2:49
  • $\begingroup$ Maybe what you mean is $\forall x \in \mathbb R, x\neq a\,\exists\,\xi$ between $a$ and $x$ such that $f_a(x)=\xi$? $\endgroup$ Nov 21 '14 at 2:51
  • $\begingroup$ I'm supposing so $\endgroup$
    – pxc3110
    Nov 21 '14 at 2:53
  • $\begingroup$ Is $f_a$ defined at the point $a$? In particular does one need to show that $f_a$ is continous on all of $\mathbb{R}$ or on $\mathbb{R}\backslash\{a\}$? $\endgroup$
    – wondermech
    Nov 21 '14 at 2:53
  • $\begingroup$ Any suggestions? $\endgroup$
    – pxc3110
    Nov 21 '14 at 2:57
0
$\begingroup$

The mean value theorem says that for any $x\ne a$, there exists $\xi \in [a,x]$ such that $$ \frac{f(x)-f(a)}{x-a} = f'(\xi). $$ I believe the question is: if the choice of $\xi$ is unique, then if we define $f_a(x)$ to be that $\xi$ for a given $x$, then is $f_a$ continuous? I also assume that "differentiable" means "the derivative exists and is continouus".

To me it seems that the assumption that the $\xi$ is always unique implies that $f'$ is monotonic. If that's true, then $(f')^{-1}$ exists and is continuous, and we have $$ f_a(x) = (f')^{-1} \bigg( \frac{f(x)-f(a)}{x-a} \bigg). $$ As a composition of continuous functions, this is clearly continuous everywhere except $x=a$. If we further define $f_a(a)=a$, then it matches its limit there and hence is continuous as well.

Can anyone (dis)prove the claim that $f'$ is monotonic under the uniqueness assumption?

$\endgroup$
1
  • $\begingroup$ I agree that you've shown that no ray emanating from $(a,f(a))$ can intersect the graph in two other points. What's the proof that this property implies that $f$ is concave (or convex) everywhere? $\endgroup$ Nov 21 '14 at 5:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.