Hi everyone,

I have a kind of big function problem i have to do... But since i have headache, i've decided to ask some help here. Of course, i do not want the whole answer; just some hints or clues so that i'll be able to finish it by myself

(I want also to apologize if i'm not writing in the good section of the forum :s I may have problems (since i'm not English) to find the exact section i want to post in)

So, here we go. Some definitions first.

Let $\displaystyle f : X \subset \mathbb{R} \to \mathbb{R}$.

$\displaystyle \forall m \in\mathbb{R}, f_m(x) = mx-f(x)$.

Let also $\displaystyle X^{\circ} = \left \{ m / \exists \lambda \in \mathbb{R}, \forall x \in X, mx-f(x) \leq \lambda \right \}$. Also $\displaystyle X^{\circ}$ is the set (maybe empty) of all the real $\displaystyle m$ so that $\displaystyle mx-f(x)$ is majorized.

Finally, let $\displaystyle \displaystyle f^{\circ}(m) = \sup_{x \in X} (mx-f(x))$

Still here ? No headache ? All right, let's continue.

I've answered a few questions. And particulary, i've prooved that : $\displaystyle \displaystyle \forall x \in X, f(x) \geq \sup_{m \in X^{\circ}} (xm - f^{\circ}(m))$ and that $\displaystyle X^{\circ}$ is an interval of $\displaystyle \mathbb{R}$.

Now, i've the following question :Ok. I've found that, knowing that $\displaystyle f$ is limited on $\displaystyle X$, $\displaystyle X^{\circ} = \mathbb{R}$. And now is the problem.We assume that $\displaystyle X$ is a segment of $\displaystyle \mathbb{R}$ and that $\displaystyle f$ is continuous on $\displaystyle X$. Derterminate $\displaystyle X^{\circ}$ and proove that $\displaystyle \forall m \in X^{\circ}, \exists x_m \in X, f\left ( x_m \right ) = mx_m - f^{\circ}(m)$.

I already know that $\displaystyle \displaystyle \exists x_m \in X, f(x_m) = \max_{x \in X} f(x) $ and i think that i have to show that $\displaystyle \displaystyle\max_{x \in X} f(x) = mx_m - f^{\circ}(m)$.

I assume that there must be some connections with the previous result : $\displaystyle \displaystyle \forall x \in X, f(x) \geq \sup_{m \in X^{\circ}} (xm - f^{\circ}(m))$ but which one ?

Maybe i'll have to use some tricks i don't quite reckognize :/ So, some clues here are welcome :') (i repeat : i don't want the answer ! ^^)

After that (hehe... what do you think ? i haven't finished yet :P ), there are other definitions :

We define $\displaystyle \left ( X^{\circ\circ}, f^{\circ\circ} \right )$ from $\displaystyle \left ( X^{\circ}, f^{\circ} \right )$ as we previously define $\displaystyle \left ( X^{\circ}, f^{\circ} \right )$ from $\displaystyle \left ( X, f \right )$.

I.E.

$\displaystyle \displaystyle \forall x \in X^{\circ\circ}, f^{\circ\circ}(x)= \sup_{m \in X^{\circ}} f^{\circ}_x(m) = \sup_{m \in X^{\circ}} \left ( xm - f^{\circ}(m) \right ) $.

(You got it ? The headache ? A real nightmare :O )

So. Now. First question of this part :

And now this is becoming difficult. For the first function, there are not few problem for the first part of the question. (I've found $\displaystyle X^{\circ} = [-1,1]$ and $\displaystyle f^{\circ} (m) = 0$). But after that, it's completely blur from my point of view. And it's becoming harder with the second function where we have 4 differents case to study !Determinate $\displaystyle \left ( X^{\circ}, f^{\circ} \right )$ and $\displaystyle \left ( X^{\circ\circ}, f^{\circ\circ} \right )$ with $\displaystyle X = \mathbb{R}$ and $\displaystyle f(x)= |x|$. Same question with $\displaystyle X = \mathbb{R}$ and $\displaystyle f(x)= |1-|x||$ .

Well, i've heard some friends telling that we have to use (i'm not quite sure if the word exists or is correct) a logarithmic differential. I mean, set $\displaystyle \ln |f|$ and differentiate so that we have $\displaystyle \frac{f'}{f}$. But i don't quite see where we are going.

Well, now i've finished :') Thank you very much for reading me, i know it mustn't have been very fun since it could be boring and my english isn't perfect :/

I'll keep searching too and if i found something, i'll tell you hear so that you won't keep searching over and over !

Again, thank you very much for helping me and giving me some hints ! I think this is the best way i could make some progress.

Hugo.