Positive functions $in C^1 $

[Camillo]

Let $f $ and $g $ be two positive functions $in C^1 $ in the $[a,b] $ interval of the real axis.
Show that if $f'(x) >=g(x) , AA x in (a,b) $ then it exists at least a point $ x^* in (a,b) $ such that :

$g(x^*) <= 1/(b-a) log [f(b)/f(a)] f(x^*) $.

Edit : corrected LHS in $g(x^*) $

[Camillo]

"ViciousGoblin":@Paolo90 Your counterexample is good - the proposition written by Camillo is false as stated.
Probably you are rigth in saying that he meant to have $x'$ in the LHS too. In this case you did it right!

Sorry, it was a typing mistake from my side ; the correct LHS is $g(x^*) <= $ etc etc

[Leonardo891]

"ViciousGoblin":s.t. = "such that"

Thanks.

[ViciousGoblin]

"Leonardo89":I agree.
By the way, is s.t. the translation of "tale che"?

s.t. = "such that"

[Leonardo891]

I agree.
By the way, is s.t. the translation of "tale che"?

[ViciousGoblin]

@Paolo90 Your counterexample is good - the proposition written by Camillo is false as stated.
Probably you are rigth in saying that he meant to have $x'$ in the LHS too. In this case you did it right!

[Paolo902]

Thank you for your posts, guys.

"ViciousGoblin":@ Paolo90 I'm afraid Leonardo89 is right
Your argument only shows that $\exists\bar x$ such that
$g(\bar x)\leq\frac{1}{b-a}\ln(\frac{f(b)}{f(a)})f'(\bar x)$.

So I think the challenge is still open ...

I thank you for your post but...

I think that the theorem (written as Camillo did) is false. Indeed, if you say that $g(x)<=\frac{1}{b-a}\ln(\frac{f(b)}{f(a)})f'(\bar x) = "number"$ this means that $g(x)$ is bounded in $[a,b]$: this is certainly true (Weierstrass) but who says that the max is exactly the RHS?
I think I've found a counter-example. Let us consider $f(x)=2x$ and $g(x)=x$ in the interval $[1,2]$.
They're $C^1$ and positive and $f'(x)=2>=x=g(x)$ forall $x in (1,2)$.

If the theorem were true, we would have $g(x)<=1/(2-1)log((2*2)/2)*2=ln4 approx 1.38$. But this is a contradiction.

So i think that the correct version is the one you have written (and I proved in my previous post): we had to prove that there exists a $bar x$ s.t.
$g(\bar x)\leq\frac{1}{b-a}\ln(\frac{f(b)}{f(a)})f'(\bar x)$.

Do you agree with me?

Another (friendly) remark: in english you should say that something "makes" sense.

Thanks a lot, you're right.

[ViciousGoblin]

@ Paolo90 I'm afraid Leonardo89 is right
Your argument only shows that $\exists\bar x$ such that
$g(\bar x)\leq\frac{1}{b-a}\ln(\frac{f(b)}{f(a)})f'(\bar x)$.

So I think the challenge is still open ...

Another (friendly) remark: in english you shoul say that something "makes" sense.

@Leonardo89 RHS= "Right Hand Side" (I guess)

[Leonardo891]

Hi Paolo90 . Sorry for the delay but in this period I'm a little busy with the exams. I still have my doubt.

"Paolo90":using the fact that $f'(x)>=g(x)$, $forall x in (a,b)$ you easily get the thesis

I don't think. You only get $g(x^*) <= f'(x^*) = 1/(b-a) log [f(b)/f(a)] f(x^*) $ not
$ AA x \in (a,b), g(x) <= f'(x) <=f'(x^*) = 1/(b-a) log [f(b)/f(a)] f(x^*) $ because you don't know if it's true that $ AA x \in (a,b), f'(x) <= f'(x^*) $.

However, what's RHS?

[Paolo902]

"Leonardo89":@Paolo90
I have a doubt but maybe I'm wrong.

Thank you for your observation, Leonardo. But...

Note that in RHS there is a $\barx$ in the argument of $f$ (Camillo didn't use the $barx$: he used instead $x^*$).
Then, using the fact that $f'(x)>=g(x)$, $forall x in (a,b)$ you easily get the thesis.

Am I wrong?

Thanks.

[Leonardo891]

@Paolo90
I have a doubt but maybe I'm wrong.

The point $barx$ is only one point. How can you determine from a single point an inequality $AA x \in (a,b)$?
You can do this if $barx$ is the maximum of $f'(x)$ but is this true?

[Paolo902]

Good exercise, thanks Camillo. Hope my solution is correct.

The main idea is Lagrange Theorem.

Indeed, since $f(x)$ is a positive function [tex]\in C^1( [a,b] )[/tex] we can consider the following composed function: $h(x)=log(f(x))$ (it has sense, since $f$ is positive).
$h(x)$ is a Lagrange function in $(a,b)$ (i.e. it satisfies the Lagrange's theorem hypotesis) therefore we can say that there exists a point $barx in (a,b)$ s.t.
$h'(barx)=(h(b)-h(a))/(b-a)$

Now, $h'(x)=1/(f(x))f'(x)$: then the last equality becomes
$1/(f(barx))f'(barx)=log((f(b))-log(f(a)))/(b-a)$

Using the property of logarithms we get
$f'(barx)=f(barx)/(b-a)log(f(b)/(f(a)))$

Using the inequality given in hypotesis we conclude the proof.[tex]\square[/tex]

Ok?