An inequality involving concave and monotone functions

[Rigel1]

Let $f: [0,1]\to RR$ be a concave function satisfying $f(0)=0$, and let $\phi: [0,1]\to RR$ be a monotone non-decreasing function.
Prove that
$\int_0^1 f(t) \phi(t) dt \leq 2[\int_0^1 t \phi(t)dt] \int_0^1 f(t) dt.$

[Leonardo891]

This

"Rigel":

Now, integrating the inequality on $[0,1]$, you get
$\int_0^1 f(t)[\phi(t)-A]dt \le \frac{f(t_0)}{t_0} \int_0^1 t [\phi(t)-A]dt = \frac{f(t_0)}{t_0} [ \int_0^1 \t \phi(t)\dt - A \int_0^1 t dt] = 0$,
since by definition $\int_0^1 t\phi(t)dt = A/2$.

is what was missing to me. Thanks very much for the problem and the solution.
When I will have time I'll try to solve your other problem "Approximation of convex functions".

[Rigel1]

"Leonardo89":Hello!
[quote="Rigel"]$\frac{f(t_0)}{t_0} t(\phi(t)-A) \le 0$ for every $t\in [0,1]$

How can you say this? You should have $f(t_0)>0$ in $]0,t_0[$ and $f(t_0)<0$ in $]t_0, 1 [ $, no?
Am I wrong? ?
[/quote]
Consider the case $t\in [0, t_0)$.
You have that $f(t) \ge \frac{f(t_0)}{t_0} t$.
If you multiply both sides of this inequality by the negative number $\phi(t) - A$, you have to reverse the inequality sign, so that
$f(t) (\phi(t)-A) \le \frac{f(t_0)}{t_0} t (\phi(t)-A).$
In the case $t>t_0$ the number $\phi(t) - A$ is positive, hence the inequality sign remains the same.

Now, integrating the inequality on $[0,1]$, you get
$\int_0^1 f(t)[\phi(t)-A]dt \le \frac{f(t_0)}{t_0} \int_0^1 t [\phi(t)-A]dt = \frac{f(t_0)}{t_0} [ \int_0^1 \t \phi(t)\dt - A \int_0^1 t dt] = 0$,
since by definition $\int_0^1 t\phi(t)dt = A/2$.

[Leonardo891]

Hello!

"Rigel":$\frac{f(t_0)}{t_0} t(\phi(t)-A) \le 0$ for every $t\in [0,1]$

How can you say this? You should have $f(t_0)>0$ in $]0,t_0[$ and $f(t_0)<0$ in $]t_0, 1 [ $, no?
Am I wrong? ?

... strictly increasing/decreasing...

Thanks.

[Rigel1]

Hello Leonardo!

From hint 1 you know that $\phi(t) - A < 0$ in $[0,t_0)$, whereas $\phi(t) -A > 0$ in $(t_0, 1]$.
Now we can use hint 2, obtaining
$f(t)(\phi(t)-A) \le \frac{f(t_0)}{t_0} t(\phi(t)-A) \le 0$ for every $t\in [0,1]$.
(You have to consider the two cases $tt_0$ separately, but in both cases you obtain the same inequality.)

Now it is enough to integrate both sides of the inequality obtained above.

"Leonardo89":
I proved also that $f(t)$ is strictly crescent/decrescent in $]t_0,1]$ if $f(t_0)>0$/$f(t_0)<0$.

... strictly increasing/decreasing...

[Leonardo891]

Hello Rigel. I was a little busy because exams and other.
I put what I did in a spoiler but it wouldn't be necessary, it's almost nothing.

I tried to use hint 3 (i already thought to it) but i solved the problem only in case that $f(t_0)=0$. Very miserable. However using the two inequalities of hints 1 and 2 I obtained:

$f(t)(\phi(t)-A)<=\frac{f(t_0)}{t_0} t(\phi(t)-A)=:h(t)$

Now, if $f(t_0)=0$ it's done. It's also done if $f(t)$ is constant in $[0,t_0]$ (this implies $f(t_0)=0$). Let us assume $f(t_0)>0$.
The function $h(t)$ is strictly negative in $]0,t_0[$ and strictly positive in $]t_0,1[$.
If, instead, $f(t_0)<0$ is the inverse: the function $h(t)$ is strictly positive in $]0,t_0[$ and strictly negative in $]t_0,1[$.
I proved also that $f(t)$ is strictly crescent/decrescent in $]t_0,1]$ if $f(t_0)>0$/$f(t_0)<0$.
However if the negative area of $h(t)$ prevail in one case it can't prevail in the other one because only the sign change between the two case. Then i can't use $h(t)$ to prove that the integral is negative.
It seems to me that this road is unproductive.

Please, decide you if give me the solution or another hint (or correct my surely many mistakes ).
And if my English is wrong, feel free of doing corrections.

[Rigel1]

Another hint for Leonardo
(You can safely read this one...)

Hint nr. 3:

Combine hints nr. 1 and 2.

[Leonardo891]

Hello Rigel!
I like this problem and I'm trying to solve it!
Then, can you suspend hints for a while, please (otherwise I would be too much tempted of use them)?
And sorry for English mistakes, in case.

[Rigel1]

Hint nr. 2:

Since $f$ is concave and $f(0) = 0$, we have that
$f(t) \ge \frac{f(t_0)}{t_0} \cdot t$, if $t\in [0,t_0]$,
$f(t) \le \frac{f(t_0)}{t_0} \cdot t$, if $t\in [t_0, 1]$.

[Rigel1]

Hint nr. 1:

Let $A:= 2 \int_0^1 t\phi(t)dt$. We have to prove that
(1) $\int_0^1 f(t) [\phi(t) - A] dt \le 0$.
If $\phi$ is constant in $(0,1)$ then $\phi(t) = A$ for every $t\in (0,1)$ and the inequality is trivial.
So, let us assume that $\phi$ is not constant in $(0,1)$.
Since
$\phi(0) \int_0^1 tdt \le \int_0^1 t\phi(t)dt\le \phi(1)\int_0^1 t dt$,
and $\phi$ is not constant in $(0,1)$, there exists $t_0\in (0,1)$ such that
$\phi(t) < A$ for every $t\in [0,t_0)$, while $\phi(t) > A$ for every $t\in (t_0, 1]$.
(In fact we do not need strict inequalities here, but they turn out to be useful in order to discuss the equality case in (1).)