How to prove the relationship of two simple functions? Given two functions $f(x)>0$ and $g(x)>0$ on $x\in[a,b]$, if they satisfy the following constraints:
\begin{cases}
f''(x) >0 \\
g''(x) = 0\\
f(a)=g(a)\\
\int_a^bf(x)\mathrm{d}x= \int_a^bg(x)\mathrm{d}x
\end{cases},
then the following conclusions hold: 


*

*$f(x)$ and $g(x)$ must have and only have one intersection on interval $(a,b)$. 

*If we denote the x-coordinate of the intersection with $s$, then: 
\begin{cases}
f(x)<g(x), \qquad \mathrm{if}\quad x \in (a,s);\\
f(x)>g(x),\qquad \mathrm{if}\quad x \in (s,b)
\end{cases}


But how to prove it?
Can you help me?
 A: Here's an idea. Notice that if there is never a point in the interval $[a,b]$ at which $f(x)>g(x)$, then the area under both curves could not be equal.
Furthermore, since these two curves have the same area under the curves in the interval $[a,b]$, it should be obvious that 
$\int_a^s g(x)-f(x)dx=\int_s^bf(x)-g(x)dx$
This means that portion of $g(x)$ above $f(x)$ is equal to the portion of $f(x)$ above $g(x)$ on the interval $[a,b]$ which holds because $f(x)$ has to "make up for" the area it was under $g(x)$.
A: $\int_{a}^{b}f(t) dt = \int_{a}^{b} g(t) dt$
$\int_{a}^{b} f(t)-g(t) dt=0$
We know that $f$ and $g$ are differentiable, hence they must be continuous. 
If $f(t)$ and $g(t)$ do not intersect, then $f(t)-g(t)$ does not change sign, then $f(t)-g(t)>0$ or $f(t)-g(t)<0$ everywhere in $(a,b)$ and the integration will not give us $0$, which is a contradiction.
Also note that $(f-g)''>0$, hence $(f-g)'$ increases. We know that $(f-g)(a)=0=(f-g)(s)$, hence $\exists w \in (a,s), (f-g)'(w)=0$. 
Since $(f-g)'$ increases, $\forall x \in (a,w), (f-g)'(x)<0  \implies \forall x \in (a,w), (f-g)(x) \leq  (f-g)(a)=0$. 
Similarly, $\forall x \in (w,s), (f-g)'(x)>0  \implies \forall x \in (w,s), (f-g)(x) \leq  (f-g)(s)=0 $. 
Hence $\forall x \in (a,s), (f-g)(x) \leq 0.$
Again, using the fact that $\forall x \in (s,b), (f-g)'(x)>0$, we can conclude that $\forall x \in (s,b), (f-g)(x) \geq (f-g)(s)=0$
A: The second condition really just says that $g$ is an affine function, i.e., a straight line. The first says that $f$ is convex, meaning it looks something like the left side in this image.
We have $f(a)=g(a)$. Now, it's easy to draw a picture of a straight line and a convex function which start from the same point, and don't intersect at all afterwards. But in that case, either $f>g$ on the whole interval, or $f<g$ on the whole interval. In either case, we definitely don't have $\int_a^b f = \int_a^b g$!
If the areas under the two graphs are to have any chance at being equal, then a necessary (but not sufficent) condition is that the convex function $f$ is sometimes under and sometimes over the straight line $g$. In fact, it's easy to see that $f$ must start out less than the straight line $g$, then rise up and cut through it at some point. It's also easy to see that they can't intersect after that point. Now it's just a matter of converting what's visually obvious into a rigorous proof. Here's an outline:


*

*Prove carefully that $g$ has constant derivative and that $f$ has strictly increasing derivative.

*Let $m$ be the (constant) derivative of $g$. Prove that if $f'(a)\geq m$, then $f>g$ on $(a, b]$. Conclude that $f'(a)<m$. Use that to prove that there exists an interval $(a, a+\epsilon]$ on which $f<g$.

*If $f$ never intersected $g$, we would therefore have $f<g$ on the whole interval and their integrals could not be equal. Let $s=\inf\{x\in(a, b)|f(x)=g(x)\}$, that is, $s$ is the first time at which $f$ and $g$ meet. Prove rigorously that $f(s)=g(s)$.

*Prove that $f'(s)>m=g'(s)$, in other words, when the convex function $f$ cuts  through the straight line $g$ from below, it must be steeper than the straight line, which is visually obvious. Conclude that $f>g$ on $(s, b]$ and that therefore $s$ is the unique intersection point of $f$ and $g$.

A: For the first question I think that the Brouwer fixed-point theorem would be enough to demonstrate it, because every point of $f(x)$ has a unique point of $g(x)$ for the same interval in the conditions that the theorem states. That means that they will cross in a single point. This is an excerpt from Wikipedia:
One-dimensional case


"In one dimension, the result is intuitive and easy to prove. The continuous function $f$ is defined on a closed interval $[a, b]$ and takes values in the same interval. Saying that this function has a fixed point amounts to saying that its graph (dark green in the figure) intersects that of the function defined on the same interval $[a, b]$ which maps $x$ to $x$ (light green).
Intuitively, any continuous line from the left edge of the square to the right edge must necessarily intersect the green diagonal. Proof: consider the function $g$ which maps $x$ to $f(x)$ - $x$. It is $\ge 0$ on $a$ and $\le 0$ on $b$. By the intermediate value theorem, $g$ has a zero in $[a, b]$; this zero is a fixed point."

