How can a convex increasing curve have an horizontal asymptote? I read that in a paper, but I just can't see how this is possible.
The situation is the the following: $a,b$ are some constants, we have a differential equation
$z^{'}(t)=(a+b z^3) f(t)$
where $f(t)$ is known.
Let $z_0$ be the initial value for $t_0$.
By seperation of variables:
$F(z)= \int\limits_{z_0}^z \dfrac{ds}{a+b s^3} = \int\limits_{t_0}^t f(\tau) d \tau =X(t)$
and we get a solution $z = F^{-1}(X)$.
Now the paper says: The graph of $F(z)$ in the sector $z \geq 0, X \geq 0$ of the Cartesian plane $(z,X)$ is a convex curve tarting at $(t_0 ,z_ 0)$ and increasing towards the asymptote
$M= \int\limits_{z_0}^{\infty} \dfrac{ds}{a+b s^3}$
I also don't understand why this is a convex graph.
If $f$ is a convex non-decreasing curve, and $f$ has a horizontal asymptote given by the line $y=c$ so that $\lim_{x\rightarrow\infty}f(x)=c$, then $f$ is a horizontal line.
Proof:
The line segment between two points on a convex curve has to lie above or on the curve.
Since $f$ is non-decreasing, there does not exist an $x$ such that $f(x)>c$.
Suppose that there exists a point $x$ such that $f(x)<c$. The height of the midpoint of the line segment from $(x,f(x))$ to $(x',f(x'))$ is at most $\frac{f(x)+c}2$. As the curve cannot go above this line segment, $f\left(\frac{x+x'}2\right)\leq \frac{f(x)+c}2$. No matter how large $x'$ gets, the curve never goes above $\frac{f(x)+c}2$ and can never approach the line $y=c$.
Therefore $f(x)=c$ for all $x$.