Proving the concavity of a function constructed from some monotonicity functions

Question

The question is, consider $x\in[0,1]$ as the function's feasible domain and I have two monotone increasing functions $p(x):[0,1]\rightarrow [0,1]$ and $q(x):[0,1]\rightarrow [0,1]$. Also, assume the function $p(x)\cdot x - q(x)$ is monotone increasing. Can we prove that $f(x): = p(x) - q(x)$ is a concave function over the feasible domain $x\in[0,1]$?

Note that we don't impose any convexity assumptions on the functions $p(x)$ or $q(x)$, you can image them as some highly nonconvex functions as long as they satisfy the monotonicity assumptions, i.e., $p(x)$, $q(x)$ and $p(x)\cdot x - q(x)$ are monotone increasing.

Here are some of my thoughts: because there are no convexity assumptions on either $p(x)$ or $q(x)$, I tend to believe we cannot prove $f(x): = p(x) - q(x)$ is a concave function. However, I am having a hard time finding such a counter-example. The reason is as follows, suppose we want $f(x)$ to be convex, which is not concave, then we might have $q(x)$ as a concave function, for example, $q(x) = \sqrt{x}$. In this case, we have $q'(x)=\frac{1}{2\sqrt{x}}$ and it is impossible to have $p(x)\cdot x - q(x)$ monotone increasing condition being satisfied, i.e., $p'(x)\cdot x + p(x) \ge q'(x)$ since $q'(x) = \infty$ when $x =0$.

It seems necessary for $q(x)$ to be a convex function in this case, which kind of indicates $f(x): = p(x) - q(x)$ will be a concave function. But a formal proof seems hard to get given only these three monotonicity conditions. On the other hand, a counter-example seems to require some highly non-trivial non-convex functions which are difficult to come up with by hand, since some naive convex/concave examples do not seem to work

Answer 1

I think that I have a counter example. Let us take

$$\begin{cases} p(x) :=e^{x} \\ q(x) := x \end{cases}. $$

You have that the monotonicity conditions for both functions are satisfied. You also have that the function $f(x): =xp(x)-q(x) $ is increasing since

$$\begin{cases} f'(x) :=e^{x} (1+x)-1>0, \forall x\in [0,1]\\ f(0) := 0\end{cases}. $$

However, you have:

$$p''(x)-q''(x) = e^x>0. $$

The function is then convex. Hope it helps !

Answer 2

For a counterexample, if $p,q:[0,1]\to[0,1]$ are given by $$ \begin{cases} p(x)=\frac{1}{3}x^2+\frac{1}{3}x+\frac{1}{3}\\[4pt] q(x)=\frac{1}{4}x^2+\frac{1}{4}x\\ \end{cases} $$ then the functions $p,q$ are strictly increasing on $[0,1]$, as is the function $$ g(x)=x{\cdot}p(x)-q(x)=\frac{1}{3}x^3+\frac{1}{12}x^2+\frac{1}{12}x $$ but the function $$ f(x)=p(x)-q(x)=\frac{1}{12}x^2+\frac{1}{12}x+\frac{1}{3} $$ is not concave (in fact it's strictly convex).