# Proving the concavity of a function constructed from some monotonicity functions

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

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 !

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).