How many solutions does equation $g(c-x) = ax+b$ have?

Question

Let $g: \mathbb{R} \to \mathbb{R}$ be convex and increasing function. How many solutions does equation $g(c-x) = ax+b$ have? where $a, b, c$ are constant.

I think the answer is at most two if $g$ is not linear function.

Answer 1

Consider $g(x) = x + \max\{0,x\}$. This is stricly increasing and convex, but not strictly convex. Then, $$g(c - x) = -2 \, x + 2\,c$$ for $c-x \ge 0$ and $$g(c-x) = -x +c$$ for $c-x \le 0$.

Hence, you have infinitely many solutions for $$a = -2, b = 2 \, c$$ and $$a = -1, b = c.$$

Answer 2

Yes, there is a maximum of two intersections.

One function can go above and then below the other function if the second derivative is decreasing.

One function can go below and then above the other function if the second derivative is increasing.

We cannot get a third intersection, because that would require a switch of the concavity of the function.

Answer 3

First, I'm not sure why we want to solve $g(c-x) = ax+b$ instead of $g(x) = ax+b$. Indeed, we could make the substitution $y=c-x$ in the first (given) equation to get $$ g(y) = a(c-y)+b = (-a)y + (b+ac) $$ which is of the form I mentioned above, so unless there are any bounds on $a$, $b$, or $c$, Let's solve $g(x) = ax+b$, where $g$ is strictly increasing and convex.

Suppose there were three solutions $x_1 < x_2 < x_3$. Since $g$ is strictly increasing, we must have $$ g(x_1) < g(x_2) < g(x_3) ,$$ which is $$ ax_1 + b < ax_2 +b < ax_3 + b ,$$ which means that $a > 0$. Now since $g$ is strictly convex, we have that for any $t \in (x_1, x_3)$, $$ g(t) < \frac{x_3-t}{x_3-x_1} g(x_1) + \frac{t-x_1}{x_3-x_1} g(x_3) .$$ In particular, for $t=x_2$, we have $$ g(x_2) < \frac{x_3-x_2}{x_3-x_1} g(x_1) + \frac{x_2-x_1}{x_3-x_1} g(x_3) ,$$ $$ ax_2 + b < \frac{x_3-x_2}{x_3-x_1} (ax_1+b) + \frac{x_2-x_1}{x_3-x_1} (ax_3+b) ,$$ $$ x_2 < \frac{x_3-x_2}{x_3-x_1} x_1 + \frac{x_2-x_1}{x_3-x_1} x_3 = x_2 ,$$ a contradiction. So when $g$ is strictly increasing and strictly convex, there can be at most two solutions.

Go through the above proof and see what you find when you replace strictly increasing with increasing, and similarly with convexity, and what edges cases then appear. The main case to expect is the case where $g$ is a linear function. This should appear as a case of the modified proof.