Quasi-concavity of profit function

Question

I'm doing a simple exercise from my textbook: suppose the revenue function is $R(p)=p^{1-\epsilon}$ with $\epsilon > 0$. Suppose the cost function is convex. Show that the profit function is quasi-concave if $\epsilon > 1$.

My attempt: Let the demand function then be $q(p)=p^{-\epsilon}$ and the cost function some $c(q(p))$.

The first-order condition is $(1-\epsilon)p^{-\epsilon}-c'(q(p))q'(p)$ and the second order condition $-\epsilon(1-\epsilon)p^{-\epsilon-1}-c''(q(p))q'(p)^2-c'(q(p))q''(p)$.

The second term $-c''(q(p))q'(p)^2$ and third terms $-c'(q(p))q''(p)$ are negative under convexity of the cost function. Hence the second order condition is unambiguously negative if the first term is negative. But that happens for $\epsilon<1$, the opposite as conjectured.

Answer 1

Your attempt is almost correct.

Let the demand function be $D(p) = p^{-\epsilon}$ and the cost function some $C(D(p))$ so that $R(p) = D(p)p = p^{1-\epsilon}$, right?

Then, I'd rewrite your first and second order conditions - as you only provided the expressions but it would be great to write them in a form of equations (or inequalities). So that:

First order condition: $R'(p) - C'(D(p))D'(p) =0$

And the second order: $R''(p) - C''\left(D(p)\right)\cdot D'(p)^2 - C'(D(p))D''(p) < 0$

This is the moment when we kind of split our approaches. I also do agree that the second term in second order condition is negative. However, it is not necessarily $R''(p)$ that is negative - rather the whole relation $$R''(p) - C'(D(p))D''(p)<0 \;(*)$$ Now, we want the second order condition to hold for every $p$ that satisfies the first order condition. Thus, I would substitute the $C'(D(p))$ from the first condition to the $(*)$ and hence, I end up with: $$R''(p) - \frac{R'(p)D''(p)}{D'(p)} <0 \;(**)$$

Now it's time for the brute force. Calculate all derivatives in $(**)$.

$$ -\epsilon(1-\epsilon)p^{-1-\epsilon} - \frac{(1-\epsilon)p^{-\epsilon} \cdot -\epsilon(-1-\epsilon)p^{-2-\epsilon}}{-\epsilon p^{-1-\epsilon}} <0 \implies \epsilon > 1$$ for both, $p$ and $\epsilon$ positive.

P.S. If you think about it, letting $\epsilon<1$ would imply that $R'(p)>0$ and thus, the first order condition would have no solutions. Why?

P.P.S. I believe a similar exercise is in Industrial Organization: Markets and Strategies by Belleflamme and Peitz in their chapter about "monopoly pricing strategies" but with the inverse demands $P(q)$.