On page 52 of What Sustains Social Norms and How They Evolve?, Assumption 1 is as follows:

Assumption 1: $d$ is continuously differentiable, $d'(x) \ge 0$ for all $x<0$ and $d'(x) \leq 0$ for all $x>0$ (it follows that $d'(0)=0$).

I don't understand the use of $\ge$ and $\le$ here. If $d'(0)=0$ implies a maximum is obtained at $x=0$, then every $x$ to the left of $0$ should have a strictly positive derivative and every $x$ to the right of $0$ should have a strictly negative derivative (i.e., $>$ and $<$ should be used).

What is the meaning behind $d'(x) \ge 0$ and $d'(x) \le 0$?

Edit: $d(g-\eta)$ is a function representing the disutility from social disapproval, where $g$ is a tip at a restaurant as a percentage of the bill and $\eta$ is the norm tip. The consumer takes as given $\eta$ (i.e., the social norm) and chooses $g$ to minimise the social disutility $d(x)$; i.e, bring it closer to $0$, where $x=g-\eta$.

Note: there is a trade-off to minimising $d$ that is probably not necessary to understand for my question.

  • $\begingroup$ The question would be somewhat more self contained if you were to include what $d$ is meant to describe. $\endgroup$
    – Giskard
    Commented Aug 3, 2022 at 6:46
  • 1
    $\begingroup$ @Giskard I have included a brief description of $d$. $\endgroup$
    – Eli J
    Commented Aug 3, 2022 at 6:57

1 Answer 1


Assumption 1 ensures that $x=0$ is a maximum, but perhaps not the only maximum. It is possible that function does not have a 'peak' at 0, but a plateau. E.g.;

$$ f(x) = \left\{ \begin{array}{cl} -(x+1)^2 & \text { if } \ x < -1 \\ 0 & \text { if } \ - 1 \leq x \leq 1 \\ -(x-1)^2 & \text { if } \ 1 < x \end{array} \right. $$

enter image description here

For reference, the Desmos code of the above piece-wise function is

y=\left{x<-1:-\left(x+1\right)^{2},-1\le x\le1:0,1<x:-\left(x-1\right)^{2}\right}


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