<p>Assumption 1 ensures that <span class="math-container">$x=0$</span> is <em>a</em> maximum, but perhaps not <em>the</em> only maximum. It is possible that function does not have a 'peak' at 0, but a plateau. E.g.;</p>
<p><span class="math-container">$$
f(x) = \left\{
\begin{array}{cl}
-(x+1)^2 &amp; \text { if } \ x &lt; -1 \\
0 &amp; \text { if } \ - 1 \leq x \leq 1 \\
-(x-1)^2 &amp; \text { if } \ 1 &lt; x 
\end{array}
\right.
$$</span></p>
<p><br><br></p>
<p><a href="https://i.sstatic.net/V0doe.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/V0doe.png" alt="enter image description here" /></a></p>
<p>For reference, <a href="https://www.desmos.com/calculator" rel="nofollow noreferrer">the Desmos</a> code of the above piece-wise function is</p>
<blockquote>
<p>y=\left{x&lt;-1:-\left(x+1\right)^{2},-1\le x\le1:0,1&lt;x:-\left(x-1\right)^{2}\right}</p>
</blockquote>

CC BY-SA 4.0
2022-08-03T07:30:42.027Z
2022-08-03T10:15:58.257Z
2022-08-03T10:15:58.257Z
Giskard
<p>On page 52 of <a href="https://www.sciencedirect.com/science/article/abs/pii/S016726810300221X" rel="nofollow noreferrer">What Sustains Social Norms and How They Evolve?</a>, Assumption 1 is as follows:</p>
<blockquote>
<p><strong>Assumption 1:</strong> <span class="math-container">$d$</span> is continuously differentiable, <span class="math-container">$d'(x) \ge 0$</span> for all <span class="math-container">$x&lt;0$</span> and <span class="math-container">$d'(x) \leq 0$</span> for all <span class="math-container">$x&gt;0$</span> (it follows that <span class="math-container">$d'(0)=0$</span>).</p>
</blockquote>
<p>I don't understand the use of <span class="math-container">$\ge$</span> and <span class="math-container">$\le$</span> here. If <span class="math-container">$d'(0)=0$</span> implies a maximum is obtained at <span class="math-container">$x=0$</span>, then every <span class="math-container">$x$</span> to the left of <span class="math-container">$0$</span> should have a strictly positive derivative and every <span class="math-container">$x$</span> to the right of <span class="math-container">$0$</span> should have a strictly negative derivative (i.e., <span class="math-container">$&gt;$</span> and <span class="math-container">$&lt;$</span> should be used).</p>
<p>What is the meaning behind <span class="math-container">$d'(x) \ge 0$</span> and <span class="math-container">$d'(x) \le 0$</span>?</p>
<p><strong>Edit</strong>: <span class="math-container">$d(g-\eta)$</span> is a function representing the disutility from social disapproval, where <span class="math-container">$g$</span> is a tip at a restaurant as a percentage of the bill and <span class="math-container">$\eta$</span> is the norm tip. The consumer takes as given <span class="math-container">$\eta$</span> (i.e., the social norm) and chooses <span class="math-container">$g$</span> to minimise the social disutility <span class="math-container">$d(x)$</span>; i.e, bring it closer to <span class="math-container">$0$</span>, where <span class="math-container">$x=g-\eta$</span>.</p>
<p>Note: there is a trade-off to minimising <span class="math-container">$d$</span> that is probably not necessary to understand for my question.</p>

CC BY-SA 4.0
2022-08-03T06:46:57.793Z
The question would be somewhat more self contained if you were to include what $d$ is meant to describe.
Giskard
CC BY-SA 4.0
2022-08-03T06:57:27.347Z
@Giskard I have included a brief description of $d$.
Eli J
<mathematical-economics>
Derivatives near optimal points
Eli J
# Derivatives near optimal points

(Asked by: username_0 on 2022-08-03)

On page 52 of [What Sustains Social Norms and How They Evolve?](https://www.sciencedirect.com/science/article/abs/pii/S016726810300221X), 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.

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: The question would be somewhat more self contained if you were to include what $d$ is meant to describe.  
> **username_0**: @Giskard I have included a brief description of $d$.  

---

## 1 answers

### Answer by username_1 on 2022-08-03. Score: 4 | Accepted answer

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](https://i.sstatic.net/V0doe.png)](https://i.sstatic.net/V0doe.png)

For reference, [the Desmos](https://www.desmos.com/calculator) code of the above piece-wise function is

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


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