<p>Assuming <span class="math-container">$f''&lt;0$</span> implies strict concavity and hence</p>
<p><span class="math-container">$$tf(y) + (1-t)f(x)&lt;f(ty+(1-t)x) \Leftrightarrow f(y) - f(x)&lt;\frac{f(ty+(1-t)x)-f(x)}{t}$$</span></p>
<p><span class="math-container">$$f(y) - f(x)&lt;\frac{f(x + t(y-x))-f(x)}{t(y-x)} (y-x),$$</span></p>
<p>now let <span class="math-container">$t$</span> go to 0 to get</p>
<p><span class="math-container">$$f(y) - f(x)&lt;f'(x) (y-x),$$</span></p>
<p>Finally, to get the result you are interested in letting <span class="math-container">$y=0$</span> and using that <span class="math-container">$f(0)=0$</span> it follows that</p>
<p><span class="math-container">$$f'(x) &lt; \frac{f(x)}{x}.$$</span></p>

CC BY-SA 4.0
2022-03-01T17:59:30.857Z
2022-03-01T17:59:30.857Z
bomadsen
<p>I was told in class that if we have a production function such that <span class="math-container">$f'(x)&gt;0$</span> and <span class="math-container">$f''(x)&lt;0$</span>, then we have that the marginal product is less than the average product. That is <span class="math-container">$f'(x)&lt;\frac{f(x)}{x}$</span>. This is not a result that I am required to prove in class, I am just proving this out of curiosity. I will show you what I have so far:</p>
<p>Proof: Suppose <span class="math-container">$f''(x)&lt;0$</span>. Let us define average product as <span class="math-container">$AP$</span>. That is,
<span class="math-container">$$AP:=\frac{f(x)}{x}$$</span>
<span class="math-container">$$\Longrightarrow AP'=\frac{f'(x)x-f(x)}{x^{2}}$$</span>
Since <span class="math-container">$sign(x^{2})&gt;0, \ \forall \ x \in \mathbb{R}$</span>, we know that <span class="math-container">$sign(AP')=sign(f'(x)x-f(x))$</span>.
Now I am stuck. I can see that if <span class="math-container">$AP'&lt;0$</span>, then our result easily follows. That is, if average product is decreasing, then <span class="math-container">$f'(x)&lt;AP$</span>. That being said, I am not able to see the connection between the assumptions I made for the proof (f'&gt;0 and f''&lt;0) and slope(AP)&lt;0. Would anyone be able to give me some insight into what I am missing. I can take the second derivative of average product and see that</p>
<p><span class="math-container">$$AP''=\frac{(f''(x)x+f'(x)-f'(x))x^{2}-2x(f'(x)x-f(x))}{x^{4}}=\frac{f''(x)x^{3}-2x^{2}f'(x)-2xf(x)}{x^{4}}$$</span></p>
<p>and sign(AP'')=sign(numerator) once again. Since <span class="math-container">$f''&lt;0$</span>, and <span class="math-container">$f',f&gt;0$</span>, then clearly the slope of the slope of AP is decreasing. That is, average product is (increasing or decreasing) at a decreasing rate. But this doesn't give me any further insight into the sign of AP.</p>

CC BY-SA 4.0
2022-03-01T16:31:10.487Z
Your calim is not true without further restriction on $f$. If $f(0)=0$ you should be able to prove it using the inequality of concavity.
Bertrand
<production-function><producer-theory><proof>
Prove that if a production function is such that f'>0 and f''<0, then f'<Average Product
Mistah White
# Prove that if a production function is such that f'>0 and f''<0, then f'<Average Product

(Asked by: username_0 on 2022-03-01)

I was told in class that if we have a production function such that $f'(x)>0$ and $f''(x)<0$, then we have that the marginal product is less than the average product. That is $f'(x)<\frac{f(x)}{x}$. This is not a result that I am required to prove in class, I am just proving this out of curiosity. I will show you what I have so far:

Proof: Suppose $f''(x)<0$. Let us define average product as $AP$. That is,
$$AP:=\frac{f(x)}{x}$$
$$\Longrightarrow AP'=\frac{f'(x)x-f(x)}{x^{2}}$$
Since $sign(x^{2})>0, \ \forall \ x \in \mathbb{R}$, we know that $sign(AP')=sign(f'(x)x-f(x))$.
Now I am stuck. I can see that if $AP'<0$, then our result easily follows. That is, if average product is decreasing, then $f'(x)<AP$. That being said, I am not able to see the connection between the assumptions I made for the proof (f'>0 and f''<0) and slope(AP)<0. Would anyone be able to give me some insight into what I am missing. I can take the second derivative of average product and see that

$$AP''=\frac{(f''(x)x+f'(x)-f'(x))x^{2}-2x(f'(x)x-f(x))}{x^{4}}=\frac{f''(x)x^{3}-2x^{2}f'(x)-2xf(x)}{x^{4}}$$

and sign(AP'')=sign(numerator) once again. Since $f''<0$, and $f',f>0$, then clearly the slope of the slope of AP is decreasing. That is, average product is (increasing or decreasing) at a decreasing rate. But this doesn't give me any further insight into the sign of AP.

**Tags:** production-function, producer-theory, proof

> **Comments:**
> **username_1**: Your calim is not true without further restriction on $f$. If $f(0)=0$ you should be able to prove it using the inequality of concavity.  

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## 1 answers

### Answer by username_2 on 2022-03-01. Score: 5

Assuming $f''<0$ implies strict concavity and hence

$$tf(y) + (1-t)f(x)<f(ty+(1-t)x) \Leftrightarrow f(y) - f(x)<\frac{f(ty+(1-t)x)-f(x)}{t}$$

$$f(y) - f(x)<\frac{f(x + t(y-x))-f(x)}{t(y-x)} (y-x),$$

now let $t$ go to 0 to get

$$f(y) - f(x)<f'(x) (y-x),$$

Finally, to get the result you are interested in letting $y=0$ and using that $f(0)=0$ it follows that

$$f'(x) < \frac{f(x)}{x}.$$


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