<p>The first-order condition would be satisfied under the assumption $c=G(l)$ if $U'(0)=0$. But this latter requirement is incompatible with the assumptions: indeed, $U'$ is nonincreasing (by the concavity of $U$) which yields $U'(x) \leq 0$ for any $x \geq 0$, which contradicts the fact that $U$ is increasing. </p>

<p>If $U'(x)&gt;0$ for all $x$, the only possibility to meet the first-order condition is indeed $\dfrac{dc}{dl}=G'(l)$.</p>

CC BY-SA 3.0
2015-08-25T11:40:34.943Z
2015-08-25T11:40:34.943Z
Oliv
<p><a href="https://en.wikipedia.org/wiki/Greenwood%E2%80%93Hercowitz%E2%80%93Huffman_preferences" rel="nofollow">Wikipedia</a> states GHH preferences to have the general form</p>

<p>$$V(c, l) =  U(c - G(l))$$</p>

<p>with $U$ increasing concave, and $G$ increasing convex.</p>

<p>In interior solution w.r.t. $l$ requires</p>

<p>$$ U'(c - G(l))(\frac{dc}{dl} - G'(l)) = 0$$ </p>

<p>As wikipedia says, this is given by </p>

<p>$$ \frac{dc}{dl} = G'(l)$$</p>

<p>However, what about the peculiar case where $c = G(l)$? Could someone elaborate a bit on the intuition </p>

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2015-08-25T10:32:32.957Z
If $c=G(l)$, the first-order condition is not satisfied, is it? Since $U'(0)>0$.
Oliv
CC BY-SA 3.0
2015-08-25T11:03:17.450Z
@Oliv With for example $U(x)  = x^{1-\sigma}$ it would be - a standard case of an increasing concave function.
FooBar
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2015-08-25T11:25:50.640Z
in that case, $U'(x)=(1-\sigma)x^{-\sigma} \rightarrow + \infty$ when $x \rightarrow 0$ (since $\sigma>0$). More generally, since $U$ is concave, $U'(0)=0$ would imply that $U'(x) \leq 0$ for $x>0$, and this contradicts the assumption that $U$ is increasing. It is probably implicitly assumed that $U'>0$ on the relevant interval.
Oliv
CC BY-SA 3.0
2015-08-25T11:28:01.730Z
Thanks, I think the two comments together suffice as an answer.
FooBar
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2015-08-25T11:30:14.260Z
You're welcome, I'm glad it answered your concern.
Oliv
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2015-08-25T11:33:08.457Z
I meant if you copy paste those two comments into an answer I can accept it, and you'll get some imaginary internet points on top of the karma ;)
FooBar
CC BY-SA 3.0
2015-08-25T11:41:23.187Z
The karma was already a great reward, can't wait for the virtual bonus ;-)
Oliv
<utility>
GHH Preferences: FOC
FooBar
# GHH Preferences: FOC

(Asked by: username_0 on 2015-08-25)

[Wikipedia](https://en.wikipedia.org/wiki/Greenwood%E2%80%93Hercowitz%E2%80%93Huffman_preferences) states GHH preferences to have the general form

$$V(c, l) = U(c - G(l))$$

with $U$ increasing concave, and $G$ increasing convex.

In interior solution w.r.t. $l$ requires

$$ U'(c - G(l))(\frac{dc}{dl} - G'(l)) = 0$$

As wikipedia says, this is given by

$$ \frac{dc}{dl} = G'(l)$$

However, what about the peculiar case where $c = G(l)$? Could someone elaborate a bit on the intuition

**Tags:** utility
> **Comments:**
> **username_1**: If $c=G(l)$, the first-order condition is not satisfied, is it? Since $U'(0)>0$.  
> **username_0**: @Oliv With for example $U(x)  = x^{1-\sigma}$ it would be - a standard case of an increasing concave function.  
> **username_1**: in that case, $U'(x)=(1-\sigma)x^{-\sigma} \rightarrow + \infty$ when $x \rightarrow 0$ (since $\sigma>0$). More generally, since $U$ is concave, $U'(0)=0$ would imply that $U'(x) \leq 0$ for $x>0$, and this contradicts the assumption that $U$ is increasing. It is probably implicitly assumed that $U'>0$ on the relevant interval.  
> **username_0**: Thanks, I think the two comments together suffice as an answer.  
> **username_1**: You're welcome, I'm glad it answered your concern.  
> **username_0**: I meant if you copy paste those two comments into an answer I can accept it, and you'll get some imaginary internet points on top of the karma ;)  
> **username_1**: The karma was already a great reward, can't wait for the virtual bonus ;-)  

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## 1 answers

### Answer by username_1 on 2015-08-25. Score: 1 | Accepted answer

The first-order condition would be satisfied under the assumption $c=G(l)$ if $U'(0)=0$. But this latter requirement is incompatible with the assumptions: indeed, $U'$ is nonincreasing (by the concavity of $U$) which yields $U'(x) \leq 0$ for any $x \geq 0$, which contradicts the fact that $U$ is increasing.

If $U'(x)>0$ for all $x$, the only possibility to meet the first-order condition is indeed $\dfrac{dc}{dl}=G'(l)$.


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