<p>Suppose <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> and <span class="math-container">$h: \mathbb{R}^n \to \mathbb{R}$</span> are functions that satisfy the <a href="https://en.wikipedia.org/wiki/Inada_conditions" rel="nofollow noreferrer">Inada conditions</a>, and also
<span class="math-container">$$
\forall i:
\lim_{x_i \to \infty} h(\mathbf{x}) = \infty.
$$</span> </p>

<p>Does the function <span class="math-container">$f(h(\mathbf{x}))$</span> also satisfy the conditions?</p>

CC BY-SA 4.0
2020-02-04T09:42:36.927Z
For $n=1$ and $g(x)\equiv f(h(x))$ we have that $g'(x)=f'(h(x))h'(x)$. I therefore guess that whether $\lim_{x\to\infty}g'(x)$ is 0 or not depends on $\lim_{x\to\infty}h(x)$ and the functional forms. My intuition is that there is a counterexample.
Elias
CC BY-SA 4.0
2020-02-04T10:54:18.547Z
@Elias I guess not having $\lim_{x \to \infty} h(x) = \infty$ is an issue. If you have that all the others seem to follow.
Giskard
CC BY-SA 4.0
2020-02-04T20:07:23.810Z
The condition of the unbounded derivative when $x$ approaches infinity seems to contradict the Inada conditions. According to your link, the derivative should converge to zero. Despite this, I don't think the result follows mainly because convexity is not preserved under composition, for example, $g(x)=-x^2$ and $f(x)=-e^{x}$ are both concave, but $f(g(x))$ is not concave. Granted $g(x)$ does not satisfy the Inada conditions, but I'm sure a counterexample exists.
Regio
CC BY-SA 4.0
2020-02-04T20:11:58.727Z
Convexity is preserved under composition if the function that serves as an argument of the other is monotonically increasing. From @Elias's comment, it seems that assuming  $h(x)$ strictly increasing and unbounded will give you the result you are looking for.
Regio
CC BY-SA 4.0
2020-02-04T20:12:41.267Z
@Regio Composition not preserving convexity does indeed seem problematic. Though the other conditions imply a sort of strictly increasing concavity, which your function $g$ is not. And indeed, over $\mathbb{R}_+$ where $g$ is strictly increasing even the composition is strictly concave. This may be a special case though. If you were to come up with a counter example, it would answer the question and I would happily upvote it. EDIT: seems like we crossposted the same thought.
Giskard
CC BY-SA 4.0
2020-02-04T20:15:03.823Z
"*if the function that serves as an argument of the other is monotonically increasing*." This is implied by the second Inada condition, is it not?
Giskard
CC BY-SA 4.0
2020-02-04T20:39:32.660Z
You are right. It is implied. I don't think you answered my first concern, did you mean to add the condition $\lim_{x_i\rightarrow\infty}h(x_i)=\infty$ for all $i$? Otherwise, the condition you currently have is contradicting the 4th condition for Inada, or am  I missing something?
Regio
CC BY-SA 4.0
2020-02-04T20:41:39.827Z
@Regio Oops! Yes, that is a mistake. Thank you for catching it!
Giskard
CC BY-SA 4.0
2020-02-05T15:33:00.877Z
@afreelunch The general argument as to why $f'$ is bounded over this particular range seems more complicated to me, but I think you are right in gist. Type up an answer for the general $(n \geq 1)$ case?
Giskard
<macroeconomics><production-function>
Combining functions that satisfy the Inada conditions
Giskard
# Combining functions that satisfy the Inada conditions

(Asked by: username_0 on 2020-02-04)

Suppose $f: \mathbb{R} \to \mathbb{R}$ and $h: \mathbb{R}^n \to \mathbb{R}$ are functions that satisfy the [Inada conditions](https://en.wikipedia.org/wiki/Inada_conditions), and also
$$
\forall i:
\lim\_{x\_i \to \infty} h(\mathbf{x}) = \infty.
$$

Does the function $f(h(\mathbf{x}))$ also satisfy the conditions?

**Tags:** macroeconomics, production-function

> **Comments:**
> **username_1**: For $n=1$ and $g(x)\equiv f(h(x))$ we have that $g'(x)=f'(h(x))h'(x)$. I therefore guess that whether $\lim_{x\to\infty}g'(x)$ is 0 or not depends on $\lim_{x\to\infty}h(x)$ and the functional forms. My intuition is that there is a counterexample.  
> **username_0**: @username_1 I guess not having $\lim_{x \to \infty} h(x) = \infty$ is an issue. If you have that all the others seem to follow.  
> **username_2**: The condition of the unbounded derivative when $x$ approaches infinity seems to contradict the Inada conditions. According to your link, the derivative should converge to zero. Despite this, I don't think the result follows mainly because convexity is not preserved under composition, for example, $g(x)=-x^2$ and $f(x)=-e^{x}$ are both concave, but $f(g(x))$ is not concave. Granted $g(x)$ does not satisfy the Inada conditions, but I'm sure a counterexample exists.  
> **username_2**: Convexity is preserved under composition if the function that serves as an argument of the other is monotonically increasing. From @username_1's comment, it seems that assuming  $h(x)$ strictly increasing and unbounded will give you the result you are looking for.  
> **username_0**: @username_2 Composition not preserving convexity does indeed seem problematic. Though the other conditions imply a sort of strictly increasing concavity, which your function $g$ is not. And indeed, over $\mathbb{R}_+$ where $g$ is strictly increasing even the composition is strictly concave. This may be a special case though. If you were to come up with a counter example, it would answer the question and I would happily upvote it. EDIT: seems like we crossposted the same thought.  
> **username_0**: "*if the function that serves as an argument of the other is monotonically increasing*." This is implied by the second Inada condition, is it not?  
> **username_2**: You are right. It is implied. I don't think you answered my first concern, did you mean to add the condition $\lim_{x_i\rightarrow\infty}h(x_i)=\infty$ for all $i$? Otherwise, the condition you currently have is contradicting the 4th condition for Inada, or am  I missing something?  
> **username_0**: @username_2 Oops! Yes, that is a mistake. Thank you for catching it!  
> **username_0**: @afreelunch The general argument as to why $f'$ is bounded over this particular range seems more complicated to me, but I think you are right in gist. Type up an answer for the general $(n \geq 1)$ case?  

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## 0 answers

