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Suppose $f: \mathbb{R} \to \mathbb{R}$ and $h: \mathbb{R}^n \to \mathbb{R}$ are functions that satisfy the 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?

Giskard
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    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 Feb 04 '20 at 09:42
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    @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 Feb 04 '20 at 10:54
  • 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 Feb 04 '20 at 20:07
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    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 Feb 04 '20 at 20:11
  • @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 Feb 04 '20 at 20:12
  • "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 Feb 04 '20 at 20:15
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    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 Feb 04 '20 at 20:39
  • @Regio Oops! Yes, that is a mistake. Thank you for catching it! – Giskard Feb 04 '20 at 20:41
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    I don't see why you need the extra condition $ \forall i: \lim_{x_i \to \infty} h(\mathbf{x}) = \infty $. To revisit Elias' example ($n = 1$), we know that the derivative of the composed function is $f'(h(x))h'(x)$. Since $h$ satisfies the Inada conditions, $h'(x) \rightarrow 0$ as $x \rightarrow \infty$. Moreover, it is not true that $f'(h(x)) \rightarrow \infty$ as $x \rightarrow \infty$: this would require that $h(x) \rightarrow 0$, which is false. Thus, the derivative of the composed function converges to 0 (without needing the condition). –  Feb 05 '20 at 11:39
  • @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 Feb 05 '20 at 15:33

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