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?

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    $\begingroup$ 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. $\endgroup$ – Elias Feb 4 '20 at 9:42
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    $\begingroup$ @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. $\endgroup$ – Giskard Feb 4 '20 at 10:54
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    $\begingroup$ 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. $\endgroup$ – Regio Feb 4 '20 at 20:11
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    $\begingroup$ 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? $\endgroup$ – Regio Feb 4 '20 at 20:39
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    $\begingroup$ 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). $\endgroup$ – user17900 Feb 5 '20 at 11:39

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