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Given the following composition of functions:

$h:\Bbb R^k\rightarrow\Bbb R$

$g:\Bbb R^n\rightarrow\Bbb R$

$f(x)=h(g_1(x),g_2(x),...,g_k(x))$

There are known rules which guarantee convexity/concavity of $f$, for example:

$f$ is convex if: $h$ is convex, $h$ is nondecreasing in each argument, and $g_i$ are convex.

It seems all such rules require $h$ to be monotonic. Are there any such convexity rules that do not require monotonicity of $h$? In other words, given a convex/concave nonmonotonic function $h$ is there any way to choose $g_i$ such that $f$ is convex/concave?

To clarify what my goal is: I have a convex function $h$ which I know, and I want to let an external "user" supply $g$. I do not know what $g$ will be but I want to be able to provide guidelines on what types of $g$ would allow $f$ to be convex function.

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  • $\begingroup$ You could take linear $g_i$. $\endgroup$ Commented May 7, 2014 at 20:04
  • $\begingroup$ Well of course. The condition you've cited is sufficient, but not necessary, to ensure convexity. So simply as a matter of logic, there exist constructions that do not satisfy the conditions, and yet result in a convex composition. What exactly are you looking for here? $\endgroup$ Commented May 7, 2014 at 20:04
  • $\begingroup$ @HansEngler: a linear $g$ is also convex, so that still satisfies the conditions. $\endgroup$ Commented May 7, 2014 at 20:06
  • $\begingroup$ @MichaelGrant I have a convex function $h$ which I know and I want to let an external "user" supply $g$. I do not know what $g$ will be but I want to be able to provide guidelines on how to choose $g$ such that $f$ will result in a convex function. $\endgroup$
    – Bitwise
    Commented May 7, 2014 at 20:15
  • $\begingroup$ Given this clarification, but without knowing the specific function $h$ (other than the fact it is non-monotonic), I frankly doubt there is going to be useful guidance you will be able to offer. Convexity is the exception, not the rule. $\endgroup$ Commented May 8, 2014 at 12:44

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Having your clarification in mind, I'll give (one case of) an example to show that every $h$ has a bad $g$. More exactly: for any convex nonmonotonic $h\colon\mathbb{R}\to\mathbb{R}$, there exists a convex $g\colon\mathbb{R}\to\mathbb{R}$ such that $h\circ g$ is neither concave nor convex.

I'll treat just the case where there exist $a,b,c$ such that $a<b<c$ and $h(b)<h(a)<h(c)$. (Other cases are similar, or can be reduced to this one with some fiddling.) Define $g(x) = a+|x|$. Then $$ h(g(\tfrac12(b-a)+\tfrac12(a-b))) = h(a) > h(b) = \tfrac12 h(g(b-a)) + \tfrac12 h(g(a-b)) $$ which shows that $h\circ g$ is not convex, and $$ h(g(\tfrac12(c-a)+\tfrac12(a-c))) = h(a) < h(c) = \tfrac12 h(g(c-a)) + \tfrac12 h(g(a-c)) $$ which shows that $h\circ g$ is not concave.

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One simple extension of the above rule for non-monotonic $h$ is if $h$ is monotonic over the range of $g$. For instance, $h(x)=x^2$ is non-monotonic; but it is increasing for positive $x$. So $h(g(x))$ is convex if $g(x)$ is non-negative and convex.

I know it's a simple one, but without further clarification on what it is you're really trying to accomplish, I don't see how to proceed. For every composition $h(g(x))$ that happens to be convex, you can probably create a rule that extends to function pairs with similar geometries. But how that would actually be practical, I don't know.

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  • $\begingroup$ Thanks. See my clarification in the question on what I am trying to do. $\endgroup$
    – Bitwise
    Commented May 7, 2014 at 20:20

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