Here we see the proof for $f(x)$ being convex where $$f(x) = h(g(x))$$given $h$ is convex and nondecreasing and $g$ is convex. But what if $$f(x) = h(g_1(x),g_2(x),g_3(x),...,g_k(x))$$ where each $g_i(x)$ is convex and $h$ is still convex and nondecreasing. Can we extend the proof to such a case?
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I would try extending the result to $k=2$. – hardmath Dec 09 '14 at 15:07
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For $n=1$, $h$ is nondecreasing. How should it be for $n\ge 2$? – ajotatxe Dec 09 '14 at 15:09
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@ajotatxe sorry, edited my post, h is always nondecreasing – Dec 09 '14 at 15:10
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@user2340818 How would you define nondecreasing for a function $h:\mathbb R^2 \to\mathbb R$? – AlexR Dec 09 '14 at 15:12
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Good question @AlexR! – Alex Silva Dec 09 '14 at 15:13
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@AlexR $y,x \in \mathbb R^k$ with $y \ge x$, then $h(y) \ge h(x)$ where $h: \mathbb R^k -> \mathbb R$ – Dec 09 '14 at 15:14
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@user2340818 Next. How would you order $\mathbb R^k$? Is $(1,0) \ge (0,1)$? – AlexR Dec 09 '14 at 15:16
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I'm not offering the proof here, but I can assure you that the claim is true: $f$ is convex. In fact, if $h$ is nondecreasing in only some of the arguments, then those arguments may be convex and the other ones are affine. – Michael Grant Dec 09 '14 at 15:18
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We define nondecreasing separately for each argument. So, for example, $h(x,y)$ is nondecreasing in $x$ if, for all fixed $y$, $x_1\geq x_2$ implies $h(x_1,y)\geq h(x_2,y)$. – Michael Grant Dec 09 '14 at 15:19
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@AlexR...Sorry, I'm not following. $h(x) \in \mathbb R$ and is nondecreasing. Why do I care about order $\mathbb R^k$? – Dec 09 '14 at 15:20
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@user2340818 $x\le y$ for $x,y\in\mathbb R^k$ is not well-defined. – AlexR Dec 09 '14 at 15:20
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@AlexR ah, right. Okay I see your point... – Dec 09 '14 at 15:21
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If we assume $h:\mathbb R^2 \to\mathbb R$ to be convex and component-wise nondecreasing, i.e. $$h(x,\cdot) \text{ and } h(\cdot, y) \text{ are nondecreasing for all }x,y\in\mathbb R\\ h(\lambda x + (1-\lambda)x', \lambda y + (1-\lambda) y') \le \lambda h(x,y) + (1-\lambda) h(x',y')$$ We can prove that $$h(g_1(x), g_2(y))$$ is convex for $g_1, g_2:\mathbb R\to\mathbb R$ convex. This allows us to generalise to $n$ dimensions: $$\begin{align*} h(g_1(\lambda x + (1-\lambda) x'), g_2(\lambda y + (1-\lambda) y')) & \le h(\lambda g_1(x) + (1-\lambda) g_1(x'), \lambda g_2(y) + (1-\lambda) g_2(y')) \\ & \le \lambda h(g_1(x), g_2(y)) + (1-\lambda) h(g_1(x'), g_2(y')) \end{align*}$$
Theorem
Let $h:\mathbb R^n \to\mathbb R$ convex and nondecreasing in each component and $g:\mathbb R^n \to\mathbb R^n$ be such that $g_i : \mathbb R^n \to\mathbb R$ is convex then $$f:\mathbb R^n \to\mathbb R, f(x) = h(g(x))$$ is convex.
Proof
$$\begin{align*}
f(\lambda x+(1-\lambda y)) & = h(g(\lambda x + (1-\lambda) y)) \\
& \le h(\lambda g(x) + (1-\lambda) g(y)) & (\ast) \\
& \le \lambda h(g(x)) + (1-\lambda) h(g(y)) \\
& = \lambda f(x) + (1-\lambda) f(y)
\end{align*}$$
In $(\ast)$ we know that $h$ is nondecreasing component-wise, i.e. $h(x + \delta e_j) \le h(x)$ for $\delta \ge 0$ and $g$ is component-wise convex so $g_j(\lambda x + (1-\lambda y)) \le \lambda g_j(x) + (1-\lambda) g_j(y)$.
AlexR
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