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I know that this function should be concave, I am working on the Hessian proof but I would rather use this property.

I know that h(a) is convex and decreasing in a, and g(x,y) is linear, specifically g(x,y)=m+k*x-l*y. Is f(x,y)=g(g(x,y)) jointly convex in x and y?

I know that any function f(x)=h(g(x)) is convex in x, if h(x) is convex and non-decreasing and g(x) is convex. I do not know, however, if this changes when x is represented as a linear combination of two variables.

I am not exactly sure how I would prove this formally, either. My training in proofs is not very good.

Thanks so much!

user130023
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1 Answers1

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Yes, and a more general fact is true:

If $h:\mathbb R\to\mathbb R$ is convex and $g:\mathbb R^n\to \mathbb R$ is linear, then the composition $h\circ g$ is convex.

Proof: The definition of convexity requires us to show that for all $\mathbf x,\mathbf y\in\mathbb R^n$ and all $t\in [0,1]$ we have $$h(g((1-t)\mathbf x+t\mathbf y)) \le (1-t)\,h(g(\mathbf x))+t\,h(g(\mathbf y)) \tag{1}$$ Since $g$ is linear, the left side of (1) can be written as $$h(g((1-t)\mathbf x+t\mathbf y))=h((1-t)g(\mathbf x)+tg(\mathbf y)) \le (1-t)\,h(g(\mathbf x))+t\,h(g(\mathbf y)) $$ Done.

For completeness, I include a related fact.

If $h:\mathbb R\to\mathbb R$ is convex nondecreasing and $g:\mathbb R^n\to \mathbb R$ is convex, then the composition $h\circ g$ is convex.

The proof is along the same lines, but the first step uses the convexity of $g$ and the monotonicity of $h$:

$$h(g((1-t)\mathbf x+t\mathbf y))\le h((1-t)g(\mathbf x)+tg(\mathbf y)) \le (1-t)\,h(g(\mathbf x))+t\,h(g(\mathbf y)) $$

user127096
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