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!
Best Answer
Yes, and a more general fact is true:
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.
The proof is along the same lines, but the first step uses the convexity of $g$ and the monotonicity of $h$:
$$h(g((1-t)\mathbf x+t\mathbf y))\le h((1-t)g(\mathbf x)+tg(\mathbf y)) \le (1-t)\,h(g(\mathbf x))+t\,h(g(\mathbf y)) $$