How to check whether a complex multivariable function is convex?

Question

Given a convex function $f : \mathbb R^2 \to \mathbb R$ and real numbers $a$ and $b$, I want to check whether the following function $g : \mathbb R^2 \to \mathbb R$ is convex.

First, I think I should use the lemma below:

So I start to write down g(αx1 + (1-α)y1 , αx2+(1-α)y2)

what I get is like this:

Then I get stuck, the function I get is just too complex and I don't know how to do next. Did I do it in the wrong way? Can anyone give me some hints?

yes of course !!! https://en.wikipedia.org/wiki/Convex_function#Convex_function_calculus — Hamza, Dec 07 '16 at 19:28
I only know that f(x) is convex if f(x) = c1f1(x) + c2f1(x) and f1(x), f2(x) are convex and c1+c2 = 1 — Parting, Dec 07 '16 at 19:33
@Chen: that is in general not true: Consider $x^2 - 2y^2$ for $(x,y)\in\mathbb R^2$. — user251257, Dec 07 '16 at 19:46
note that $-2y$ is a concave function your coeficient need to be non negative. — Hamza, Dec 07 '16 at 19:47

score 2 · Answer 1 · answered Dec 07 '16 at 19:49

2

Hints:

Sum of convex functions are convex.
Non-negative multiple of a convex function is convex.
Composition of a convex function and an affine function is convex.
Maximum of convex functions is convex.

answered Dec 07 '16 at 19:49

user251257

9,229

score 2 · Accepted Answer · edited Dec 16 '16 at 16:58

$$ g(x,y)=a(x,y)+b(x,y)+c(x,y) $$ where $$ a(x,y)= 5(x^2+y^2+xy)=\frac{5x^2}{4}+5 \left(\frac{x}{4}+y \right)^2=a_1(x,y)+a_2(x,y)\\ b(x,y)=f(x-y,ax+by)=f(A(x,y)) \textrm { where } A=\left(\begin{matrix}1 &-1\\ a &b \end{matrix} \right) \\ c(x,y)=\max\{|x-y|,2y-x\}=\max\{c_1(x,y),c_2(x,y)\} $$ we have $a_1$ is convex ($x^p$ is convex if $p\geq 1$), and $a_2$ is a composite of convex ($x \mapsto x^2$) and linear function ($x\mapsto x/4+y$), so convex :

if $f$ is convex and $g$ is affine then $f \circ g$ is convex : \begin{eqnarray*} f(g(\alpha x+ (1-\alpha)y))&=& f(\alpha g(x)+ (1-\alpha)g(y))\\ &\leq & \alpha f(g(x))+(1-\alpha)f(g(y)) \end{eqnarray*}

so $a$ is convex.

The same argument works for $b$, so $b$ is convex too.

Finally, the maximum of two convex function is also convex, which proves that $c$ is convex.

We conclude that $g$ is convex.

no for all matrix $A$ because the important thing is that $A$ is linear. — Hamza, Dec 07 '16 at 22:37

How to check whether a complex multivariable function is convex?

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