# Convexity of a function of two variables given convexity in one variable.

Suppose $f:\mathbb R^2\to \mathbb R$ is twice continuously differentiable, such that

$f(x,y)=f(y,x);\ f(0,0)=0;\ t>0\Rightarrow f(tx,ty)=tf(x,y);\ \text {and}\ g(x)=f(x,1)$ is convex.

The claim is then that $f$ itself is convex.

The homogeneity of $f$ implies that $f(a,b)=af_x(a,b)+bf_y(a,b).$ Also, $f_x(a,b)=f_y(b,a)$ and $h(x)=f(1,x)$ is convex.

I thought this was going to be a simple calculation, using the above facts. But I tried various substitutions into the expression $f(t\vec x+t(1-t)\vec y)-tf(\vec x)-(1-t)f(\vec y),$ and different paths relating $f(x,y)$ with $f(x,1)$ and $f(1,x)$ without getting the answer.

How should I proceed?

The function defined as $f(x,y)=\sqrt{x^2+y^2}$ when $x\ge0$ or $y \ge0$ and $f(x,y)=\sqrt{x^2+4xy+y^2}$ when $x, y \lt 0$ seems to be a counter example.
• I guess we suppose that $f$ is twice continuously differentiable. I found the problem in this paper – Matematleta Dec 19 '17 at 1:07