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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?

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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.

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  • $\begingroup$ I guess we suppose that $f$ is twice continuously differentiable. I found the problem in this paper $\endgroup$ – Matematleta Dec 19 '17 at 1:07
  • $\begingroup$ There would still be counter examples. The important difference with the paper is that the function there is only defined for (u, v) in the first quadrant, which makes the result visually trivial. $\endgroup$ – random Dec 19 '17 at 10:22

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