The conjugate of conjugate of the function $f=yx^2$

Question

Suppose $f=yx^2$ and the arguments are $x\in\mathbb{R},y>0$. According to the definition of conjugate, $f^*(x,y)=\max\limits_{y',x'}xx'+yy'-y'x'^2$ Because $f$ is not a convex function, I cannot use the first optimal condition. Furthermore, I prefer to enlarge the dimension of $x$. This is the first time I try to solve the conjugate of a non-convex function. Thus, I try to post a question. I do not know this question is a very easy one or a difficult one. But if it is a impossible task, please let me know.

Answer 1

Because of $y'>0$ you can complete the square

$xx′+yy′−y′x′^2=yy′-y'(x'-\tfrac{x}{2y'})^2+\frac{x^2}{4y'}$

Now since this only contains $x'$ in the second term, this can always be maximized by setting $x'=\tfrac{x}{2y'}$. It remains to find the maximum of

$yy′+\frac{x^2}{4y'}=(\sqrt{yy'}+\tfrac{x}{2y'})^2-x\sqrt{y}$

which is a convex function. Unfortunately, it is not bounded above.