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Prove $f(x, y) = frac{x^2}{y}$ is a convex function on the set ${(x,y) in mathbb{R}^2 : y > 0}$.

Attempt:

I start with the basic convexity, i.e.,
begin{align}
f( alpha_1 x_1 + alpha_2 x_2) leq alpha_1 f(x_1) + alpha_2f(x_2) ,
end{align}
where $alpha_1 + alpha_2 = 1$.

Let $z = (x,y)$, then
begin{align}
f( alpha z_1 + (1-alpha) z_2) leq alpha f(z_1) + (1-alpha)f(z_2) ,
end{align}

How to proceed from here? Thank you so much.

Question: Can this be proved by perspective function? If yes, how to prove that.

A twice continuously differentiable function of several variables is convex on a convex set if and only if its Hessian matrix of second partial derivatives is positive semidefinite on the interior of the convex set.
$$f(x,y) = frac{x^2}{y}$$
Partial derivatives:
$$frac{partial f(x,y)}{partial x} = frac{2x}{y}, frac{partial f(x,y)}{partial y} = -frac{x^2}{y^2}$$

$$frac{partial^2 f(x,y)}{partial x^2} = frac{2}{y}, frac{partial^2 f(x,y)}{partial x partial y} = -frac{2x}{y^2}$$

$$frac{partial^2 f(x,y)}{partial y partial x} = -frac{2x}{y^2}, frac{partial^2 f(x,y)}{partial y^2} = frac{2x^2}{y^3}$$

Hessian matrix:
$$H =
begin{bmatrix}
frac{partial^2 f(x,y)}{partial x^2} & frac{partial^2 f(x,y)}{partial x partial y} \
frac{partial^2 f(x,y)}{partial y partial x} & frac{partial^2 f(x,y)}{partial y^2}
end{bmatrix}
$$

$$H =
begin{bmatrix}
frac{2}{y} & -frac{2x}{y^2} \
-frac{2x}{y^2} & frac{2x^2}{y^3}
end{bmatrix}
$$

Sylvester's criterion is used to prove that matrix is positive semidefinite:
matrix is positive semidefinite if and only if all leading minors are non-negative.

$$Delta_{(1)} = frac{partial^2 f(x,y)}{partial x^2} = frac{2}{y} >= 0$$
$$Delta_{(2)} = frac{partial^2 f(x,y)}{partial y^2} = frac{2x^2}{y^3} >= 0$$
$$Delta_{(1,2)} =
begin{vmatrix}
frac{2}{y} & -frac{2x}{y^2} \
-frac{2x}{y^2} & frac{2x^2}{y^3}
end{vmatrix} =
frac{2}{y} * frac{2x^2}{y^3} - (-frac{2x}{y^2}) * (-frac{2x}{y^2}) =
frac{4x^2}{y^4} - frac{4x^2}{y^4} = 0$$

So, all leading minors are non-negative on $lbrace (x,y) in mathbb{R}^2 : y>0 rbrace$.

Hence, Hessian matrix is positive semidefinite.

Hence, function $f(x,y)$ is a convex function on the set $lbrace (x,y) in mathbb{R}^2 : y>0 rbrace$.

Hessian matrix is positive semidefinite, therefore it is convex:
$$H=begin{pmatrix} f_{xx}& f_{xy}\ f_{yx}&f_{yy}end{pmatrix}=
begin{pmatrix} frac{2}{y}& -frac{2x}{y^2}\ -frac{2x}{y^2}& frac{2x^2}{y^3}end{pmatrix}\
H_1=frac2y>0\
H_2=0.$$

If you want something in the spirit of perspective function rather than showing the Hessian is positive semidefinite:

The epigraph of this function can be reduced to a Second Order Cone, which is convex. You can reverse engineer the details by studying CVX's quad_over_lin function https://github.com/cvxr/CVX/blob/master/functions/%40cvx/quad_over_lin.m . I leave you to do that work.