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

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.

• Inasmuch as copying the first line to the second line, it is the "right path", yes. – uniquesolution Mar 21 at 22:32

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 $$\lbrace (x,y) \in \mathbb{R}^2 : y>0 \rbrace$$.

• Thank you. +1.On the other hand, I am wondering if this can be proved via a perspective function? Do you have any idea how to proceed? – learning Mar 22 at 7:20

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

• Thank you. +1.On the other hand, I am wondering if this can be proved via a perspective function? Do you have any idea how to proceed? – learning Mar 22 at 7:20

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.