-1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Inasmuch as copying the first line to the second line, it is the "right path", yes. $\endgroup$ – uniquesolution Mar 21 at 22:32
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – learning Mar 22 at 7:20
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – learning Mar 22 at 7:20
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.