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

Question

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? Can this be proved by perspective function? If so, how to prove that?

Inasmuch as copying the first line to the second line, it is the "right path", yes. — uniquesolution, Mar 21 '19 at 22:32

score 3 · Answer 1 · answered Mar 24 '19 at 12:05

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.

Georgii · Accepted Answer · 2019-03-21T23:41:08.187

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 '19 at 07:20

score 2 · Answer 3 · answered Mar 21 '19 at 23:14

2

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

answered Mar 21 '19 at 23:14

farruhota

31,482

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 '19 at 07:20

Jeremy Chu · Answer 4 · 2023-05-20T12:42:47.610

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definition: the perspective of a function $g : \mathcal{R}^n \to \mathcal{R}$ is the function $f : \mathcal{R} ^n × \mathcal{R} \to \mathcal{R}$, $$f(x, y) = y\cdot g\left(\frac{x}{y}\right), \textbf{dom}~f = \left\{(x, y)~:~\frac{x}{y} ∈ \textbf{dom}~g,~y > 0\right\}$$ Then $f$ is convex if $g$ is convex. $\square$

Since $g(x) = x^2$ is convex for $x\in\mathcal{R}$. So $f(x,y)=y \cdot g\left(\frac{x}{y}\right)=y\left(\frac{x}{y}\right)^2=\frac{x^2}{y}$ is convex for $y>0$.

The answer can be found in [BV04] Example 3.18 or slide.

By checking Hessian matrix is positive semidefinite.

This proof can also be found in [BV04] $\S$3.1.5 or P.73 Quadratic-over-linear function

The epigraph of this function can be reduced to a Second Order Cone, which is convex.

This proof can be found in [BV04] Example 3.4

edited May 20 '23 at 12:42

answered May 19 '23 at 16:15

Jeremy Chu

21
3

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Rodrigo de Azevedo · Answer 5 · 2023-05-20T14:31:30.157

Complementing Mark's answer, introduce variable $z$ and define the epigraph of $f$ via $\frac{x^2}{y} \leq z$ or, since $y > 0$, via $x^2 \leq y z$. These inequalities (plus $z \geq 0$) can be rewritten as the following linear matrix inequality (LMI)

$$ \begin{bmatrix} y & x \\ x & z\end{bmatrix} \succeq 0 $$

Thus, the epigraph of $f$ is a spectrahedon and, thus, convex. Hence, $f$ is convex.

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

5 Answers5

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