Convexity of $f(x,y)=\frac{x}{y^2}$

Question

I would like to ask the convexity of function

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

where $x\geqslant0, y>0$.

Since $f(x,y)$ is differentiable but not twice differentiable, I used the first order condition and have

$$\frac{x_2(y_1^2-y_2^2)}{y_1^2y_2^2}-\frac{2x_1(y_1-y_2)}{y_1^3}$$

Assume $x_1>>x_2, y_1>y_2$, the expression above is less than 0.

So is it correct that $f(x,y)$ is nonconvex?

Thank you.

Dylan

You should note that your function is twice differentiable on ${(x,y) \in \mathbb{R}^2 : y > 0 }$. — gerw, Mar 21 '16 at 12:06

score 5 · Accepted Answer · answered Mar 19 '16 at 02:18

5

If you look at the characterizations of convexity in the following link, at page 11 ( http://www2.econ.iastate.edu/classes/econ500/hallam/documents/Convex_Opt_000.pdf ) you will see that convexity requires $$ \frac{\partial^2 f(x,y)}{\partial x^2}\frac{\partial^2 f(x,y)}{\partial y^2} - \left(\frac{\partial^2 f(x,y)}{\partial x\partial y}\right)^2\geq 0. $$ However, we have $$ \frac{\partial^2 f(x,y)}{\partial x^2} = 0, $$ and $$ \left(\frac{\partial^2 f(x,y)}{\partial x\partial y}\right)^2 = \frac{2^2}{(y^3)^2}>0. $$ Thus, we can never satisfy the first inequality, hence the function cannot be convex.

answered Mar 19 '16 at 02:18

Frank

188

Is the requirement a necessary condition or sufficient and necessary condition? – Dylan Lan Mar 19 '16 at 02:44
For twice differentiable functions, it is both necessary and sufficient. – Frank Mar 19 '16 at 02:45
But the function $f(x,y)=\frac{x}{y^2}$ is not twice differentiable. – Dylan Lan Mar 19 '16 at 02:53

Convexity of $f(x,y)=\frac{x}{y^2}$

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