Convexity of $y\geqslant\frac{1}{x}$ and $xy\geqslant 1$

Question

I would like to ask the convexity of $y\geqslant\frac{1}{x}$ and $xy\geqslant1$.

We know that $y\geqslant\frac{1}{x}$ is convex for $x>0$. But if we transform the inequality into $$xy\geqslant1(x,y>0)$$ or $$1-xy\leqslant0(x,y>0)$$

The inequality becomes nonconvex as the Hessian of the inequality is:

$$ \begin{bmatrix} 0 & -1\\ -1 & 0\\ \end{bmatrix} $$

which has eigenvalues of 1 and -1.

So is it correct that $y\geqslant\frac{1}{x}$ is convex and $xy\geqslant1$ is nonconvex?

Thank you.

Dylan

Answer 1

Hint

The epigraph is convex $\iff$ the boundary is convex. Notice that they ask you the convexity of $$\left\{(x,y)\mid xy\geq 1\text{ and }y\geq \frac{1}{x}\right\}=\left\{(x,y)\mid x>0\text{ and }y\geq \frac{1}{x}\right\}.$$

Answer 2

If I am not wrong, then your question is how to show that-

$$\left\{(x,y)\mid xy\geq 1\text{ and } x,y> 0 \right\}=\left\{(x,y)\mid x>0\text{ and }y\geq \frac{1}{x}\right\}$$.

Hint:

1. The difference between function and its sublevel set.
2. Check $f(x,y)=xy-1$ and $f(x,y)=1-xy$ when $x,y > 0$.
3. Use the defination of convex set.

Answer 3

You are confusing the convexity of sets and the convexity of functions.

Don't.

They're closely related, tightly linked, but are by no means identical.

The fundamental link between convex functions and convex sets is this: a function $f$ is convex if and only if its epigraph $\{(x,y)\,|\,f(x)\leq y\}$ is a convex set.

So if you have a convex function---and $f(x) = 1/x$ is indeed convex for $x>0$---then its epigraph $\{(x,y)\,|\,x>0,~1/x\leq y\}$ is a convex set.

The important thing to note is that above if-and-only-if relationship starts with a function. It doesn't start with a set. You can't necessarily extract a convex function from a convex set.

For instance, there are lots of ways to represent the set, including $\{(x,y)\,|\,x,y\geq 0,~xy\geq 1\}$. But just because this is a convex set doesn't mean that $f(x,y)=xy$ is a convex function.

In fact, not every convex set is the epigraph of a convex function; for instance, $C=\{x\in\mathbb{R}^2\,|\,\|x\|\leq 1\}$. Is there a convex function in there? Sure is; $f(x)=\|x\|$ is convex. But the set is not the epigraph of $f$.

So again: don't confuse convex sets and convex functions.