3

I am to prove that $\sqrt{x^2+y^2}$ is convex for $x,y>0$.

Intuitively, if I look at the derivatives, $\frac{x}{\sqrt{x^2+y^2}}$, $\frac{y}{\sqrt{x^2+y^2}}$, they are increasing in every positive direction. However, that isn't a very formal argument (or even correct?)

Of course, one could compute the Hessian, but that seems like a pain and since this is a minor subquestion from an optimisation exam I am preparing for, there must be a simpler way.

I thought about looking at it as a composite function, however the lemmas known to me require the outer function to be convex.

Thank you for help.

Dahn
  • 5,574

3 Answers3

6

Your function is actually $f(x,y) = \sqrt{x^2 + y^2} = \Vert (x,y) \Vert_2$, i.e. a norm.

Using the triangle inequality (and homogeneity), it is easy to see that norms are always convex.

PhoemueX
  • 35,087
  • Thanks, that indeed is a quick and efficient way, if one knows this property of norms. Given the background though, this is definitely not what the examiner would expect the person to answer, so I'd guess there still has to be another way. – Dahn May 26 '14 at 08:39
3

In 3D space $z=\sqrt{x^2+y^2}$ is a cone. Every 'vertical' planar section of it is either a parabola hyperbola or a V-shape like the $|\cdot|$ modulus graph. Given that, it's easy to prove convexity.

CiaPan
  • 13,049
2

Alternately, there are many ways to prove the inequality below (e.g. consider a triangle with vertices $(0, 0), (x_1, y_1), (-x_2, -y_2)$ and use the triangle inequality)

$$\frac{\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}}2 \ge \sqrt{\left(\frac{x_1+x_2}2\right)^2+\left(\frac{y_1+y_2}2\right)^2} $$

shows convexity of $f(x, y)$...

Macavity
  • 46,381