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I am trying to prove that the following is a convex set:

$$\{x \in \mathbb{R}^5: \sum_i^5ix_i^2\le1\}$$

I know that this is a convex set, as this is very similar to the equation of a sphere in 3 dimensions, or a circle in two dimensions, but I am having trouble representing this in a way that I can prove it is convex.

My approach is to let $A = \left[ \begin{matrix} 1 & 0 & 0 & 0 & 0\\ 2 & 0 & 0 & 0 & 0\\ 3 & 0 & 0 & 0 & 0\\ 4 & 0 & 0 & 0 & 0\\ 5 & 0 & 0 & 0 & 0\\ \end{matrix} \right]$

and $v = Ax$ so we can achieve the same result as $\sum_i^5ix_i^2$ with $\langle v,x\rangle$

But if i do this i feel like am am losing the representation of one of the $x$'s and it is not able to be used in proving the original set to be convex. Am I on the right track or is there a better way to approach this?

Steven White
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In two dimensions this is the figure bounded by the ellipse $$ x^2 + 2y^2 = 1. $$ In space the boundary is the analogous ellipsoid.

In $n$-space just scale the $i$th axis by $1/i$ to change the ellipsoid into the convex sphere. That scaling preserves convexity.

Your matrix $A$ is trying to do that. Just put $1/i$ on the diagonal.

Ethan Bolker
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  • ok but how does that help me prove it is convex beyond intuition? It it sufficient to say "if you scale this space by A you have an ellipse in 5 dimensions which is convex, because ellipses are convex and scaling maintains convexity?

    My original plan was to show $$\langle v, \lambda x + (1-\lambda)y\rangle$$ is convex through the definition of convexity, by my worry was that this was incorrect due to v being treated as if it was a constant, and I would be effectively ignoring an x

     – Steven White Apr 18 '17 at 00:53
  • I would accept that as a proof. (But say "scaling converts the ellipsoid to a sphere, which is convex."). I would even accept "ellipsoids are convex" unless the whole point of the exercise was to practice with the definition of convex. Then I'd want you to show why scaling preserves it (not hard). – Ethan Bolker Apr 18 '17 at 00:58