# About the convexity of Ky Fan's norm

As we know, the Ky Fan norm is convex, and so is the Ky Fan k-norm. My question is, does this imply that the difference between them is a non-convex function, since it results from "difference between two convex" functions ?

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The man's name was Ky Fan (capital K, capital F). Show some respect, please. I fixed it. – bubba Mar 2 '13 at 3:38
Thanks for fixing it. – user64607 Mar 2 '13 at 5:05
No problem. He was my thesis advisor, and I liked him very much, so maybe I'm a little too sensitive about this. – bubba Mar 2 '13 at 9:31

First of all, every norm is a convex function, so the mention of the Ky Fan norms specifically is puzzling, unless you think these have additional relevant properties (which you didn't tell us about).

The difference between two convex functions could be convex or not.

Examples of both cases are easy to construct.

Also, if your proposed "difference theorem" were true, then, given two convex functions $f$ and $g$, both $f-g$ and $g-f$ would be convex.

So, no, nothing is implied by the convexity of the two norms.

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If we consider the case of singular values of a matrix, is \sum_{i=k+1}^N \sigma_i = \sum_{i=1}^N \sigma_i - \sum_{i=1}^k \sigma_i (Ky Fan N-norm - Ky Fan k-norm) a convex function ? Thanks – user64607 Mar 2 '13 at 5:04

As bubba remarked, being the difference of two convex functions does not imply either convexity or non-convexity. We have to actually look at the function. The simplest nontrivial example is $k=1$ for $2\times 2$ diagonal matrices with positive entries on the diagonal: $$M=\begin{pmatrix}a & 0 \\ 0 & d \end{pmatrix}, \quad \sigma_1(M)=\max(a,d), \ \sigma_2(M)=\min(a,d) \tag1$$ The function $\sigma_2(M)=\min(a,d)$ is not convex, as you can check by considering its values at $(a,d)=(1,3)$, $(a,d)=(2,2)$, and $(a,d)=(3,1)$.

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