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Let $X$ be a normed vector space, $x\in X$ and $Z\subseteq X$. Then we define the point-to-set distance function as: $$ \|x\|_Z = \inf_{z\in Z} \| x-z\| $$ I use the notation $\|\cdot\|_Z$ for convenience without implying that the operation is a norm. Assuming that $Z$ is convex we may not conclude that: $$ \|x+y\|_Z \leq \|x\|_Z + \|y\|_Z $$ What sort of properties should $Z$ possess so that the triangle inequality holds for all $x\in X$? For example, $Z$ being a singleton brings about the desired inequality but I'm looking for a less trivial example.

Additionally, let $A\in M_n(\mathbb{R})$ be a matrix and let $Z\neq X$. I noticed that we may define:

$$ \|A\|_Z:= \sup_{x\in\mathbb{R}^n,x\notin Z}\frac{\|Ax\|_Z}{\|x\|_Z} $$

while $\|A\|_X:=0$.

Then, it should be true that: $$ \|Ax\|_Z \leq \|A\|_Z\cdot\|x\|_Z $$

So, this is one norm-like property for $\|\cdot\|_Z$.

What other properties of $\|\cdot\|_Z$ can we deduce based on assumed properties of $Z$?

Update: As Davide Giraudo pointed out, the triangle inequality holds true in case $Z$ is a linear subspace of $X$. It seems that it holds for convex cones too (for nonconvex cones it's easy to find a counter example on the plane). Let us assume that $Z$ is a convex cone. Then:

$$ \|x+y\|_Z=\inf_{z\in Z}\|x+y-z\|\leq \|x+y-(z_1+z_2)\| $$

where $z_1,z_2\in Z$ and we proceed as in David's answer. Here, we used the fact that whenever $Z$ is a convex cone, then $z_1,z_2\in Z \Rightarrow z_1+z_2\in Z$ so any vector in $Z$ can be decomposed into a sum of vectors of $Z$ and vice versa.

Question: Would be nice to identify other classes of sets for which the induced point-to-set distance satisfies the triangle inequality.

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If $Z$ is a vector subspace then for $x,y\in X$, $z_1,z_2\in Z$ we have $$||x+y||_Z\leq ||x+y-(z_1+z_2)||\leq ||x-z_1||+||y-z_2|||,$$ and taking the infimum with respect to $z_1$ we get: $||x+y||_Z\leq ||x||_Z+||y-z_2||$. Now taking the infimum with respect to $z_2$ we get $||x+y||_Z\leq ||x||_Z+||y||_Z$.

In fact when $Y$ is a closed subspace of $X$, we can consider the equivalence relation $x\sim y$ iff $x-y\in Y$ and denoting $[x]$ the class of $X$, we can put $||[x]||:= ||x||_Y$. It defines a norm on $X/Y$ and their are many links between the properties of $Y$, $X$, $X/Y$, named "three spaces property". Namely, consider a property (P), like separability, completeness, etc... The three spaces property means that if between $X,Y,X/Y$ two have the property (P) then the third will also have the property (P).

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I have the feeling that the triangle inequality also holds for cases in which $Z$ is a cone... – Pantelis Sopasakis Feb 9 '12 at 22:05
Yes, you're right. – Davide Giraudo Feb 9 '12 at 22:06

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