Is infimum distributive for norms? I am trying to prove that the point to set distance when the set $S$ is a convex cone, is sub-additive:
$K$ is a cone, and $x,y\in K$
$dist(x+y|K) \le dist(x|K) + dist(y|K)$
where $dist(y|K)=\inf_{z\in K} ||y-z||_2$
So I said $dist(x+y|K)=\inf_{z\in K}||x+y-z||_2$, and because $K$ is a cone, then any vector in $K$ can be decomposed into elements of $K$: $z_1+z_2=z$, $z_1,z_2 \in K$:
$\inf_{z\in K}||x+y-z||_2=\inf_{z \in K}||x+y-(z_1+z_2)||_2$
Then by the triangle inequality of the norm, $\inf_{z \in K}||x+y-(z_1+z_2)||_2 =\inf_{z \in K}||(x-z_1)+(y-z_2)||_2\le \inf_{z_1 \in K}||(x-z_1)||+\inf_{z_2 \in K}||(y-z_2)||_2$
and if I am allowed to distribute infimum then 
$\inf_{z \in K}||(x-z_1)+(y-z_2)||_2=\inf_{z_1 \in K}||(x-z_1)||+\inf_{z_2 \in K}||(y-z_2)||_2$
...and this means 
$dist(x+y|K) \le dist(x|K) + dist(y|K)$.
Can I distribte the infimum?
 A: I think in order to take infimum you need to have something like 
$$\inf_{z \in K}||x+y - z||  \leq ||x-z_1|| + ||y-z_2||$$
for all $z_1 , z_2 \in K$  and not 
$$\inf_{z \in K}||x+y -z|| \leq ||x-z_1|| + ||y-z_2||$$
for some $z_1,z_2 \in K$ which is what you have in your case. Since in this case their might exist $z* \in K$ such that $||x-z^*|| < \inf_{z \in K}||x+y -z||$ , but (eventually) in our case this not going to happen since $K$ is convex cone.
To prove the desired inequality lets start in a different way, 
let $z_1 , z_2 \in K$ two arbitrary points in $K$ then since $K$ is convex cone $z_1 + z_2 \in K$ so , 
$$\inf_{z\in K}||x+y-z|| \leq ||x+y-(z_1+z_2)|| \leq ||x-z_1|| + ||y-z_2||$$
in particular 
$$\inf_{z \in K}||x+y -z || \leq ||x-z_1|| + ||y-z_2||$$
and the above inequality holds for all $z_1, z_2 \in K$. So ,
if we fix $z_2 \in K$ then 
$$\inf_{z \in K}||x+y-z|| - ||y-z_2|| \leq ||x-z_1|| $$ 
for all $z_1 \in K$ , now we can take inf in respect to $z_1$ so ,
$$\inf_{z \in K}||x+y-z|| \leq ||y-z_2|| + \inf_{z_1 \in K}||x-z_1||$$ 
for all $z_2 \in K$ and exactly with the same argument as before we get 
$$\inf_{z \in K} ||x+y-z|| \leq \inf_{z_2 \in K}||y-z_2|| + \inf_{z_1 \in K}||x-z_1||$$
A: No, you cannot distribute the infimum. Here is the correct argument: let $z_1,z_2 \in K$. Then $z=z_1+z_2 \in K$ so $dist(x+y,K) \leq ||x+y-(z_1+z_2)|| \leq ||x-z_1||+||y-z_2||$. Now take infimum over all possible $z_1$ and $z_2$ to complete the proof. 
