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?


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

$$\inf_{z \in K} ||x+y-z|| \leq \inf_{z_2 \in K}||y-z_2|| + \inf_{z_1 \in K}||x-z_1||$$


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.


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