On an article I'm reading, I find that: if $v$ is a vector, the projection of of $v$ on the unit ball is: $$p(v)=\frac{v}{\max\{1,\|v\|\}}$$ I know that a projection of a point $v$ into a space is the nearest point to $v$ inside the space..why the expression above?
$\endgroup$ 82 Answers
$\begingroup$I presume the setting is a Hilbert space $\mathbb{H}$.
Then the projection onto the closed unit ball $\bar{B}$ is given by a solution to $\min_{ x \in \bar{B}} \|x-v\|$.
If $v=0$, it is clear that the solution is $x=0$, so we will assume $v \neq 0$ in the sequel.
We can write any $x \in \mathbb{H}$ as $x = \lambda v + w$, where $w \bot v$. In particular, we have $\|x\|^2 = \lambda^2 \|v\|^2 + \|w\|^2$, and so $\|x-v\|^2 = (1-\lambda)^2 \|v\|^2 + \|w\|^2$. Hence if $x = \lambda v + w \in \bar{B}$, we see that $\lambda v \in \bar{B}$, and $\|\lambda v-v\|^2 \le \|x-v\|^2$.
If we let $V = \operatorname{sp} \{v\}$, we see that $\min_{ x \in \bar{B}} \|x-v\| = \min_{ \lambda v \in \bar{B}} (1-\lambda)^2 \|v\|^2 $, which is a one dimensional problem.
Since $\lambda v \in \bar{B}$ iff $|\lambda| \le {1 \over \|v\|}$, we see that the problem is solved by $\lambda = \min(1,{1 \over \|v\|} )$, that is, $x= \min(1,{1 \over \|v\|} )v$.
It is straightforward to see that this is the same as $p(v) $ above.
$\endgroup$ 2 $\begingroup$You can do it using KKT Conditions.
Here is something I wrote once doing so (Solution to Home Work exercise I had):
I wrote MATLAB code which implements them both at Mathematics StackExchange Question 2338491 - GitHub.
There is a test which compares the result to a reference calculated by CVX.
