# Orthogonal Projection onto the ${L}_{2}$ Unit Ball

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?

• If $v$ is outside the closed unit ball, then $\|v|| > 1$, and hence $p(v) = \tfrac{1}{\|v\|}v$, which is on the unit sphere, and hence in the closed unit ball. If, instead, $v$ is inside the closed unit ball, then $\|v\| \leq 1$, and hence $p(v) = v$. So, $p$ maps vectors outside the closed unit ball to their normalisations, which lie on the unit sphere, and does nothing to vectors that are already in the closed unit ball. – Branimir Ćaćić Jan 4 '14 at 15:46
• and it's the same of find $\arg_x \min {||x-v||}$ right? – volperossa Jan 4 '14 at 15:58
• For $x$ restricted to the closed unit ball, yes, I think so. – Branimir Ćaćić Jan 4 '14 at 16:05
• all clear! thank you :) – volperossa Jan 4 '14 at 16:06
• Orthogonal Projection onto the ${\ell}_{\infty}$ / l Infinity Ball - math.stackexchange.com/questions/1825747. – Royi Jun 18 '17 at 16:34

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.

• can you give a hint for a projection onto a ball around some point $c$ with radius r? I thought maybe setting the function to be $min(||y-x-c||_2^2:||y-c||\leq r)$ – convxy Jun 8 '20 at 17:27
• @ronkurman: Show that $v$ solves the above problem iff $c+rv$ solves your problem. Look at the problem geometrically. – copper.hat Jun 8 '20 at 17:32

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.

• can you give a hint for a projection onto a ball around some point $c$ with radius r? I thought maybe setting the function to be $min(||y-x-c||_2^2:||y-c||\leq r)$ – convxy Jun 8 '20 at 17:24
• If you look on the code it supports any radius $r$. Shifting is no issue. Remove the shift form the input, project and them return it to the output. – Royi Jun 8 '20 at 19:36