# 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. Jan 4, 2014 at 15:46
• and it's the same of find $\arg_x \min {||x-v||}$ right? Jan 4, 2014 at 15:58
• For $x$ restricted to the closed unit ball, yes, I think so. Jan 4, 2014 at 16:05
• all clear! thank you :) Jan 4, 2014 at 16:06
• Orthogonal Projection onto the ${\ell}_{\infty}$ / l Infinity Ball - math.stackexchange.com/questions/1825747.
– Royi
Jun 18, 2017 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 $min(||y-x-c||_2^2:||y-c||\leq r)$ Jun 8, 2020 at 17:27
• @ronkurman: Show that $v$ solves the above problem iff $c+rv$ solves your problem. Look at the problem geometrically. Jun 8, 2020 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 $min(||y-x-c||_2^2:||y-c||\leq r)$ Jun 8, 2020 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, 2020 at 19:36

An alternative solution that works in any inner product space: if $$v$$ is already in the unit ball, then clearly $$p(v)=v$$. We can thus assume that $$\lVert v\rVert>1$$. Given an arbitrary $$a$$ in the unit ball, it suffices to show that

$$\lVert v-a\rVert^2 \geq \left\lVert v-\frac v{\lVert v\rVert}\right\rVert^2.$$

Note that the RHS rewrites as $$(\lVert v\rVert-1)^2$$, and expanding the squares we find the equivalent inequality:

$$\lVert v \rVert^2 - 2\langle v,a \rangle + \lVert a \rVert^2 \geq \lVert v\rVert^2 - 2\lVert v\rVert +1,$$ which simplifies as $$\lVert a \rVert^2 - 2\langle v,a \rangle + 2\lVert v \rVert -1 \geq 0.$$

To prove this last inequality, note that \begin{align} \lVert a \rVert^2 - 2\langle v,a \rangle + 2\lVert v \rVert -1 &\geq \lVert a \rVert^2 - 2\lVert v \rVert\lVert a \rVert + 2\lVert v \rVert -1 \tag{1} \\ &= \lVert a \rVert^2 + 2\lVert v \rVert(1-\lVert a \rVert) -1 \\&\geq \lVert a \rVert^2 + 2(1-\lVert a \rVert) -1 \tag{2} \\&= (\lVert a \rVert-1)^2 \geq 0.\end{align}

$$(1)$$: Cauchy-Schwarz inequality
$$(2)$$: multiply both sides of $$\lVert v \rVert \geq 1$$ by the nonnegative scalar $$1-\lVert a \rVert$$.