1

It is well-known that for a convex closed set $K\subset\mathbb{R}^D$, the projection operator $\Pi:\mathbb{R}^D\rightarrow K$ given by $$ \Pi(x)=\arg \min_{y\in K} \| x-y\|_2 $$ is non-expansive in the norm $\ell_2$, that is, $$ \| \Pi(x) - \Pi(x')\|_2 \leq \| x - x'\|_2, \forall x,x'\in \mathbb{R}^D. $$ See for example this question asked before.

My question is, does the modified projection $$ \bar{\Pi}(x)=\arg \min_{y\in K} \| x-y\|_1 $$ satisfy non-expansiveness in the $\ell_1$ norm? In particular, does it hold that $$ \| \bar{\Pi}(x) - \bar{\Pi}(x')\|_1 \leq \| x - x'\|_1, \forall x,x'\in \mathbb{R}^D, $$ and if not, is there a "natural" projection operator satisfying $\bar{\Pi} \circ \bar{\Pi} = \bar{\Pi}$ that is non-expansive in the $\ell_1$ norm (or more generally, in any $\ell_p$ norm)?

A related discussion seems to be this, but non-expansiveness as above does not seem to follow from the $\nabla \| \cdot \|_1$-resolvent of the normal cone operator $N_K$.

ntrstd11
  • 239

1 Answers1

2

The projection in the $l^1$-norm is in general not uniquely determined. That is the problem $$ \min_{y\in K} \|x-y\|_1 $$ may have multiple solutions. Then in your inequality we could choose $x=x'$ but $\Pi(x)\ne \Pi(x)$.

For example, the solution of the above problem for $K=\{y: \|y\|_1\le 1\}$ and $x=(1,1)$ are all the points on the line between $(1,0)$ and $(0,1)$.

daw
  • 54,637
  • 2
  • 44
  • 85