Why is the epigraph of Moreau-Yosida Regularization a projection of a convex set?

179 Views Asked by At

The Moreau-Yosida Regularization is given by

\begin{equation} f_\mu(x) = \inf_y \left( f(y) + \frac{1}{2\mu} \| x - y \|^2 \right). \end{equation}

We know that $L(x, y) = f(y) + \frac{1}{2\mu} \| x - x \|^2$ is jointly convex in $x$ and $y$.

Source: https://statweb.stanford.edu/~candes/math301/Lectures/Moreau-Yosida.pdf

Why is the epigraph of $f_\mu(x)$ a projection of a convex set?

1

There are 1 best solutions below

0
On BEST ANSWER

To the best of my knowledge, this is a common mistake. It might be, of course, a projection of some convex set onto the $(x, t)$ coordinates, but it is not "the" convex set you might first think of.

Like any other 'infimal convolution', the epigraph of the Moreau-Yosida regularization is a sum of epigraphs. To be precise, it is the Minkowski sum of the epigraph of $f$ and the epigraph of $\frac{1}{2\mu}\|x\|_2^2$: $$ \operatorname{epi}(f_\mu) = \operatorname{epi}(f)+\operatorname{epi}(\tfrac{1}{2\mu}\|\cdot\|_2^2) \equiv \{ (x+y, t+s) : (x, t) \in \operatorname{epi}(f),~(y, s) \in \operatorname{epi}(\tfrac{1}{2\mu}\|\cdot\|_2^2) \} $$ See the excellent paper on Epigraphical Analysis by Attouch and Wets for more details. The sum of convex sets is convex, and that is one reason why $f_\mu$ is convex.