The Moreau-Yosida Regularization

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Motivation

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Definitions

Let $(X,d)$ be a metric space. A function $g:X\to (-\infty ,+\infty ]$ is said to be proper if it is not identically equal to $+\infty$, that is, if there exists $x\in X$ such that $g(x)<+\infty$.

For a given function $g:X\to (-\infty ,+\infty ]$ and $k\geq 0$, its Moreau-Yosida regularization $g_{k}:X\to [-\infty ,+\infty ]$ is given by

$g_{k}(x):=\inf \limits _{y\in X}\left[g(y)+kd(x,y)\right].$

Note that

$g_{k}(x)=\inf \limits _{y\in X}\left[g(y)+kd(x,y)\right]\leq g(x)+kd(x,x)=g(x)$.

Examples

• If $k=0$, then by definition $g_{0}$ is constant and $g_{0}\equiv \inf \limits _{y\in X}g(y)$.
• If $g$ is not proper, then $g_{k}=+\infty$ for all $k\geq 0$.

Take $(X,d):=(\mathbb {R} ,|\cdot |)$. If $g$ is finite-valued and differentiable, we can explicitly write down $g_{k}$. Then for a fixed $x\in \mathbb {R}$, the map $g_{k,x}:y\mapsto g(y)+k|x-y|$ is continuous everywhere and differentiable everywhere except for when $y=x$, where the derivative does not exist due to the absolute value. Thus we can apply standard optimization techniques from Calculus to solve for $g_{k}(x)$: find the critical points of $g_{k,x}$ and take the infimum of $g_{k,x}$ evaluated at the critical points. One of these values will always be the original function $g$ evaluated at $x$, since this corresponds to the critical point $y=x$ for $g_{k,x}$.

• Let $g(x):=x^{2}$. Then

$g_{k}(x)=\min \left\{x^{2},{\frac {k^{2}}{2}}+k\left|x\pm {\frac {k}{2}}\right|\right\}.$

Plot of $g(x)=x^{2}$ and $g_{k}(x)$ for $k=0,1,2,3$.

Results

Proposition.

• If $g$ is proper and bounded below, so is $g_{k}$. Furthermore, $g_{k}$ is continuous for all $k\geq 0$.
• If, in addition, $g$ is lower semicontinuous, then $g_{k}(x)\nearrow g(x)$ for all $x\in X$.
• In this case, $g_{k}\wedge k:=\min(g_{k},k)$ is continuous and bounded and $g_{k}(x)\wedge k\nearrow g(x)$ for all $x\in X$.

Plot of $g(x)=x^{2}$ and $g_{k}(x)\wedge k$ for $k=0,3,6,9$.

Proof.

• Since $g$ is proper, there exists $y_{0}\in X$ such that $g(y_{0})<+\infty$. Then for any $x\in X$

$-\infty <\inf \limits _{y\in Y}g(y)\leq g_{k}(x)\leq g(y_{0})+kd(x,y_{0})<+\infty .$

Thus $g_{k}$ is proper and bounded below. Next, for a fixed $y\in X$, let $h_{k,y}(x):=g(y)+d(x,y)$. Then as

$h_{k,y}(x_{1})-h_{k,y}(x_{2})=kd(x_{1},y)-kd(x_{2},y)\leq kd(x_{1},x_{2})$ ,

the family $\{h_{k,y}\}_{y\in X}$ is uniformly Lipschitz and hence equicontinuous. Thus $g_{k}=\inf \limits _{y\in Y}h_{k,y}$ is Lipschitz continuous.

• Suppose that $g$ is also lower semicontinuous. Note that for all $k_{1}\leq k_{2}$, $g_{k_{1}}(x)\leq g_{k_{2}}(x)\leq g(x)$. Thus it suffices to show that $\liminf \limits _{k\to \infty }g_{k}(x)\geq g(x)$. This inequality is automatically satisfied when the left hand side is infinite, so without loss of generality assume that $\liminf \limits _{k\to \infty }g_{k}(x)<+\infty$. By definition of infimum, for each $k\in \mathbb {N}$ there exists $y_{k}\in X$ such that

$g(y_{k})+kd(x,y_{k})\leq g_{k}(x)+{\frac {1}{k}}$.

Then

$+\infty >\liminf \limits _{k\to \infty }g_{k}(x)\geq \liminf \limits _{k\to \infty }\left[g(y_{k})+kd(x,y_{k})\right].$

$g(y_{k})$ is bounded below by assumption, while the only way $kd(x,y_{k})$ is finite in the limit is for $d(x,y_{k})$ to go to zero. Thus $y_{k}$ converges to $x$ in $X$, and by lower semicontinuity of $g$,

$\liminf \limits _{k\to \infty }g_{k}(x)\geq \liminf \limits _{k\to \infty }\left[g(y_{k})+kd(x,y_{k})\right]\geq g(x)$.

• By definition, $g_{k}\wedge k\in C_{b}(X)$. Since $g_{k}(x)\nearrow g(x)$ for all $x\in X$, $g_{k}(x)\wedge k\nearrow g(x)$ for all $x\in X$.