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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k(x) = \min \left\{ x^2 , \frac{k^2}{2} + k \left| x \pm \frac{k}{2} \right| \right\}.}

Plot of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) = x^2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k(x)} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k = 0, 1, 2, 3} .

Results

Proposition.

• If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is proper and bounded below, so is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k} . Furthermore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k} is continuous for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \geq 0} .
• If, in addition, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is lower semicontinuous, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k(x) \nearrow g(x)} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in X} .
• In this case, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k \wedge k := \min(g_k,k)} is continuous and bounded and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_k(x) \wedge k \nearrow g(x)} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in X} .

Proof.

• Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is proper, there exists Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_0 \in X} such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(y_0) < +\infty} . Then for any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in X}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty < \inf\limits_{y \in Y} g(y) \leq g_k(x) \leq g(y_0) + k d(x,y_0) < +\infty .}

Thus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $x\in X$.

References

Possible list of references, will fix accordingly

Bauschke-Combette Ch 12.; Santambrogio (6); Ambrosio-Gigli-Savare (59-61)