A problem about Moreau-Yosida approximation

Question

Let X be a metric space with metric $d(\cdot,\cdot)$, and let $F:X\rightarrow \mathbf R\cup\{\infty\} $be a functional. For $\lambda\gt 0$, we define the Moreau-yosida approximation $F^\lambda$ of $F$ as

$F^\lambda(x):=\inf_{y\in X}(\lambda F(y)+d^2(x,y))$

for $x\in X$. (I know this definition can be a little different from others, like enter link description here)

The following lemma is already knew:

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LEMMA $1$

Let $H$ be a Hilbert space, Let $F:H\rightarrow \mathbf R\cup\{\infty\}$be convex, bounded from below, and lower semicontinuous. Then for every $x\in H$ and $\lambda\gt 0$, there exists a unique

$y^\lambda:=J^\lambda(x)$

with

$F^\lambda(x)=\lambda F(y^\lambda)+d^2(x,y^\lambda)$

The metric $d(\cdot,\cdot)$ is the one induced by the inner product of $H$.

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Now I'm trying to use this Lemma 1 to prove the following lemma:

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LEMMA $2$

Let F and $y^\lambda=J^\lambda(x)$ be as in lemma 2. Let x be in the closure of $D(F)$ (here $D(F)$ is the domain of $F$, and for the $x\notin D(F)$, $F(x)=\infty$). Then

$x=\lim_{\lambda\to 0}J^\lambda(x)$

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I tried to prove this by contradiction but I had no idea. Any help would be appreciated!

Answer 1

Suppose for a contradiction that $\exists \lambda_n \to 0$ so that $||J^{\lambda_n}(x)-x||\geq \epsilon$ for all $n$. Then pick a sequence $y_n \in D(F)$ with $y_n \to x$ and $F(y_n)\lambda_n \to 0$. (Note this can always be done. Indeed, if $z_n \to x$, let for instance $y_1=z_1$, and let $y_n=z_1$ until $N_1=\inf\{n:F(z_1)\lambda_k \leq \frac{1}{2} \ and \ F(z_2)\lambda_k \leq \frac{1}{2} \ \forall k \geq n\}$, then let $y_{N_1+n}=z_2$ until $N_2=\inf \{n >N_1:F(z_2)\lambda_k \leq \frac{1}{4} \ and \ F(z_3)\lambda_k \leq \frac{1}{4} \ \forall k \geq n\}$ etc.)

Let now $a=\inf F$. Then $F^{\lambda_n}(x) \geq a\lambda_n +d^2(x,y^{\lambda_n}) \geq a\lambda_n+\epsilon^2$ for all $n$ by assumption. But note that $ \lambda_n F(y_n) + d^2(x,y_n) \to 0 $ so in particular, for n large enough $\lambda_n F(y_n) + d^2(x,y_n) < a\lambda_n+\epsilon^2\leq F^{\lambda_n}(x)$. This contradicts minimality, and thus $J^{\lambda_n}(x)\to x$.