Prove that $\mathbb{R}$ is 1-Lipschitz injective (Non linear Hahn Banach Theorem).

Question

Let $(X,d)$ be a metric space. We say $(X,d)$ is Lipschitz injective if $\forall$ metric spaces $(Y,\rho)$ and $Z\subset Y$ and any Lipschitz map $f:Z \rightarrow X$, we have an extension of $f$ to Lipschitz map $F: Y \rightarrow X$ s.t. $F_{|Z}=f$.

Further, if there exists $p>0$ such that, $L.C(F)\leq L.C(f)$ for all such functions, then $(X,d)$ is called $p-$ Lipschitz injective.(where L.C(f) is the Lipschitz constant of f)

The objective is to show that $(R, usual)$ is 1-Lipschitz injective.

Answer 1

Let $(X,d)$ be a metric space and $Y\subsetneq X$.\ ($w.l.o.g$ Assume $Y$ is closed, else $f$ can be extended uniquely to $cl(Y)$.)\ Suppose $f:Y \rightarrow \mathbb{R}$ be a Lipschitz map, with Lipschitz constant $=L$. Let $x_0\in X\setminus Y$ and $Y_1=Y\cup {x_0}$.\ Let $x_1,x_2 \in Y$, then $$ \\ f(x_1)-f(x_2)\leq L\cdot d(x_1,x_2)\leq L \cdot d(x_1,x_0)+L\cdot d(x_0,x_2)$$ $$f(x_1)-L \cdot d(x_1,x_0)\leq f(x_2)+L\cdot d(x_0,x_2) \hspace{5mm} \forall \hspace{2mm}x_1,x_2\in Y$$ $$ \therefore A= \sup_{x_1 \in Y}f(x_1)-L \cdot d(x_1,x_0)\leq \inf_{x_2 \in Y} f(x_2)+L\cdot d(x_0,x_2)=B$$

Choose $C \in \mathbb{R}$ such that $A\leq C \leq B$.\ Define [ f_1(x) = \begin{cases} f(x) & x\in Y \\ C & x=x_0\\ \end{cases} ] Then $ |f_1(x)-f_1(x_0)| \leq L \cdot d(x,x_0)$.\ Thus, $f_1: Y_1 \rightarrow \mathbb{R}$ is a Lipschitz function.\\ Define $\mathcal{F}=\{(Y_1,f_1): Y\subseteq Y_1\subseteq X, f_{1_{|Y}}=f, f_1 \text{Lipschitz with Lipschitz constant} L\}$.\ Define an order on $\mathcal{F}$ as $(Y_1,f_1)\preceq(Y_2,f_2)$ $\iff$ $Y_1 \subset Y_2$, $f_{2_{|Y_{1}}}=f_1$. Then $(\mathcal{F},\preceq)$ is a poset which is non empty as $(Y,f) \in \mathcal{F}$.\ Let $\{(Y_\alpha,f_\alpha)\}_{\alpha \in \Lambda}$ be a chain in $\mathcal{F}$. Then define $Y_0=\cup_{\alpha \in \Lambda}Y_\alpha$ and for $y\in Y$ (say $y \in Y_{\alpha_0}$) $f_0(y)=f_{\alpha_0}(y)$. Because $\{(Y_\alpha,f_{\alpha})\}_{\alpha \in \Lambda}$ is a chain, the above function is well defined and Lipschitz with Lipschitz constant L. Thus $(Y_0,f_0) \in \mathcal{F}$ and $(Y_\alpha,f_\alpha)\preceq (Y_0,f_0)$ $\forall$ $\alpha \in \Lambda$.\

Thus every chain has a maximal element and hence, by Hausdorff maximalilty principle, $(\mathcal{F},\preceq)$ has a maximal element say $(Y',F)$.\ Claim: $Y'=X$.\ If not $\exists $ $x_0 \in X \setminus Y'$ and from above exercise the function $F$ can be extended to a Lipschitz function with Lipschitz constant $L$, contradicting the maximality of ${Y',F}$.

Thus we have a Lipschitz function $F: X\rightarrow \mathbb{R}$ with Lipschitz constant $L$.

$i.e.$ $\mathbb{R}$ is 1- injective Lipschtiz.\

Answer 2

$\textbf{Lemma :}$ Let $\{f_i:X \rightarrow \mathbb{R}\}_{i\in J}$ be a collection of Lipschitz function with Lipschitz constant $L$. Then the function $F: X \rightarrow \mathbb{R}$, defined as $F(x)=\inf_{i \in J}f_i(x)$ is also a Lipschitz function (with Lipschitz constant L) provided $\inf_{i \in J}f_i(x_0)$ is finite for atleast one of the $x_0\in X$.

Proof of Lemma : Let $x,y \in X$, then $$ f_i(x)-f_i(y)\leq L\cdot d(x,y)$$ $$\inf_{j \in J}f_j(x) \leq f_i(x) \leq f_i(y)+L \cdot d(x,y) \hspace{3mm} \forall \hspace{2mm} i \in J$$
$$\inf_{j \in J}f_j(x)\leq \inf_{i\in J}f_i(y)+L\cdot d(x,y)$$

Since this is true for every $x,y \in X$, $-\infty< \inf_{j \in J}f_j(x_0)-L\cdot d(x_0,y)\leq \inf_{i\in J}f_i(y)$ $\forall$ $y \in X$. Thus, $$F(x) \leq F(y)+L \cdot d(x,y)$$ Similarly, $$F(y) \leq F(x)+L \cdot d(x,y)$$ $$\text{i.e.} |{F(x)-F(y)}|\leq L\cdot d(x,y)$$

Solution (Without using Axiom of Choice): Let $(X,d)$ be a metric space and $Y\subset X$. Let $f: Y \rightarrow \mathbb{R}$ be a Lipschitz map with Lipschitz constant $L$.

For each $y \in Y$, define $f_y: X \rightarrow \mathbb{R}$ as $$f_y(x)=f(y)+L \cdot d(x,y)$$ Then $f_y(x_1)-f_y(x_2)=L(d(x_1,y)-d(x_2,y))\leq L \cdot d(x_1,x_2)$. $i.e.$ $f_y$ is a Lipschitz function with Lipschitz constant $L$.\

Also, $f$ is Lipschitz $\implies f(y)-f(y_1)\leq L \cdot d(y,y_1)$ i.e. $f(y)\leq f(y_1)+L\cdot d(y,y_1)$ $\forall$ $y_1\in Y$.(So, $\inf_{y \in Y} f_y(x)$ is finite for all $x \in Y$)\

By $Lemma$, $F: X \rightarrow \mathbb{R}$ defined as $F(x)=\inf_{y \in Y} f_y(x)$ is a Lipschitz function with Lipschitz constant $L$ and $F(y)=f_y(y)=f(y)\hspace{2mm}(\leq f_{y_1}(y)=f(y_1)+Ld(y,y_1))$.\

Thus $F: X \rightarrow \mathbb{R}$ is a Lipschitz extension of $f: Y \rightarrow \mathbb{R}$.

$i.e.$ $\mathbb{R}$ is 1- injective Lipschtiz.