Is there a metric proof for the Caristi theorem?

Question

I'm writing a paper about hyperconvexity in metric spaces and came across Caristi's theorem:

Let $(X, d)$ be a complete metric space and $\phi\colon X \to \mathbb{R}^+$ a continuous function. An arbitrary function $f\colon X \to X$ has a fixpoint if $\forall x \in X\colon d\big(x, f(x)\big) \leq \phi(x)-\phi\big(f(x)\big).$

I'm looking for an easy proof for this theorem. The original proof uses transfinite induction and is therefore too complicated. I found a shorter version in "An Introduction to Metric Spaces and Fixed Point Theory" (M. A. Khamsi) that uses nets. (I have added screenshots at the end of my question.)

However, I'm not familiar with the concept of nets as I've only worked with filters in topology. I was therefore wondering whether there exists a metric proof of Caristi's theorem, such that more advanced topology isn't necessary.

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Answer 1

I believe that you need to have $\varphi:X\to[0,\to)$ (or at least have its range bounded below). I don’t immediately see a substantially different argument, but I can get rid of the nets for you.

Let $X,d,\varphi$, and $f$ be as given in the statement of the theorem. Define a partial order $\preceq$ on $X$ by setting $x\preceq y$ iff $d(x,y)\le\varphi(x)-\varphi(y)$, just as in the Khamsi proof (as specialized to Caristi’s theorem). For each $x\in X$ let $T_x=\{y\in X:x\preceq y\}$.

$T_x=\{y\in X:\psi_x(y)\ge 0\}$, where

$$\psi_x:X\to\Bbb R:y\mapsto\varphi(x)-\varphi(y)-d(x,y)$$

is continuous, so $T_x$ is closed in $X$. Now suppose that $x,y\in X$ with $x\preceq y$, and let $z\in T_y$. Then $x\preceq y\preceq z$, so $z\in T_x$, and therefore $T_y\subseteq T_x$.

Now let $C$ be a chain in $\langle X,\preceq\rangle$, and let $r=\inf\varphi[C]$. For $x\in C$ let $S_x=\{y\in T_x:\varphi(y)\ge r\}$; clearly $S_x$ is closed. If $y\in S_x$, then $d(x,y)\le\varphi(x)-\varphi(y)\le\varphi(x)-r$, so

$$\operatorname{diam}S_x\le 2\big(\varphi(x)-r)\big)\;.$$

Thus, $\mathscr{S}=\{S_x:x\in C\}$ is a nested family of closed sets in the complete metric space $X$, and it has members of arbitrarily small diameter, so $\bigcap\mathscr{S}=\{z\}$ for some $z\in X$. For each $x\in C$ we then have $z\in S_x\subseteq T_x$, so $x\preceq z$, and $z$ is an upper bound for $C$. By Zorn’s lemma $X$ has a $\preceq$-maximal element $y$. By hypothesis $d\big(y,f(y)\big)\le\varphi(y)-\varphi\big(f(y)\big)$, so $y\preceq f(y)$, and the maximality of $y$ then implies that $y=f(y)$.

Answer 2

Because $\varphi$ is proper, there exists $u\in X$ such that $\varphi(u)<\infty$. Let $$C=\{x\in X :d(u,x)\leq\varphi(u)-\varphi(x)\}.$$ Then $C$ is a nonempty closed subset of $X$. We show that $C$ in invariant under $f$.

For each $x\in C$, we have $$d(u,x)\leq\varphi(u)-\varphi(x)$$ and hence from $d(x,f(x))\leq\varphi(x)-\varphi(f(x))$, $$\begin{align} \varphi(fx)\leq \varphi(x)-d(x,fx) &\leq \varphi(x)-d(x,fx)+\varphi(u)-\varphi(x)-d(u,x) \\ &= \varphi(u)-[d(x,fx)+d(u,x)]\\ &\leq \varphi(u)-d(u,fx), \end{align}$$ and it follows that $fx \in C$. Suppose, for contradiction, that $x\neq fx$ for all $x\in C$, then for each $x\in C$, there exists $w\in C$ such that $x\neq w$ and $d(x,w)\leq\varphi(x)-\varphi(w)$. There exists an $x_{0}\in C$ with $\varphi(x_{0})=\inf_{x\in C}\varphi(x)$. Hence for such an $x_{0}\in C$ we have $$ 0<d(x_{0},fx_{0})\leq\varphi(x_{0})-\varphi(fx_{0}) \leq \varphi(fx_{0})-\varphi(fx_{0}) = 0, $$ a contradiction.