Theorem 1. (Caristi fixed point theorem) Let $(X, d)$ be a complete metric space. Let $T: X \to 2^X \setminus \{\emptyset\}$ and $f: X \to \mathbb{R}_+$ be a l.s.c function. Suppose that, for all points $x$ in $X$, $$\exists y \in Tx: f(y) + d(x, y) \le f(x).$$
Then $T$ has a fixed point in $X$, i.e. a point $\bar x$ such that $\bar x \in T \bar x$.
Theorem 2. (Ekeland's variational principle) Let $(X, d)$ be a complete metric space, and $f: X \to \mathbb{R}_+$ be a l.s.c. function. Then for every $\varepsilon$-optimal point $x_\varepsilon$ in $X$ ($f(x_\varepsilon) \le \inf_X f + \varepsilon$) and every real $\lambda > 0$ there exists a point $\bar x$ in $X$ such that the following conditions hold:
- $f(\bar x) \le f(x_\varepsilon)$;
- $d(x_\varepsilon, \bar x) \le \lambda$;
- $f(\bar x) < f(x) + \tfrac{\varepsilon}{\lambda} d(x, \bar x)$ for ever $x \in X \setminus \{\bar x\}$.
James Caristi states in his 1976 paper "Fixed point theorems for mappings satisfying inwardness condition" that these theorems are equivalent, and I found several more papers stating this result, but none of them contains a proof.
My idea to show that Theorem 1 implies Theorem 2 is to first set $\lambda = 1$ by rescaling the metric $d$, and then take $$Tx = \{ y \in X, y \ne x: f(y) + \varepsilon d(x, y) \le f(x) \}.$$ It clearly has no fixed point, meaning that some of the conditions of Theorem 1 must fail, namely $T\bar x = \emptyset$ for some $\bar x$, implying the third inequality of Theorem 2.
However, I do not understand where the first two inequalities come from.
Update: I found the proof of one-way implication in 1989 article "Fixed point theorems for multivalued mapping on complete metric spaces" by Mizoguchi and Takahasho. We start with selecting a $\varepsilon$-optimal point $x_\varepsilon$. We then put $$X' = \{x \in X: f(x) \le f(x_\varepsilon) - \varepsilon d(x, x_\varepsilon)\}.$$ It is nonempty (contains $x_\varepsilon$) and closed (because $f$ is l.s.c.) subset of a complete metric space, hence a complete metric space itself. We then indeed proceed by defining $T$ as I did. First two inequalities follow from the definition of $x_\varepsilon$ and $X'$: $$f(\bar x) \le f(x_\varepsilon) - \varepsilon d(\bar x, x_\varepsilon) \le f(x_\varepsilon)$$ and $$d(\bar x, x_\varepsilon) \le \frac{f(x_\varepsilon) - f(\bar x)}{\varepsilon} \le 1.$$
But I still don't know how to prove the converse implication. Any suggestions?