Caristi fixed-point theorem

Remarks

The theorem follows from Ekeland's variational principle. With $\varepsilon = 1$, it gives a point $x_0$ with $\varphi(x_0) < \varphi(x) + d(x, x_0)$ for every $x \ne x_0$. If $T(x_0) \ne x_0$, then taking $x = T(x_0)$ gives $\varphi(x_0) < \varphi(T(x_0)) + d(T(x_0), x_0) \le \varphi(x_0)$, a contradiction.

Example: the Banach fixed point theorem is a special case. If $d(T(x), T(y)) \le k\, d(x, y)$ with $0 \le k < 1$, put $\varphi(x) = d(x, T(x))/(1 - k)$, which is continuous. The Caristi condition becomes $d(T(x), T(T(x))) \le k\, d(x, T(x))$, which holds.

Completeness is needed: on $X = (0, 1]$ with the usual metric, $T(x) = x/2$ and $\varphi(x) = x$ satisfy $d(x, T(x)) = x/2 = \varphi(x) - \varphi(T(x))$, but $T$ has no fixed point in $X$.

Going deeper

For a student

The theorem does not ask $T$ to be continuous, which makes it more flexible than the Banach fixed point theorem. It is equivalent in strength to Ekeland's variational principle, and each can be derived from the other. It is used in nonlinear analysis for maps that are only known to decrease a lower semicontinuous potential, such as some descent schemes in optimisation.