Ekeland's variational principle

Define a function ${\displaystyle G:X\times X\to \mathbb {R} \cup \{+\infty \}}$by $${\displaystyle G(x,y)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~f(x)+\varepsilon \;d(x,y)}$$which is lower semicontinuous because it is the sum of the lower semicontinuous function ${\displaystyle f}$and the continuous function ${\displaystyle (x,y)\mapsto \varepsilon \;d(x,y).}$Given ${\displaystyle z\in X,}$denote the functions with one coordinate fixed at ${\displaystyle z}$by $${\displaystyle G_{z}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~G(z,\cdot ):X\to \mathbb {R} \cup \{+\infty \}\;{\text{ and }}}$$$${\displaystyle G^{z}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~G(\cdot ,z):X\to \mathbb {R} \cup \{+\infty \}}$$and define the set $${\displaystyle F(z)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{y\in X:G^{z}(y)\leq f(z)\right\}~=~\{y\in X:f(y)+\varepsilon \;d(y,z)\leq f(z)\},}$$which is not empty since ${\displaystyle z\in F(z).}$An element ${\displaystyle v\in X}$satisfies the conclusion of this theorem if and only if ${\displaystyle F(v)=\{v\}.}$It remains to find such an element.

It may be verified that for every ${\displaystyle x\in X,}$

1. ${\displaystyle F(x)}$is closed (because ${\displaystyle G^{x}\,{\stackrel {\scriptscriptstyle {\text{def}}}{=}}\,G(\cdot ,x):X\to \mathbb {R} \cup \{+\infty \}}$is lower semicontinuous);
2. if ${\displaystyle x\notin \operatorname {dom} f}$then ${\displaystyle F(x)=X;}$
3. if ${\displaystyle x\in \operatorname {dom} f}$then ${\displaystyle x\in F(x)\subseteq \operatorname {dom} f;}$in particular, ${\displaystyle x_{0}\in F\left(x_{0}\right)\subseteq \operatorname {dom} f;}$
4. if ${\displaystyle y\in F(x)}$then ${\displaystyle F(y)\subseteq F(x).}$

Let ${\displaystyle s_{0}=\inf _{x\in F\left(x_{0}\right)}f(x),}$which is a real number because ${\displaystyle f}$was assumed to be bounded below. Pick ${\displaystyle x_{1}\in F\left(x_{0}\right)}$such that ${\displaystyle f\left(x_{1}\right)<s_{0}+2^{-1}.}$Having defined ${\displaystyle s_{n-1}}$and ${\displaystyle x_{n},}$let $${\displaystyle s_{n}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\inf _{x\in F\left(x_{n}\right)}f(x)}$$and pick ${\displaystyle x_{n+1}\in F\left(x_{n}\right)}$such that ${\displaystyle f\left(x_{n+1}\right)<s_{n}+2^{-(n+1)}.}$For any ${\displaystyle n\geq 0,}$${\displaystyle x_{n+1}\in F\left(x_{n}\right)}$guarantees that ${\displaystyle s_{n}\leq f\left(x_{n+1}\right)}$and ${\displaystyle F\left(x_{n+1}\right)\subseteq F\left(x_{n}\right),}$which in turn implies ${\displaystyle s_{n+1}\geq s_{n}}$and thus also $${\displaystyle f\left(x_{n+2}\right)\geq s_{n+1}\geq s_{n}.}$$So if ${\displaystyle n\geq 1}$then ${\displaystyle x_{n+1}\in F\left(x_{n}\right){\stackrel {\scriptscriptstyle {\text{def}}}{=}}\left\{y\in X:f(y)+\varepsilon \;d\left(y,x_{n}\right)\leq f\left(x_{n}\right)\right\}}$and ${\displaystyle f\left(x_{n+1}\right)\geq s_{n-1},}$which guarantee $${\displaystyle \varepsilon \;d\left(x_{n+1},x_{n}\right)~\leq ~f\left(x_{n}\right)-f\left(x_{n+1}\right)~\leq ~f\left(x_{n}\right)-s_{n-1}~<~{\frac {1}{2^{n}}}.}$$

It follows that for all positive integers ${\displaystyle n,p\geq 1,}$$${\displaystyle d\left(x_{n+p},x_{n}\right)~\leq ~2\;{\frac {\varepsilon ^{-1}}{2^{n}}},}$$which proves that ${\displaystyle x_{\bullet }:=\left(x_{n}\right)_{n=0}^{\infty }}$is a Cauchy sequence. Because ${\displaystyle X}$is a complete metric space, there exists some ${\displaystyle v\in X}$such that ${\displaystyle x_{\bullet }}$converges to ${\displaystyle v.}$For any ${\displaystyle n\geq 0,}$since ${\displaystyle F\left(x_{n}\right)}$is a closed set that contain the sequence ${\displaystyle x_{n},x_{n+1},x_{n+2},\ldots ,}$it must also contain this sequence's limit, which is ${\displaystyle v;}$thus ${\displaystyle v\in F\left(x_{n}\right)}$and in particular, ${\displaystyle v\in F\left(x_{0}\right).}$

The theorem will follow once it is shown that ${\displaystyle F(v)=\{v\}.}$So let ${\displaystyle x\in F(v)}$and it remains to show ${\displaystyle x=v.}$Because ${\displaystyle x\in F\left(x_{n}\right)}$for all ${\displaystyle n\geq 0,}$it follows as above that ${\displaystyle \varepsilon \;d\left(x,x_{n}\right)\leq 2^{-n},}$which implies that ${\displaystyle x_{\bullet }}$converges to ${\displaystyle x.}$Because ${\displaystyle x_{\bullet }}$also converges to ${\displaystyle v}$and limits in metric spaces are unique, ${\displaystyle x=v.}$${\displaystyle \blacksquare }$Q.E.D.