TITLE: Geometric applications of Ekeland's variational principle
QUESTION [11 upvotes]: I'm looking for geometric applications of Ekeland's variational principle in order to see it at work in a context I'm familiar with. Let me recall the principle itself:
Definition. Let $(X,d)$ be a metric space, let $f$ be a real-valued function on $X$, and let $\epsilon > 0$. A point $m \in X$ is an $\epsilon$-minimizer if (1) $f(m) \leq \inf f + \epsilon$ and (2) $m$ is the unique minimizer of the perturbed function $x \mapsto f(x) + \epsilon d(m,x)$. 
Ekeland's weak principle. If $f$ is lower semi-continuous and $(X,d)$ is complete, there is an $\epsilon$-minimizer for every value of $\epsilon > 0$. 
Ekeland's strong principle. Assume $f$ is lower semi-continuous and $(X,d)$ is complete. If for a given $\epsilon > 0$ the point $y \in X$ satisfies $f(y) \leq \inf f + \epsilon$, then there is an $\epsilon$-minimizer $m \in X$ such that $f(m) \leq f(y)$ and which is at distance $\leq 1$ from the point $y$.
There are some applications in Ekeland's paper, but I'd like to see something more geometric using, for example, the Hausdorff distance on the space of convex sets. Actually, this principle caught my eye and I'm just curious as to what a geometer can do with it.

REPLY [5 votes]: Here is a quick application of the Ekeland's Variational Principle to Spectral Theory. Let $A$ be a bounded  linear symmetric operator on a Hilbert space $H$, and let $\mathbb{S}$ be the unit sphere of $H$. Then, an elementary result states:

$$\inf_{x\in\mathbb{S}}(Ax\cdot x)=\min \sigma(A)$$ 
  $$\sup_{x\in\mathbb{S}}(Ax\cdot x)=\max \sigma(A)$$

The standard proof is not complicated (it relies on the spectral radius formula, the identity $\|A^2\|=\|A\|^2$ for symmetric operators, plus some translation argument). Here is a completely different proof via the  Ekeland's principle, that also gives a nice geometrical insight.
A well-known basic fact is that  $(\lambda,x)\in \mathbb{R}\times\mathbb{S}$ is a pair eigenvalue-eigenvector for $A$ iff it is a pair critical value-critical point for the quadratic form of $A$ restricted on the unit sphere, namely the bounded and smooth function $q: \mathbb{S}\ni x\mapsto (Ax\cdot x)$. This just because $\nabla_{\mathbb{S}}q(x)=2(Ax-q(x)x)$. On the same lines, $\lambda\in\mathbb{R}$ is a spectral value iff it is a Palais-Smale level of $q$, that is, there exists a sequence $(x_j)_{j\ge0}\subset \mathbb{S}$ with $\nabla_{\mathbb{S}}q(x_j)=o(1)$ and $q(x_j)=\lambda+o(1)$ (Indeed, this is equivalent to $(A-\lambda)x_j=o(1)$ for a sequence of norm-one vectors, which exactly means that the symmetric operator $A-\lambda$ is not invertible). The proof via the EVP is now clear: take a minimizing resp. maximizing sequence for $q$ on $\mathbb{S}$. By the EVP, one can assume it is a Palais-Smale sequence, ending the proof.