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.

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.