# Mountain Pass Theorem on a non-vector space

(Reposted from Math StackExchange because this may be more suited for professional mathematicians.)

The Mountain Pass Theorem roughly says the following. Consider a differentiable functional $I: H \rightarrow \mathbb{R}$ from a Hilbert space $H$ to the reals which satisfies an appropriate compactness/properness condition (the Palais-Smale condition). Assume the following are true:

• $I[0] = 0$;
• There exist positive constants $r$ and $a$ such that $I[q] \geq a$ for all $\|q\| = r$;
• There exists a $p$ (with $\|p\| > r$) at which $I[p] \leq 0$.

Then $I$ has a critical point $c$ at which $I[c] > 0$ (in fact $I[c] \geq a$).

My question is the following: can an analogous statement be made for functionals whose domain is not a vector space? For a concrete example (which is the one I care about), let $\mathcal{M}$ be the space of all differentiable Riemannian metrics on the (two-dimensional) sphere, and let $I: \mathcal{M} \rightarrow \mathbb{R}$ be a functional from $\mathcal{M}$ to the reals. Obviously $\mathcal{M}$ is not a vector space and so the Mountain Pass Theorem can't be applied. But say the following properties hold:

• $I$ obeys some appropriate properness condition;
• $I[g^\mathrm{round}] = 0$, where $g^\mathrm{round}$ is the metric of the round sphere;
• There exists a neighborhood $U$ of $g^\mathrm{round}$ in $\mathcal{M}$ such that $I[g] > 0$ for any $g \in U \setminus \{g^\mathrm{round}\}$;
• There exists a metric $\bar{g} \in \mathcal{M}$ at which $I[\bar{g}] < 0$.

Then it seems like one should still be able to apply the spirit of the proof of the Mountain Pass Theorem to conclude that $I$ must have some critical point $g^\mathrm{crit}$ at which $I[g^\mathrm{crit}] > 0$. Obviously making this statement precise would require understanding what the "appropriate properness condition" above is, but in principle it seems to me like the spirit of the theorem should still apply.

Does anyone know if this is indeed the case?

• $\mathcal M$ (metrics, say of class $C^k$) may not be a vector space. But if you fix a background metric $g_0$ (why not the standard one) and let $H$ denote the the vector space of $g_0$-selfadjoint sections of $\operatorname{End}(TS^2)$ of class $C^k$, there is an isomorphism $H\to\mathcal M$ given by $A\mapsto g_0(e^A-,-)$. Now $H$ is a vector space, but you want a Hilbert space, say, a Sobolev space of order $k$. This means, your critical point a priori corresponds to a singular metric. You next need some arguments to prove it is regular. – Sebastian Goette May 7 '17 at 10:30