[proofplan]
Every admissible path begins at $0$ and ends at a point outside the sphere of radius $\rho$, so continuity of the path forces it to meet that sphere; the mountain-pass lower bound gives $c\geq \alpha$. To prove existence of a critical point, we argue by contradiction: if no critical point occurs at level $c$, the Palais-Smale condition forces the derivative norm to be uniformly bounded away from zero near that level. A quantitative deformation lemma then pushes every almost optimal path strictly below level $c$, while keeping its endpoints fixed, contradicting the definition of $c$.
[/proofplan]
[step:Show every admissible path crosses the sphere of radius $\rho$]
Fix $\gamma\in\Gamma$. Define the continuous map $r_\gamma:[0,1]\to\mathbb{R}$ by
\begin{align*}
r_\gamma(t)=\|\gamma(t)\|_X.
\end{align*}
Since $\gamma(0)=0$, we have $r_\gamma(0)=0$. Since $\gamma(1)=e$, we have $r_\gamma(1)=\|e\|_X>\rho$. By the intermediate value property for the continuous real-valued map $r_\gamma$, there exists $t_\gamma\in(0,1)$ such that
\begin{align*}
\|\gamma(t_\gamma)\|_X=\rho.
\end{align*}
The mountain-pass hypothesis therefore gives
\begin{align*}
I[\gamma(t_\gamma)]\geq \alpha.
\end{align*}
Hence
\begin{align*}
\max_{t\in[0,1]} I[\gamma(t)]\geq \alpha.
\end{align*}
Taking the infimum over all $\gamma\in\Gamma$ gives
\begin{align*}
c\geq \alpha.
\end{align*}
[/step]
[step:Derive a uniform derivative lower bound near level $c$ if no critical point exists]
Let $X^*$ denote the continuous dual [Banach space](/page/Banach%20Space) of $X$, equipped with its operator norm $\|\cdot\|_{X^*}$. Assume, toward a contradiction, that there is no $u\in X$ such that $I[u]=c$ and $I'[u]=0$ in $X^*$. We claim that there exist constants $\varepsilon_0>0$ and $\delta_0>0$ such that
\begin{align*}
|I[v]-c|\leq 2\varepsilon_0 \implies \|I'[v]\|_{X^*}\geq \delta_0
\end{align*}
for every $v\in X$.
Indeed, if no such constants existed, then for every $k\in\mathbb{N}$ there would be $u_k\in X$ such that
\begin{align*}
|I[u_k]-c|\leq \frac{1}{k}
\end{align*}
and
\begin{align*}
\|I'[u_k]\|_{X^*}<\frac{1}{k}.
\end{align*}
Thus $I[u_k]\to c$ and $\|I'[u_k]\|_{X^*}\to 0$. By the Palais-Smale condition at level $c$, there exists a subsequence $(u_{k_j})_{j=1}^{\infty}$ and an element $u\in X$ such that $u_{k_j}\to u$ strongly in $X$. Since $I:X\to\mathbb{R}$ and $I':X\to X^*$ are continuous, we obtain
\begin{align*}
I[u]=\lim_{j\to\infty} I[u_{k_j}]=c
\end{align*}
and
\begin{align*}
I'[u]=\lim_{j\to\infty} I'[u_{k_j}]=0 \quad \text{in } X^*.
\end{align*}
This contradicts the assumption that no critical point exists at level $c$. Therefore the claimed constants $\varepsilon_0$ and $\delta_0$ exist.
[/step]
[step:Use the deformation lemma to push an almost optimal path below $c$]
Choose $\varepsilon>0$ satisfying
\begin{align*}
0<\varepsilon<\varepsilon_0
\end{align*}
and
\begin{align*}
2\varepsilon<c.
\end{align*}
The second condition is possible because $c\geq \alpha>0$.
We use the following standard quantitative deformation lemma, stated here in the precise form needed because its standalone Androma theorem is not yet available: if $I\in C^1(X;\mathbb{R})$ and $\|I'[v]\|_{X^*}\geq \delta_0$ for every $v$ with $|I[v]-c|\leq 2\varepsilon$, then there exists a continuous map
\begin{align*}
\eta:X\to X
\end{align*}
such that $\eta(v)=v$ whenever $I[v]\leq c-2\varepsilon$, and such that
\begin{align*}
I[v]\leq c+\varepsilon \implies I[\eta(v)]\leq c-\varepsilon.
\end{align*}
Because $I[0]=0$ and $2\varepsilon<c$, we have
\begin{align*}
I[0]\leq c-2\varepsilon.
\end{align*}
Because $I[e]<0$ and $2\varepsilon<c$, we also have
\begin{align*}
I[e]\leq c-2\varepsilon.
\end{align*}
Thus the deformation fixes both endpoints:
\begin{align*}
\eta(0)=0
\end{align*}
and
\begin{align*}
\eta(e)=e.
\end{align*}
[guided]
The contradiction will come from taking a path whose height is only slightly above $c$ and deforming it into a path whose height is strictly below $c$. The deformation lemma is allowed here because the previous step produced a uniform lower bound for $\|I'[v]\|_{X^*}$ on the whole energy band $|I[v]-c|\leq 2\varepsilon$. This is exactly the hypothesis that prevents the flow from getting stuck near level $c$.
Choose $\varepsilon>0$ with $\varepsilon<\varepsilon_0$ and $2\varepsilon<c$. Since $c\geq\alpha>0$, such a choice is possible. The quantitative deformation lemma applied at level $c$ gives a continuous map
\begin{align*}
\eta:X\to X
\end{align*}
with two relevant properties. First, $\eta$ fixes all points whose energy is at most $c-2\varepsilon$. Second, every point whose energy is at most $c+\varepsilon$ is pushed to energy at most $c-\varepsilon$:
\begin{align*}
I[v]\leq c+\varepsilon \implies I[\eta(v)]\leq c-\varepsilon.
\end{align*}
We must check that applying $\eta$ to an admissible path keeps the endpoints admissible. Since $I[0]=0$ and $2\varepsilon<c$, we have
\begin{align*}
I[0]\leq c-2\varepsilon.
\end{align*}
Therefore $\eta(0)=0$. Also $I[e]<0$ and $2\varepsilon<c$, so
\begin{align*}
I[e]\leq c-2\varepsilon.
\end{align*}
Therefore $\eta(e)=e$. Thus the deformation changes the middle of the path but leaves the required endpoints unchanged.
[/guided]
[/step]
[step:Contradict the definition of the minimax level]
By the definition of $c$ as an infimum, there exists $\gamma\in\Gamma$ such that
\begin{align*}
\max_{t\in[0,1]} I[\gamma(t)]\leq c+\varepsilon.
\end{align*}
Define the deformed path $\widetilde{\gamma}:[0,1]\to X$ by
\begin{align*}
\widetilde{\gamma}(t)=\eta(\gamma(t)).
\end{align*}
Since $\gamma$ and $\eta$ are continuous, $\widetilde{\gamma}$ is continuous. Since $\eta(0)=0$ and $\eta(e)=e$, we have
\begin{align*}
\widetilde{\gamma}(0)=0
\end{align*}
and
\begin{align*}
\widetilde{\gamma}(1)=e.
\end{align*}
Hence $\widetilde{\gamma}\in\Gamma$.
For every $t\in[0,1]$, the choice of $\gamma$ gives $I[\gamma(t)]\leq c+\varepsilon$. The deformation lemma therefore gives
\begin{align*}
I[\widetilde{\gamma}(t)]=I[\eta(\gamma(t))]\leq c-\varepsilon.
\end{align*}
Consequently
\begin{align*}
\max_{t\in[0,1]} I[\widetilde{\gamma}(t)]\leq c-\varepsilon.
\end{align*}
Since $\widetilde{\gamma}\in\Gamma$, the definition of $c$ gives
\begin{align*}
c\leq \max_{t\in[0,1]} I[\widetilde{\gamma}(t)].
\end{align*}
Combining the last two inequalities yields $c\leq c-\varepsilon$, a contradiction.
[/step]
[step:Extract the critical point at level $c$]
The contradiction shows that the assumption of no critical point at level $c$ is false. Therefore there exists $u\in X$ such that
\begin{align*}
I[u]=c
\end{align*}
and
\begin{align*}
I'[u]=0 \quad \text{in } X^*.
\end{align*}
Together with the lower bound $c\geq\alpha$ proved above, this completes the proof.
[/step]
