Since the curve selection lemma is central to many proofs concerning semianalytic sets on finite dimensional spaces, it's interesting to consider when a similar result might hold for sets defined through inequalities involving analytic functions on infinite dimensional spaces.
The curve selection lemma often functions as a kind of compactness result that allows us to restrict attention to a one-dimensional curve. Like the Lojasiewicz inequality, it won't hold in general in infinite dimensions and this failure can be linked to the non-compactness of the unit sphere. For example, suppose we have $\mathcal{E}(u) = \|u\|^3 - c(u)\|u\|^2$. For an orthonormal basis $\{e_i\}$, we can arrange that the coefficient $c(e_i) \to 0$ as $i \to \infty$, as we cycle through the infinite number of dimensions available. Thus, the set $\{ \mathcal{E}(u) > 0 \}$ contains a sequence approaching the origin but contains no analytic curve emanating from the origin.
First, note that the desired curve exists if and only if there exists at least one sequence $x_n \to 0$ with $x_n \in N \subset X$, where $N$ is a finite dimensional analytic manifold, since the ordinary curve selection lemma can then be applied.
We consider now the special case of a function with a Hessian that is elliptic.
Let $V$ be a Hilbert space and let $U \subseteq V$ be an open subset. Let $\mathcal{E} \in C^2(U)$ be an analytic function and assume the $0 \in U$ is a critical point, i.e. $\mathcal{E}'(0) = 0$. We suppose that $\mathcal{E}''(0)$ is a Fredholm operator, that is, it has finite-dimensional kernel and cokernel, and closed range. We also assume for convenience that $\mathcal{E}(0) = 0$. We define the set $W^\varepsilon = \{u:\mathcal{E}(u)\neq 0, \varepsilon\|\mathcal{E}_\theta\| \leq |\mathcal{E}_r| \}$.
Let $P$ be the orthogonal projection onto $\ker \mathcal{E}''(0)$ and $P'$ the adjoint projection. We define the finite dimensional analytic manifold \[ S = \{u \in U| (I-P')\mathcal{E}'(u)=0 \}, \] and denote by $Q$ the nonlinear projection onto $S$ (see [1] for details). We have the following Taylor series. \[ \mathcal{E}(u) = \mathcal{E}(Qu) + \frac{1}{2}\langle \mathcal{E}''(Qu)(u-Qu),u-Qu \rangle + o(\|u-Qu\|^3). \] $\bf{Lemma}$ Define the set $K \subseteq U$ by $K = \mathcal{E}(u) + \mathcal{H}(u) \ \sigma \ 0$ where $\sigma \in \{<,\leq,>,\geq\}$, and $\mathcal{H}$ is an analytic function consisting only of terms of order 3 and higher. Suppose $0 \in Cl(K \cap W^\varepsilon)$. Then there exists an analytic curve $\gamma(t):[0,\varepsilon) \to K \cap W^\varepsilon$ with $\gamma(0)=0$.
$\bf{Proof}$ Since $\mathcal{H}$ consists only of higher order terms which can be incorporated into the higher order terms of $\mathcal{E}$, we assume that $\mathcal{H}=0$. We also assume for readability that $\sigma$ is $>$, since the other cases are analogous. We have \begin{align*} K & = \{u \in U | \mathcal{E}(Qu) + (\mathcal{E}(u) - \mathcal{E}(Qu)) > 0\} \\ & = \{u \in U | \mathcal{E}(Qu) + \frac{1}{2}\langle \mathcal{E}''(Qu)(u-Qu),u-Qu \rangle + o(3) > 0\}. \end{align*} From 12.15 of [1], we know that $\|(I-P')\mathcal{E}'\| \geq c||u-Qu||$. Then from the triangle inequality and the definition of $W^\varepsilon$, we know that \[ |\mathcal{E}_r| \geq c||\mathcal{E}'|| \geq c||u-Qu|| \;\; (*). \] We can assume that $\mathcal{E}(Qu) \leq 0$ in a neighbourhood of $0$, since otherwise we can apply the usual curve selection lemma to the finite dimensional manifold $S$. We can write the quadratic term as \[ \frac{1}{2}\langle \mathcal{E}''(Qu)\hat{u},\hat{u} \rangle ||u-Qu||^2, \] where $\hat{u} = (u - Qu) / \| u - Qu \|$. If there exists $u_0$ such that the quadratic term is positive, then it is trivial to find the required curve. Thus, we may assume that \[ \frac{1}{2}\langle \mathcal{E}''(Qu)\hat{u},\hat{u} \rangle \leq 0 \] in $W^\varepsilon$. By assumption there exists a sequence $u_n \in K \cap W^\varepsilon$ with $u_n \to 0$. Since $\mathcal{E}(u_n) > 0$, The only remaining case is \[ \frac{1}{2}\langle \mathcal{E}''(Qu_n)\hat{u}_n,\hat{u}_n \rangle \to 0. \] Since the derivative must grow linearly along $V_1$, this can only happen if the radial component of $\mathcal{E}''(Qu_n)(u_n-Qu_n)$ is going to zero. This however violates $(*)$, since we are inside $W^\varepsilon$.
We remark that unlike in the finite dimensional case a curve selection lemma will not hold for the set $S$ outside of $W^\varepsilon$, as the Hessian cannot control the behaviour of the higher order terms where the linear growth in the derivative has no radial component. However, a curve selection lemma may hold for other expressions such as those involving the derivative $\mathcal{E}'$.
[1] Chill, R., Fasangova, E., Gradient Systems
