Let
- $H$ be a $\mathbb R$-Hilbert space;
- $E$ be a $\mathbb R$-Banach space;
- $\Omega\subseteq H$ be open;
- $c:\Omega\to E$ be continuously Fréchet differentiable;
- $M:=\left\{c=0\right\}$;
- $x\in M$ such that ${\rm D}c(x)$ is surjective;
- $u_0\in\ker{\rm D}c(x)$.
How can we show that there is a $\rho>0$ and $\gamma\in C^1((-\rho,\rho),M)$ with $\gamma(0)=x$ and $\gamma'(0)=u_0$?
Since ${\rm D}c(x)$ is bounded, $U:=\ker{\rm D}c(x)$ is closed and $$H=U\oplus U^\perp\tag1$$. Since ${\rm D}c(x)$ is surjective, $$L:=\left.{\rm D}c(x)\right|_{U^\perp}$$ is bijective and, by the bounded inverse theorem, $$L^{-1}\in\mathfrak L(E,U^\perp).\tag2$$ Since $x\in\Omega$ and $\Omega$ is open, there is a $\varepsilon>0$ with $$B_\varepsilon(x)\subseteq\Omega\tag3.$$ Let $B:=B_{\varepsilon/2}(0_H)$. Then $$B^2\ni(u,v)\mapsto x+u+v\in B_\varepsilon(x)\tag4$$ and hence $$\Phi(u,v):=c(x+u+v)\;\;\;\text{for }u\in B\cap U\text{ and }v\in B\cap U^\perp$$ is continuously Fréchet differentiable. Note that $B\cap U$ is open in $U$, $B\cap U^\perp$ is open in $U^\perp$, $$\Phi(0,0)=c(x)=0\tag5$$ and $${\rm D}_2\Phi(0,0)={\rm D}c(x)\circ\operatorname{id}_{U^\perp}=L\tag6.$$ By the implicit function theorem, there is a $\delta\in(0,\varepsilon/2]$ and a continuously differentiable $\varphi:B_\delta(0_H)\cap U\to B_\delta(0_H)\cap U^\perp$ with $$\varphi(0)=0\tag7$$ and $$\Phi(u,\varphi(u)=0\tag8,$$ $${\rm D}\varphi(u)=-\left({\rm D}_2\Phi(u,\varphi(u))\right)^{-1}\circ{\rm D}_1\Phi(u,\varphi(u))\tag9$$ for all $u\in B_\delta(0_H)\cap U$. Assume $u_0\ne 0$ (otherwise the claim is trivially satisfied with $\gamma(t):=x$ for $t\in\mathbb R$). Let $$\rho:=\frac\delta{\left\|u_0\right\|_H}$$ and $$\gamma(t):=x+tu_0+\varphi(tu_0)\;\;\;\text{for }t\in(-\rho,\rho).$$ Note that $$tu_0\in B_\delta(0_H)\cap U\tag{10}$$ and hence $$(c\circ\gamma)(t)=\Phi(tu_0,\varphi(tu_0))=0\tag{11}$$ for all $t\in(-\rho,\rho)$, i.e. $$\gamma((-\rho,\rho))\subseteq M\tag{12}.$$ Moreover, $$\gamma(0)=x\tag{13},$$ $${\rm D}_1\Phi(0,0)={\rm D}c(x)\circ\operatorname{id}_U\equiv0\tag{14},$$ $${\rm D}\varphi(0)=-\left({\rm D}_2\Phi(0,0)\right)^{-1}\circ{\rm D}_1\Phi(0,0)=0\tag{15}$$ and hence $$\gamma'(0)=u_0+{\rm D}\varphi(0)=u_0.\tag{16}$$