Let
- $H,E$ be $\mathbb R$-Hilbert spaces;
- $f\in C^1(\Omega)$;
- $c\in C^1(\Omega,E)$;
- $M:=\left\{c=0\right\}$;
- $x\in M$ be a local minimum of $f$ constrained on $M$, i.e. $$f(x)\le f(y)\;\;\;\text{for all }M\cap N\tag1$$ for some open neighborhood $N$ of $x$.
I want to show that if $x$ is a regular point of $c$ (i.e. ${\rm D}c(x)$ is surjective), then there is a $\lambda\in E$ with $$\left.{\rm D}f(x)\right|_{U^\perp}=\langle\lambda,{\rm D}c(x)\rangle_E\tag2.$$
As shown here, $U:=\ker{\rm D}c(x)$ is closed, $L:=\left.{\rm D}c(x)\right|_{U^\perp}$ is bijective and $L^{-1}\in\mathfrak L(E,U^\perp)$. Thus, $$g:={\rm D}f(x)\circ L^{-1}\in E'\tag3$$ and hence $$g=\langle\lambda,\;\cdot\;\rangle_E$$ by Riesz' representation theorem. So, $$\left.{\rm D}f(x)\right|_{U^\perp}=\langle\lambda,L\rangle_E\tag4.$$
We are left to show that if $u_0\in U$, then $${\rm D}f(x)u_0=0\tag5.$$
Again, as shown here, there is a $\varepsilon>0$ and a $\gamma\in C^1((-\varepsilon,\varepsilon),M)$ with $\gamma(0)=x$ and $\gamma'(0)=u_0$. Moreover, $$(f\circ\gamma)'(0)={\rm D}f(x)u_0.\tag6$$
So, all we need to do is showing that $0$ is a local minimum of $f\circ\gamma$. How can we do this?
We need to use $(1)$. And since $\gamma$ maps into $M$, $$B:=\gamma^{-1}(M\cap N)=\gamma^{-1}(N)\tag7$$ is open. But is $B$ a neighborhood of $0$ with $$(f\circ\gamma)(0)\le(f\circ\gamma)(t)\tag8$$ for all $t\in B$?
We have $$f(x)\le f(y)\;\;\;\text{for all }y\in M\cap N\tag1.$$ Since $\gamma(0)=x\in M\cap N$, the set $B=\gamma^{-1}(M\cap N)$ is an open neighbourhood of $0$. Notice that $\gamma(t)\in M\cap N$ for all $t\in B$ and by $(1)$ we can conclude$$f(x)=f\circ\gamma(0)\le f\circ\gamma(t)$$ for all $t\in B$.