Let
- $H,E$ be $\mathbb R$-Hilbert spaces;
- $f\in C^2(\Omega)$;
- $c\in C^2(\Omega,E)$;
- $M:=\left\{c=0\right\}$;
- $x\in M$ 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 $x$.
Now, let $$\mathcal L(x,\lambda):=f(x)-\langle\lambda,c(x)\rangle_E\;\;\;\text{for }\lambda\in E.$$ As shown here, $${\rm D}_1\mathcal L(x,\lambda)={\rm D}f(x)-\langle\lambda,{\rm D}c(x)\rangle_E=0\tag2$$ for some $\lambda\in E$ and, under the identification $\mathfrak L(H,\mathbb R)=H'\cong H$, $${\rm D}f(x)\in\left(\ker{\rm D}c(x)\right)^\perp=\overline{\operatorname{im}\left(({\rm D}c(x))^\ast\right)}\tag3.$$
I would like to conclude $$\langle{\rm D}_1^2\mathcal L(x,\lambda)u,u\rangle_H\ge0\;\;\;\text{for all }u\in\ker({\rm D}c(x)).\tag4$$ (Note that ${\rm D}_1^2\mathcal L(x,\lambda)\in\mathfrak L(H,H')\cong\mathfrak L(H)$.)
We should be able to argue in the following manner: Let $u\in\ker({\rm D}c(x))$. We know that there is a $\varepsilon>0$ and a $\gamma\in C^2((-\varepsilon,\varepsilon),M)$ with $\gamma(0)=x$ and $\gamma'(0)=u$. By definition of $x$, $0$ is a local minimum of $f\circ\gamma$ and hence $$0\le(f\circ\gamma)''(0)=\left({\rm D}^2f(x)\gamma'(0)\right)\gamma'(0)+{\rm D}f(x)\gamma''(0)\tag5.$$ On the other hand, $${\rm D}_1^2\mathcal L(x,\lambda)={\rm D}^2f(x)-\langle\lambda,{\rm D}^2c(x)\rangle_E.\tag6$$
Now we somehow need to incorporate $(2)$ and $\gamma'(0)\in\ker({\rm D}c(x))$. How can we do that?
You know $c(\gamma(t))=0$. Differentiating twice with respect to $t$ yields $$ D^2c(\gamma(t))(\gamma'(t),\gamma'(t)) + Dc(\gamma(t))\gamma''(t)=0. $$ Setting $t=0$ gives $$ D^2c(x)(u,u) + Dc(x)\gamma''(0)=0. $$ Then from (2) $$ Df(x) \gamma''(0) = \langle \lambda, Dc(x) \gamma''(0)\rangle = -\langle \lambda, D^2c(x)(u,u) \rangle. $$