# Second derivative test in the Hilbert space case

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 should improve the notation: $Dc(x)\in L(E,E)$, so $\langle \lambda,Dc(x)\rangle_E$ makes no sense. Same for the second derivatives. – daw Nov 4 at 14:43
• @daw To me it makes perfectly sense. Writing $\langle\lambda,{\rm D}c(x)\rangle_E$ instead of $E\ni x\mapsto \langle\lambda,{\rm D}c(x)x\rangle_E$ is the same as writing $f$ instead of $x\mapsto f(x)$ for any function $f$. – 0xbadf00d Nov 4 at 15:00

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) $$Df(x) \gamma''(0) = \langle \lambda, Dc(x) \gamma''(0)\rangle = -\langle \lambda, D^2c(x)(u,u) \rangle.$$
• You obtain $D^2c(\gamma(t))(\gamma'(t),\gamma'(t)) + Dc(\gamma(t))\gamma''(t)=0$ by arguing that $c\circ\gamma$ is identically $0$ in a neighborhood of $t$, right? – 0xbadf00d Nov 5 at 18:08
• I'm asking cause I wonder which assumptions we really need. In order for $(2)$ to hold it's sufficient that $f$ is Fréchet differentiable (not necessarily continuously Fréchet differentiable) and $c$ is continuously Fréchet differentiable. Moreover, in order to conclude $(f\circ\gamma)''(0)\ge0$, we only need that $f$ is twice Fréchet differentiable at the particular $x$ (not necessarily on all of $Ω$). Now in order for the identity in my previous comment to hold, it should suffice to further assume that $c$ is twice Fréchet differentiable at the particular $x$ as well. What do you think? – 0xbadf00d Nov 5 at 18:32