1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – daw Nov 4 at 14:43
  • $\begingroup$ @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 $f$. $\endgroup$ – 0xbadf00d Nov 4 at 15:00
0
$\begingroup$

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) $$ Df(x) \gamma''(0) = \langle \lambda, Dc(x) \gamma''(0)\rangle = -\langle \lambda, D^2c(x)(u,u) \rangle. $$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – 0xbadf00d Nov 5 at 18:08
  • $\begingroup$ 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? $\endgroup$ – 0xbadf00d Nov 5 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.