Proof of necessary condition for a constrained extremum

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Let

  • $E_i$ be a $\mathbb R$-Banach space
  • $\Omega_1\subseteq E_1$ be open
  • $g\in C^1(\Omega_1,E_2)$
  • $M_1:=\{g=0\}$
  • $f:\Omega_1\to\mathbb R$
  • $x_1\in M_1$ be a local extremum of $f$ on $M_1$$^1$

I want to show that if $f$ is differentiable at $x_1$ and ${\rm D}g(x_1)$ is bijective, then $${\rm D}f(x_1)=\lambda_2{\rm D}g(x_1)\tag1$$ for some $\lambda_2\in E_2'$.

Note that $$B_\varepsilon(x_1)\subseteq\Omega_1\tag2$$ for some $\varepsilon>0.$

I'm only able to prove the claim under the assumption that $E_1$ is a $\mathbb R$-Hilbert space: Let $$H_1:=\operatorname{ker}{\rm D}g(x_1),$$ $U_1:=H_1\cap B_\varepsilon(x_1)$ and $V_1:=H_1^\perp B_\varepsilon(x_1)$. Now let $$\tilde f(u_1,v_1):=f(x_1+u_1+v_1)\;\;\;\text{for }$$ and $$\tilde g(u_1,v_1):=g(x_1+u_1+v_1)$$ for $(u_1,v_1)\in U_1\times V_1$ and note that $\tilde f$ is differentiable at $(0,0)$ with $${\rm D}_2\tilde f(0,0)=\left.{\rm D}f(x_1)\right|_{H_1^\perp}$$ and $\tilde g\in C^1(U_1\times V_1,E_2)$ with $${\rm D}_2\tilde g(0,0)=\left.{\rm D}g(x_1)\right|_{H_1^\perp}$$ being bijective. By the implicit function theorem, there is an open neighborhood $A_1$ of $0\in H_1$, an open neighborhood $B_1$ of $0\in H_1^\perp$ and a $h\in C^1(A_1,B_1)$ with $$\tilde g(a_1,h(a_1))=0\;\;\;\text{for all }a_1\in A_1\tag3,$$$$\forall(a_1,b_1)\in A_1\times B_1:\tilde g(a_1,b_1)=0\Rightarrow b_1=h(a_1)\tag4$$ and $${\rm D}h(a_1)=-\left({\rm D}_2\tilde g(a_1,h(a_1))\right)^{-1}{\rm D}_1\tilde g(a_1,h(a_1))\;\;\;\text{for all }a_1\in A_1\tag5.$$ Note that $$F(a_1):=\tilde f(a_1,h(a_1))=\tilde f(a_1,b_1)\;\;\;\text{for all }(a_1,b_1)\in(A_1\times B_1)\cap\{\tilde g=0\}\tag6$$ and $$h(0)=0\tag7.$$ Now $F$ is differentiable at $0$ with $${\rm D}F(0)={\rm D}_1\tilde f(0,0)-{\rm D}_2\tilde f(0,0)({\rm D}_2\tilde g(0,0))^{-1}{\rm D}_1\tilde g(0,0)\tag8.$$ $F$ has a local extremum at $0$ and hence ${\rm D}F(0)=0$ and hence $${\rm D}_1\tilde f(0,0)={\rm D}_2\tilde f(0,0)({\rm D}_2\tilde g(0,0))^{-1}{\rm D}_1\tilde g(0,0)\tag9.$$ On the other hand, $${\rm D}_2\tilde f(0,0)={\rm D}_2\tilde f(0,0)({\rm D}_2\tilde g(0,0))^{-1}{\rm D}_2\tilde f(0,0)\tag{10}.$$ Letting $$\lambda_2:=-{\rm D}_2\tilde f(0,0)\left({\rm D}_2\tilde g(0,0)\right)^{-1}$$ we obtain the claim.

Can we use a similar argumentation when $E_1$ is a general $\mathbb R$-Banach space?


i.e., in the case of a local minimum, $$f(x_1)\le f(y_1)\;\;\;\text{for all }y_1\in M_1\cap N_1$$ for some neighborhood $N_1$ of $x_1$ and analogously in the case of a local maximum.