Lagrange multiplier in Banach space

Question

I am trying to understand a proof of the Lagrange multiplier theorem for Banach spaces and there is some point I do not understand. Let me recall the setting.

Let $I, J: E \to \mathbb R$ two functionnal of class $C^1$ on a Banach space $E$. If there exists $u_0 \in A$ such that $I(u_0) = \text{min}_{u \in A} I(u)$ with $$A = \{u \in E~|~J(u) = 0\},$$ and $DJ(u_0) \neq 0$ (the Fréchet derivative of $J$) as a map from $E^*$ to $\mathbb R$, then there exists $\lambda \in \mathbb R$ such that $$DI(u_0) = \lambda DJ(u_0).$$

To prove the statement, the author shows that, for $\forall h \in \text{ker } DJ(u_0)$, we have $$DI(u_0) (h) = 0$$ so that $\text{ker } DJ(u_0) \subset\text{ker } DI(u_0).$ And from that he directly deduces that the existence of a $\lambda \in \mathbb R$ with $$DI(u_0) = \lambda DJ(u_0).$$ How does he do that? Is there an obscure theorem of functional analysis that allows us to deduce such result?

Answer 1

These two kernels are hyperplanes. Since one is contained in the other one, they are equal. And two linear forms having the same kernel are proportional.

Answer 2

I will write $a$ instead of $u_0$. First of all, let us clarify the world where $DI(a)$, $DJ(a)$ live in. These are "linearized objects", so that the expressions $$ \begin{aligned} &I(a+h)-I(a)-DI(a)h\ ,\\ &J(a+h)-J(a)-DJ(a)h\ , \end{aligned} $$ make sense (and in norm are "controlled" by the norm in $E$ of $h$), so $DI(a),DJ(a)$ map elements $h$ from $E$ (linearly) to elements in the codomain of $I,J$, which is $\Bbb R$, so $$ DI(a)\ ,\ DJ(a)\in E^*\ . $$ Since $DJ(a)$ is not the zero map in $E^*$, there is some vector $v\in E$ with $DJ(a)v$ not zero, and after norming we may and do suppose $DJ(a)v=1$. Now take any other $x\in E$, compute the number $DJ(a)x$, and subtract from $x$ a scalar multiple of $v$ to get an element in the kernel of $DJ(a)$. So the functional $DJ(a)$ has a simple construction, considering how it acts on $E$ splitted as $$ E = \operatorname{Ker} DJ(a)\oplus \Bbb R v\ . $$ Now the same applies for $DI(a)$, which has either the same kernel, or $v$ is also in the kernel.

Take now $\lambda$ to fit the two functionals in the $\Bbb Rv$ direction.