The constraint subset of $H_0^1(\Omega)$ is a $C^1$-submanifold.

Question

This problem comes from the constraint problem in CoV. (the lagrange-multiplier case)

Let $\Omega\subset \mathbb R^N$ be open bounded, smooth boundary. We define the sub-manifold $$ M:=\{u\in H_0^1(\Omega),\,\,\int_\Omega g(x,u,\nabla u)dx\equiv 0\} $$ where $g(x,u,\xi)$, from $\Omega\times\mathbb R\times\mathbb R^N\to \mathbb R$, is $C^2$.

We denote $$ g_u(x,u,\xi):=\frac{d}{du}g(x,u,\xi) $$ and $$ g_\xi(x,u,\xi):=\nabla_\xi g(x,u,\xi) $$

I want to conclude that the set $M$ is of $C^1$-submanifold so that I could apply Lagrange multiplier rule on it. The book, by Struwe, page 15 has a quick but not clear prove for a very specific example but I want to prove a general version.

Here is what I tried and where I got stuck.

First of all, if both $g_u(x,u,\nabla u)$ and $g_\xi(x,u,\nabla u)$ are a.e. $0$ on $\Omega$, then we have set $M$ is entire $H_0^1(\Omega)$ and it is not interesting.

Now, assume one of $g_u(x,u,\nabla u)$ or $g_\xi(x,u,\nabla u)$ is not a.e. $0$ on $\Omega$, then I want to conclude that for every $u\in M$, I have $$\int_\Omega g_u(x,u,\nabla u)\cdot u\,dx+\int_\Omega g_\xi(x,u,\nabla u)\cdot\nabla u \,dx\neq 0 \tag 1$$

I got stuck on proving $(1)$... I tried the contradiction but it does not work... please help me about this.

Lastly, book states that if $(1)$ hold, then the set $M$ is a $C^1$-submanifold of $H_0^1(\Omega)$ by implicit function theorem. I know what is implicit function theorem but I don't quit see how we applied it here... Please provided me some details. Thank you!

Answer 1

Let $H$ be a Hilbert space and define $M$ by $$M=\{u\in H:\ F(u)=0\},$$

where $F:H\to\mathbb{R}$ is a $C^1(H)$ function.

Theorem: Suppose that for all $u\in M$, $F'(u)\neq 0$. Then, $M$ is a $C^1$ Hilbert Manifold of $H$.

To prove it, fix $u\in H$. Remember that $F':H\to H^\star$, so $F'(u)\neq 0$ means that the linear function $F'(u)$ has non trivial kernel, which we will call $K$. Let $e\in H$ be such that $$\{e\}\oplus K =H.$$

Define $G:\{e\}\oplus K\to\mathbb{R}$ by $G(t,k)= F(te+k)$. Write $u=t'e+k'$ and note that $$\frac{\partial G}{\partial t}(t',k')=F'(u)e\neq 0,$$

and $$G(t',k')=F(u),$$

hence, from the Implicit Function Theorem, there is a $C^1$ function $\varphi: U_{k'}\to (-\delta+t',\delta+t')$, where $(-\delta+t',\delta+t')$ is an open neighborhood of $t'$ and $U_{k'}$ is an open neighborhood of $k'$, such that $$G(\varphi(k),k)=F(\varphi(k)e+k)=0,\ \forall\ k\in U_{k'}.$$

Therefore, there is an open neighborhood $V_{F(u)}$ such that $$V_{F(u)}=\{\varphi(k)e+k:\ k\in U_{k'}\}.$$

Can you conclude from here?

Remark: If $X$ is a Banach space then, a similar procedure shows that $M$ is a Banach Manifold of class $C^1$.