Let $X$ be a separable and reflexive Banach space and consider the $C^1$ functional $I:X\to\mathbb{R}$. We define the Nehari manifold corresponding to $I$ by $$ N=\{u\in X:I'(u)u=0\}. $$ Then the set $N$ is very useful in the existence theory of PDE's and $N$ is referred to as the Nehari manifold.
I am confused whether $N$ is actually a topological manifold? I guess this is not a manifold. Can someone please confirm with a proper reason.
Thanks.
Consider the map $$ F: X \mapsto \mathbb{R}, u \mapsto \langle I'(u),u \rangle_{X^* \times X} $$ All you would have to do is show that $0$ is a regular value of $F$, i.e. that it has a surjective differential with splitting kernel if I remember correctly. In order to show that, you need some additional assumptions.