Let $X, Y$ be Hilbert spaces and $F:X \rightarrow Y$ smooth. Assume that $M := F^{-1}(0) \subset X$ is a smooth submanifold. Is it true that for any $x\in M$, the tangent space $T\_xM$ is a Hilbert subspace of $\mathrm{ker} D\_xF$?


Of course, if $0$ is a regular value of $F$, then by the implicit function theorem, $M\subset X$ is a smooth submanifold and $T\_xM = \mathrm{ker} D\_xF$.


What if $0$ is not a regular value?



Tangent space of smooth Hilbert submanifolds
True: via a local chart we can assume $F^{-1}(0)$ is a closed linear subspace $N$ of $X$, and since $F\_{|N}=0$, we also have $N\subset \text{ker} DF(x) $.


(A formal explanation of the latter: if we denote $i\_N:N\to X$ the (bounded, linear) inclusion map, $F\_{|N}=F\circ i\_N:N\to Y$ is the null map and by the chain rule $0=D(F\_{|N})(x)= D(F\circ i\_N)(x) = DF(i\_N(x))\circ i\_N=DF(x)\_{|N}$ that is $N\subset \text{ker} DF(x) $ ).


