Why does the adjoint of a Fréchet derivative has closed range?

Question

Let $H,E$ be $\mathbb R$-Hilbert spaces, $\Omega\subseteq H$ be open, $c\in C^1(\Omega,E)$ and $x\in\Omega$.

How can we show that $\mathcal N({\rm D}c(x))^\perp=\mathcal R(({\rm D}c(x))^\ast)$?

Clearly, $\mathcal N({\rm D}c(x))^\perp=\overline{\mathcal R(({\rm D}c(x))^\ast)}$ and hence we only need to show that $({\rm D}c(x))^\ast$ has closed range. This should be an application of the closed range theorem, but why and how can we apply it?

EDIT: I guess we need to assume that ${\rm D}c(x)$ has closed range. Then the claim follows fromthe closed range theorem. However, I've seen the claim in the finite-dimensional case $H=\mathbb R^d$,$E=\mathbb R^k$ without this further assumption. Is there anything which simplifies the situation in finite-dimension?

Answer 1

Take $H=E=\Omega$ be a separable Hilbert space. Let $c\in B(H)$ be a selfadjoint operator with non-closed range. Then $Dc(x)=c$, so the range of the adjoint of $Dc(x)$ is not closed.

In finite dimension the assertion is true because all operators have closed range.