A standard result in bifurcation theory due to Keller, known as ABCD Lemma, asserts that if $V$ is a Banach space, $M\,:\,V\times \mathbb{R} \to V\times \mathbb{R}$ is such that $$ M:=\begin{pmatrix} A & b \\ \langle c , \cdot \rangle & d \end{pmatrix}, $$ where $A\,:\,V \to V$, $b \in V\setminus\{0\}$, $c\in V'\setminus\{0\}$ and $d \in \mathbb{R}$, then in particular
(1) if $A$ is an ismorphism on $V$, then $M$ is an isomorphism $V\times \mathbb{R}$ if and only if $d - \langle c, A^{-1}b\rangle \neq 0$.
All references to this result point to a paper by Keller from 1977, in which he doesn't actual provide proof, merely stating it will be provided elsewhere and giving a reference to some unpublished lecture notes.
How does one prove this result in the case of an infinite-dimensional space?
The part '$\impliedby$' is easy, one can look at $\|M(u,k)\|$ when $u = -kA^{-1}b$ to see that $d - \langle c,A^{-1}b\rangle \neq 0$ is a necessary condition. Can somebody provide me with a hint as to how to prove the part '$\implies'$? It seems like it shouldn't be too hard, we should establish existence of some constant $C>0$ such that $$ \|M(u,k)\| \geq C(\|u\| + |k|) $$ and I suppose, especially when $V$ is a Hilbert space, the trick is to decompose $u = \alpha A^{-1}b + v$, where $v \in \{v' \in H\,|\,(v',A^{-1}b)_V = 0\}$, where $(\cdot,\cdot)_V$ is the inner product? Although I do not quite see how this would work. If it is not a Hilbert space, then it is even less clear to me. Any hints will be greatly appreciated!
We solve the system $M(u,k)=(f,g)$ explicitly: $$ Au + b k = f, \ cu + dk = g. $$ Now, $A$ is invertible, hence $u=A^{-1}(f-bk)$. Plugging this into the second equation yields $$ cA^{-1}(f-bk) + dk = (d-cA^{-1}b)k + cA^{-1}f $$ which should equal $g$. This is an equivalence transformation: $M(u,k)=(f,g)$ if and only if $k$ solves $$ (d-cA^{-1}b)k = g- cA^{-1}f. $$ This shows that the equivalence in the desired statement: the system is solvable if and only if $d-cA^{-1}b\ne0$.