Fredholm and Banach continuation interfaces

Assume the Axiom of Choice. Let $X,Y$ be real Banach spaces: real normed vector spaces complete for their norm metrics. A bounded linear map $L:X\to Y$ is Fredholm if its kernel is finite dimensional, its range is closed, and its cokernel $Y\mathrm{/}\mathrm{ran}L$ is finite dimensional. Its index is $\mathrm{ind}L=\dim \ker L-\dim (Y\mathrm{/}\mathrm{ran}L)$.

Equip $\mathbb{R}\times X$ with the maximum product norm $\mathrm{\parallel }(q,w){\mathrm{\parallel }}_{\max }=\max \{\mathrm{\mid }q\mathrm{\mid },\mathrm{\parallel }w{\mathrm{\parallel }}_X\}$, for which it is a real Banach space. A local continuation problem consists of an open set $U\subseteq \mathbb{R}\times X$ for this norm and a $C^1$ map $\mathcal{H}:U\to Y$. Here $C^1$ means Fréchet differentiable with derivative continuous in operator norm. A zero is $(q,w)\in U$ with $\mathcal{H}(q,w)=0$. At such a point the partial derivative $D_w\mathcal{H}(q,w):X\to Y$ is $D\mathcal{H}(q,w)$ restricted to vectors of the form $(0,v)$. A bounded isomorphism is a bounded linear bijection with bounded linear inverse.

If $D_w\mathcal{H}(q_0,w_0)$ is a bounded isomorphism at a zero, the implicit-function theorem provides open neighborhoods $A$ of $q_0$ and $B$ of $w_0$ and a unique $C^1$ map $w:A\to B$ whose graph is exactly the zero set of $\mathcal{H}$ in $A\times B$. This criterion requires invertibility; Fredholm index zero alone does not supply it.

For an endpoint $q_{\ast }$, a branch $w(q)$ of zeros can be continued to a zero at $q_{\ast }$ by a limit argument when a sequence $q_n\to q_{\ast }$ has $w(q_n)\to w_{\ast }$ in $X$, $(q_{\ast },w_{\ast })\in U$, and any additional admissibility or graph conditions required by the particular problem hold at the limit. A set of branch values is relatively compact when its closure in $X$ is compact for the norm metric. Relative compactness of the branch values can supply a convergent subsequence; it does not by itself ensure that the limit remains in $U$ or satisfies those additional conditions.

If $L$ is Fredholm and $K:X\to Y$ is a compact linear operator, meaning that it maps bounded sets to relatively compact sets, then $L+K$ is Fredholm with the same index. This statement alone does not identify the kernel of $L+K$.

The Fredholm terminology and index are those of Fredholm operator cokernel and index, and the differentiability convention is that of C k map between Banach spaces. By The reals are complete, every Cauchy sequence of real numbers for the absolute-value metric converges to a real number. Consequently $\mathbb{R}$ with its absolute-value norm is Banach by Banach space. The chosen norm is the maximum product norm of The standard product norms on a finite product of normed spaces. Since $\mathbb{R}$ with its absolute-value norm and $X$ are Banach, Finite products of Banach spaces are Banach makes $\mathbb{R}\times X$ Banach. The product parameter space and the partial derivative fit the hypotheses of Implicit function theorem for Banach spaces, with parameter space $\mathbb{R}$, unknown space $X$, and target $Y$; the stated graph conclusion follows under its bounded-isomorphism hypothesis and Choice. Index zero is insufficient for invertibility: the zero map $\mathbb{R}\to \mathbb{R}$ has one-dimensional kernel and cokernel, hence index zero, but is not injective. For the endpoint assertion, the differentiability-implies-continuity argument in the Remarks of Fréchet derivative between Banach spaces applies to the $C^1$ map $\mathcal{H}$ on the open subset $U$ of the real Banach space $\mathbb{R}\times X$, so $\mathcal{H}$ is continuous on $U$. Give $U$ the restricted maximum-norm metric and $Y$ its norm metric. These are metric spaces, so the choice-free implication (a) to (d) of Metric continuity characterisations, with countable choice for the sequential converse makes $\mathcal{H}:U\to Y$ sequentially continuous. The assumed component limits give $\mathrm{\parallel }(q_n,w(q_n))-(q_{\ast },w_{\ast }){\mathrm{\parallel }}_{\max }=\max \{\mathrm{\mid }{q}_n-q_{\ast }\mathrm{\mid },\mathrm{\parallel }w(q_n)-w_{\ast }{\mathrm{\parallel }}_X\}\to 0$: for each positive tolerance, both terms are smaller than it once $n$ exceeds their two thresholds. Since the sequence consists of zeros and $(q_{\ast },w_{\ast })\in U$, sequential continuity yields $\mathcal{H}(q_{\ast },w_{\ast })={\lim }_n\mathcal{H}(q_n,w(q_n))=0$; preservation of any extra conditions must be checked separately. If the branch values have compact closure $C\subseteq X$, In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, assertions 1 and 3, applies to $C$ with its restricted norm metric: compactness implies countable compactness and then sequential compactness. Thus the sequence of branch values has a subsequence converging to a point of $C\subseteq X$. This extraction alone says nothing about membership of the associated limit pair in $U$ or preservation of additional admissibility conditions. The compact-perturbation assertion is exactly Fredholm index is stable under compact perturbations under its stated hypotheses.