Family of complemented subspaces

Question

Let $X$, $Y$, $A$, $B$ be topological vector spaces. Given two jointly continuous families of linear injective maps $P: Y \times A \rightarrow X$ and $R: Y \times B \rightarrow X$, such that for $y=0$ we have the topological complementation $X = Im\, P_0 \oplus Im\, R_0$. Here and in the following $P_y$ denotes the contracted map $A \rightarrow X, a \rightarrow P(y,a)$.

I would like to have criteria on $P$ and $R$ for the following claim: There exists a open subset $W$ in $Y$ around $0$ such that $X = Im\, P_y \oplus Im\, R_y$ for all $y \in W$.

I might have a (fuzzy) proof for the case when $A$ (or $B$) is finite dimensional, but are there other cases?

Partial answers for $X$ being a Banach space are also ok, but I prefer at least Fréchet spaces.

Answer 1

I will assume $A,B, X$ are Banach spaces, so that I can use Banach isomorphism theorem (BIT) and the fact that $GL(X)$ (the bounded invertible operators on $X$) is open in $B(X)$ (the bounded linear operators on $X$). I only assume that $Y$ is a topological space, and I fix the initial condition at some $y_0$ in $Y$.

Note that $P_0:A\longrightarrow \mbox{Im} P_0$ is injective continuous onto a closed subspace of $X$, whence a Banach space. Thus $P_0$ is linear homeomorphism onto its range by BIT. Likewise, $R_0$ is a linear homeomorphism from $B$ onto its range. So up to composing with the isomorphism from $X$ onto $A\times B$ given by $(P_0^{-1},R_0^{-1})$, we can assume that $X=A\times B$ and $P_0$ (resp. $R_0$) is the canonical projection onto $A$ (resp. $B$).

Now consider $\overline{P_y}:A\times B\longrightarrow A\times B$ defined by $\overline{P_y}(a,b):=P_y(a)$. Likewise set $\overline{R_y}(a,b):=R_y(b)$. By your assumptions, the map $$ y\longmapsto S_y:=\overline{P_y}+\overline{R_y}:(a,b)\longmapsto P_y(a)+R_y(b) $$ is continuous from $Y$ to $B(X)=B(A\times B)$ the bounded linear operators on $X=A\times B$. The initial condition is $S_{y_0}=\mbox{Id}_X$. This is in particular invertible. Since $GL(X)$ is open when $X$ is a Banach space, it follows that there exists an open neighborhood $W$ of $y_0$ in $Y$ such that $S_y$ be invertible for every $y\in W$.

If $S_y$ is invertible, in particular it is surjective whence $X=A\times B=\mbox{Im} S_y=\mbox{Im}P_y+\mbox{Im} R_y$ for every $y\in W$. And also $S_y$ is injective. So if $x\in \mbox{Im}P_y\cap\mbox{Im} R_y$, we get $x=P_y(a)=R_y(b)$ whence $S_y(a,-b)=P_y(a)-R_y(b)=x-x=0$. Thus $a=b=0$ and $x=0$. Therefore $\mbox{Im}P_y\cap\mbox{Im} R_y=\emptyset$ for every $y\in W$. Since $S_y^{-1}$ is bounded, note that both $\mbox{Im}P_y$ and $\mbox{Im} R_y$ are closed for every $y\in W$.

Conclusion: there does exist an open neighborhood $W$ of $y_0$ in $Y$ such that we have the topological complementation $X=\mbox{Im}P_y\oplus \mbox{Im} R_y$ for every $y\in W$. Removing the Banach space assumptions on $A,B,X$ seems impossible, since we really need $GL(X)$ to be open, which is not true in a general Fréchet space. Unless we make the ad hoc assumption on $P_y$ and $R_y$ from the beginning that $S_y$ be invertible near $y_0$. Then the argument works the same with $A,B,X$ Fréchet spaces, as BIT holds.