Let $F$ be a Fréchet space and $A:F\rightarrow F$ a continuous linear operator such that $A\left( F\right) $ is closed and there is some closed subspace $E\subset F$ such that $F=A\left( F\right) \oplus E$. Is there some neighborhood $V$ of $0$ such that $F=\left( A-\lambda \right) \left( F\right) +E$ for all $\lambda \in V$ ? I know that $\ker \left( A\right) $ may not have a closed complementary subspace. I'm not yet familiar with Fréchet spaces so any help please ?
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I think this is not true: Take the space $F=\omega(\mathbb{N})$ of all real (or complex) sequences endowed with the Fréchet-metric $$ d(x,y)= \sum_{n=1}^\infty \frac{1}{2^n}\frac{|x_n-y_n|}{1+|x_n-y_n|}, $$ and the linear operator $A:F \to F$, $$ Ax=(\frac{1}{n}x_n) \quad (x=(x_n) \in F). $$ Then $A$ is continuous (since convergence in $\omega(\mathbb{N})$ means coordinatewise convergence) and invertible ($A^{-1}x=(nx_n)$). Hence $A(F)=F$ and $F=A(F)\oplus \{0\}$. Thus $E=\{0\}$. But for each $k \in \mathbb{N}$ we have $$ (A-\frac{1}{k}{\rm id}_F)(F)+E=(A-\frac{1}{k}{\rm id}_F)(F) = \{(y_n)\in F: y_k=0\} \not=F. $$
Gerd
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Thank you @Gerd – Djalal Ounadjela Dec 25 '24 at 16:58
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Please see https://mathoverflow.net/questions/484575/operators-with-closed-complemented-ranges-and-surjective-operators?_gl=1vea7z3_gaMTc3Mzk3MDIxNS4xNjU2Mjg3MDMx_ga_S812YQPLT2*MTczNTE0NTcxNy41NzIuMS4xNzM1MTQ1OTc2LjAuMC4w @Gerd – Djalal Ounadjela Dec 25 '24 at 17:00