Let $\left( S_{n}^{1}\right) $ and $\left( S_{n}^{2}\right) $ two sequences of operators in $\mathcal{L}(E_{1},F_{1})$ and $\mathcal{L}(E_{2},F_{2})$ where $E_{i},F_{i},i=1,2$ are Hilbert spaces such that $\left\Vert S_{n}^{1}x-S^{1}x\right\Vert _{F_{1}}\underset{n\rightarrow \infty }{% \rightarrow }0,$ $\forall x\in E_{1}$ for some $S^{1}\in \mathcal{L}% (E_{1},F_{1})$ and such that $$ \ker S_{n}^{1}=\{0\}\Leftrightarrow \ker S_{n}^{2}=\{0\},\text{ }\forall n\geq 0. $$ Assume that $\ker S^{1}=\{0\}.$ Does this implies that $\ker S^{2}=\{0\}?.$ Thank you in advance.
1 Answer
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No, even if $\|S^2_n x-S^2x\|_{F_2}\to0$ for some $S^2\in\mathcal L(E_2,F_2)$ and all $x\in E_2$.
Indeed, let e.g. $E_1=F_1=E_2=F_2=H:=\ell^2$, $S^1_n=I$, $S^1=I$, $S^2=0$, and $S^2_n x=(e^{-|k-2n|}x_k)_{k=1}^\infty$ for $x=(x_k)_{k=1}^\infty\in\ell^2$. Then all the conditions hold. In particular, $$\|S^2_n x\|_H^2\le e^{-n}\sum_{k=1}^n|x_k|^2+\sum_{k=n+1}^\infty|x_k|^2\to0$$ as $n\to\infty$.
However, the conclusion $\text{ker}S^2=\{0\}$ does not hold.
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1$\begingroup$ @Gustave : Are you satisfied with the answer? $\endgroup$ Commented Apr 22, 2022 at 4:16
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