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This is a question I was thinking of some time ago.


Suppose $\mathbf{X} \equiv (X, \|\cdot\|_X)$ is a (real or complex) Banach space, $U$ is a dense subspace of $\mathbf{X}$, and $\phi$ is a bounded linear operator $(U,\|\cdot\|_X) \to \mathbf{X}$. We know from the B.L.T. theorem that $\phi$ can be uniquely extended to a bounded linear operator $\Phi: \mathbf{X} \to \mathbf{X}$.

Question. Provided $x \in X$, does there exist a sequence, $\{x_n\}_{n=1}^\infty$, in $U$ such that $\lim_n x_n = x$ in $\mathbf{X}$ and, for each $n \in \mathbb{N}^+$, $\{\Phi^k(x_n)\}_{k=1}^n \subseteq \phi(U)$?

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1 Answer 1

up vote 4 down vote accepted

Let $\Phi:L^2(\mathbb R)\to L^2(\mathbb R)$ be the (continuous extension of the) Fourier transform. Let $U$ be the dense subspace of compactly supported functions; we can just take $\phi=\Phi\vert_U$.

Note that $\Phi$ is injective and $\Phi^2(U)=U$, while $\phi(U)\cap U=\{0\}$, so the existence of such sequences is impossible unless $x=0$. For $x_n\in U\setminus\{0\}$, $\Phi^2(x_n)\in U\setminus \{0\}$, so $\Phi^2(x_n)\not\in\phi(U)$.

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So nice! Just you may want to edit your answer to fix that typo with tranform. :) –  Salvo Tringali Dec 7 '11 at 10:38
    
Thanks, fixed :) –  Jonas Meyer Dec 7 '11 at 15:34

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