Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n \in \mathbb{N}^+$, $\{\Phi^k(x_n)\}_{k=1}^n \subseteq \phi(U)$?

share|cite|improve this question
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 $\Phi^2(x_n)\not\in\phi(U)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.