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Let X,Y be normed spaces and $T: X \to Y$ a linear operator. If $(T_n)$ is a sequence of linear operators also from $X \to Y$ then prove $$A := \{x \in X : T_nx \not\to Tx \}$$ is dense or empty.

So essentially, I think what I'm supposed to do is to pick an x where there isn't convergence, and so $||T_nx - Tx||> \epsilon$ for the usual ways. But I'm not sure where to go from here. If I take a Cauchy sequence around x, this doesn't exactly work because if the $T_n,T$'s are not continuous then the image may not be Cauchy.

Thanks for any help.

Michael
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1 Answers1

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Hint: Prove that the complement of any proper subspace of $X$ is dense.

Mike F
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  • Thank you, could I ask for another hint on where to begin with that? Do you mean set-theoretic complement or orthogonal complement? The reason I'm asking now is that for Hilbert Spaces at least we have a characterization of if the orthogonal complement is null the we have density, but we're not necessarily in a Hilbert space. – Michael May 14 '21 at 03:50
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    @Michael I think he means set-theoretic compliment. To prove this claim, it suffices to show that for any element of the proper subspace, you can find an arbitrarily close element that lies outside the subspace. – angryavian May 14 '21 at 04:03
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    Yes I meant the set-theoretic complement, thanks @angryavian. – Mike F May 14 '21 at 04:42
  • Thank you both. I was able to solve it pretty quickly with this help. – Michael May 15 '21 at 03:05