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Let $V$ a closed infinite dimensional subspace of $\ell^2$

Is it possibile that $V$ is closed in $\ell^1$ too ?

Thanks for any suggestion

Matey Math
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    Generally speaking, $\ell^2$ isn't even a subspace of $\ell^1$, so there's no way this is true in general. – Michael L. Sep 19 '17 at 13:45
  • Indeed the question does not seem to make sence..instead of subspace,don't mean subset? – Marios Gretsas Sep 19 '17 at 13:48
  • A subspace inherits the topology of the space itself..but the subspce $(V,d_{l_2})$ is not a subspace of $(l_1,d_1)$ unless the two metrics define the same topology...which i do not believe it is the case – Marios Gretsas Sep 19 '17 at 13:54
  • My question is if exist a closed infinite dimensional subspace of $\ell^2$ such that it is a closed subspace (not subset) of $\ell^1$ – Matey Math Sep 19 '17 at 13:58
  • I do not belive that exists... – Marios Gretsas Sep 19 '17 at 13:58
  • i'm not sure that not exists, for finite dimensional it's trivial that exists – Matey Math Sep 19 '17 at 14:06
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    You should have focused the question on whether an infinite dimensional subspace of $\ell^1$ can be closed in the $\ell^2$ norm. –  Sep 19 '17 at 15:44

1 Answers1

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It seems that closed in $\ell^2$ implies closed in $\ell^1$ for any subspace of $\ell^1$.

Assume that $M \le \ell^1 \subset \ell^2$ is a closed subspace of $\ell_2$.

We shall prove that $M$ is closed in $\ell^1$ as well.

Take a sequence $(x_n)_{n=1}^\infty$ in $M$ such that it is convergent in $\ell^1$ and set $x_n \xrightarrow{\|\cdot\|_1} x \in \ell^1$.

Since we have $\|\cdot\|_2 \le \|\cdot\|_1$, we get that $x_n \xrightarrow{\|\cdot\|_2} x$ also holds:

$$\|x_n - x\|_2 \le \|x_n - x\|_1 \xrightarrow{n\to\infty} 0$$

Since $M$ is closed in $\ell^2$, we have $x \in M$.

Thus, $M$ is closed in $\ell^1$.

In fact, the situation seems to be the same for any $p < q \in [1, +\infty\rangle$ since:

$$1 = \sum_{n=1}^\infty\frac{|x_n|^q}{\|x\|_q^q} = \sum_{n=1}^\infty\left(\underbrace{\frac{|x_n|}{\|x\|_q}}_{\le 1}\right)^q \le \sum_{n=1}^\infty\left(\frac{|x_n|}{\|x\|_q}\right)^p = \frac{\|x\|_p^p}{\|x\|_q^p} \implies \|x\|_q \le \|x\|_p$$

mechanodroid
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    Okay, but this leaves open a question of whether such a subspace exists to begin with (an infinite-dimensional subspace of $\ell^1$ that is closed in the $\ell^2$ norm). –  Sep 19 '17 at 15:43
  • @Michelle Yes, I haven't been able to come up with such a subspace. – mechanodroid Sep 19 '17 at 15:45
  • @mechanodroid thanks for your answer – Matey Math Sep 19 '17 at 15:55
  • @Michelle thanks for your useful comments – Matey Math Sep 19 '17 at 16:14
  • @MateyMath You could consider posting a new question about the existence of such a subspace. It seems interesting. – mechanodroid Sep 19 '17 at 16:19
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    I don't think there could be one. If there were, the formal identity would be an isomorphism from a closed subspace of $\ell_1$ to a closed subspace of $\ell_2$. But every infinite dimensional closed subspace of $\ell_1$ contains a copy of $\ell_1$ while no subspace of $\ell_2$ does. – David Mitra Sep 19 '17 at 19:42
  • @DavidMitra I'm not familiar with that property, how would you prove it? – mechanodroid Sep 19 '17 at 20:24
  • You can construct a sequence in the subspace that is a small perturbation of a block basis of the standard unit vector basis of $\ell_1$ (it will b equivalent to the standard basis of $\ell_1$). The details are in Diestel's Sequences and Series in Banach Spaces (and other texts, I imagine). There may be a simpler way of seeing why we can't have such a subspace, but nothing comes to my mind at the moment. – David Mitra Sep 19 '17 at 20:31