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It is easy to show that $\ell^p \subset \ell^q$ when $1 \leq p<q \leq + \infty$, but is the injection continuous?

If so is $\ell^{\infty}$ the direct limit $\lim\limits_{\rightarrow} \ \ell^p$ as topological space?

NB: If some result depends on the set we are working on, I am working on $\mathbb{R}$ or $\mathbb{C}$.

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If $x=(x_n) \in \ell^p$ with $\|x\|_p = 1$, then $|x_n| \le 1$ for all $n$, so $|x_n|^q \le |x_n|^p$ for all $n$, and so $\|x\|_q = (\sum |x_n|^q)^{1/q} \le (\sum |x_n|^p)^{1/q} \le 1^{1/q}=1 = \| x\|_p$. Using $x=(1,0,0,\ldots)$, it is easy to see that the operator norm of the embedding of $\ell_p$ into $\ell_q$ is actually equal to $1$.

As to the second question, the sequence $(1,1,1,\ldots) \in \ell^\infty$ is not in the direct limit of the $\ell^p$-spaces, which suggests that the answer is no. In order to show that they are not homeomorphic, you can use the fact that $\ell^\infty$ is not separable, but each $\ell^p$ for $1\le p < \infty$ is, so the direct limit is separable, too. (Just take the image of countable dense sets in $\ell^1, \ell^2, \ell^3, \ldots$ in the direct limit, giving you a countable dense set in the direct limit.)

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  • $\begingroup$ Thanks, I just figured that out myself a second before you gave the answer. :) $\endgroup$ – Lukas Geyer Oct 28 '12 at 15:04
  • $\begingroup$ I see :) +1. The same argument also yields that $\ell^{\infty}$ is not the inductive limit as Banach space ($\ell^1$-sum modulo usual relations). $\endgroup$ – commenter Oct 28 '12 at 15:07
  • $\begingroup$ The argument for $\|x\|_q\leq\|x\|_p$ is not sufficient since the norms involve a power of $1/q$ and $1/p$, respectively. A proof can be found here How do you show that $\ell^p\subset \ell^q$ for $p\le q$? $\endgroup$ – AD. Feb 21 '14 at 6:43
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It can be seen as a consequence of closed graph theorem as:

  • $\ell^p$ is a Banach space for each $1\leq p\leq \infty$;
  • if $x^{(n)}\to x$ in $\ell^p$ then $x^{(n)}_k\to x_k$ for each integer $k$.
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