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Let $X = \{(x_n) \mid \sum_{n = 1}^\infty |x_n| < \infty\}$, and let $\|\cdot\|_1$ and $\|\cdot\|_\infty$ be the $L^1$ and $L^\infty$ norms on $X$, respectively. Are the normed vector spaces $(X, \|\cdot\|_1)$ and $(X, \|\cdot\|_\infty)$ homeomorphic? If so, is there a linear homeomorphism between these spaces?

One thing I do know is that $\|\cdot\|_1$ and $\|\cdot\|_\infty$ do not induce the same topology on $X$. This is easily seen from this criterion—in our case, we do not have the bound $\|x\|_1 \leq C\|x\|_\infty$. For instance, the vector $x = (1, 1 / 2, 1 / 3, \dots, 1 / n, 0, 0, \dots)$ always satisfies $\|x\|_\infty = 1$, but $\|x\|_1$ can be made arbitrarily large.

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    $\begingroup$ I think a linear homeomorphism must induce an equivalence of norms, which would be a contradiction. $\endgroup$
    – FShrike
    Commented Aug 26, 2022 at 22:25
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    $\begingroup$ They cannot be homeomorphic. $X$ with $L^1$ is separable and $X$ with $L^\infty$ is not. $\endgroup$
    – Severin Schraven
    Commented Aug 26, 2022 at 22:27
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    $\begingroup$ @SeverinSchraven Is $X$ with $L^\infty$ inseparable? I believe the set of finite rational sequences is dense: for any $(x_n) \in X$, there is some $N \geq 1$ such that $\sup_{n \geq N} |x_n| < \varepsilon$, and then the head can be approximated by some finite rational sequence. $\endgroup$
    – Frank
    Commented Aug 26, 2022 at 22:38
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    $\begingroup$ @Frank Indeed, you are right, I had the full $L^\infty$ in mind. I'll keep the comment such that otherwise will not fall for the same stupidity. $\endgroup$
    – Severin Schraven
    Commented Aug 26, 2022 at 22:45
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    $\begingroup$ There cannot exist a linear homeomorphism. That would mean that we have a bijective bi-lipschitz map between those two spaces. However, such a map conserves completeness (as Cauchy sequences get mapped to Cauchy sequences) and $(X, \Vert \cdot\Vert_1)$ is complete and $(X, \Vert \cdot \Vert_\infty)$ is not (consider truncations of $(1/n)_n$). $\endgroup$
    – Severin Schraven
    Commented Aug 27, 2022 at 0:51

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There cannot exist a homeomorphism between the two spaces. This is explained here https://mathoverflow.net/questions/256623/is-a-normed-space-which-is-homeomorphic-to-a-banach-space-complete. Namely, in that mathoverlow post it is established that if a normed space is homeomorphic to a Banach space, then the normed space itself must be a Banach space. However, in our case $(X, \Vert \cdot \Vert_\infty)$ is not complete (consider truncations of $(1/n)_{n\in \mathbb{N}_{\geq 1}}$), while $(X, \Vert \cdot \Vert_1)$ is complete.

As mentioned in the comments, it is easier to show that there exists no linear homeomorphism $\varphi: (X, \Vert \cdot \Vert_\infty) \rightarrow (X, \Vert \cdot \Vert_1)$. Again completeness is the property that comes to our help. A linear homeomorphism between normed spaces is a bi-lipschitz map and thus preserves Cauchy sequences (lipschitz maps send Cauchy sequences to Cauchy sequences). Hence, if we have a Cauchy sequence $(x_n)_n \subseteq (X, \Vert \cdot \Vert_\infty)$, then so is $(\varphi(x_n))_{n}$ in $(X, \Vert \cdot \Vert_1)$. However, $(X, \Vert \cdot \Vert_1)$ is complete and therefore $(\varphi(x_n))_n$ converges in $(X, \Vert \cdot \Vert_1)$. But $\varphi$ is homeomorphism and therefore $\varphi^{-1}$ is continuous. As continous functions send converging sequences to converging sequences, we get that $(x_n)_n = (\varphi^{-1}(\varphi(x_n)))_n$ converges in $(X, \Vert \cdot \Vert_\infty)$. But as mentioned in the previous paragraph, $(X, \Vert \cdot \Vert_\infty)$ is not complete and we obtain the desired contradiction.

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