Are these normed vector spaces homeomorphic? 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.
 A: 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.
