Let $p\in(1,\infty)$ a $\ell^p$ be the space of sequences $x=\{x_n\}$ such that $\|x\|^p=\sum_{n=1}^\infty |x_n|^p<\infty$. Are spaces $\ell^2$ and $\ell^3$ linearly isomorphic (as normed spaces) ?
Asked
Active
Viewed 103 times
0
-
3isomorphic as what? – Henno Brandsma Dec 05 '19 at 08:49
-
3As normed spaces, they are not isomorphic: $\ell^2$ satisfies the parallelogram identity while $\ell^p$ does not if $p\ne 2$. – Berci Dec 05 '19 at 08:51
-
@Berci Parllelogram identity is not preserved by isomorphisms. I suppose you wanted to say that the spaces are not isometrically isomorphic. – Kavi Rama Murthy Dec 05 '19 at 08:54
-
Yes, that's what I wanted to mean. – Berci Dec 05 '19 at 09:17
-
1They are homeomorphic as topological spaces. (All separable locally convex metrisable TVS's are, but this special case is simpler.) – Henno Brandsma Dec 05 '19 at 10:16
-
@HennoBrandsma, thanks, I corrected my question. – elliptic Dec 05 '19 at 11:37
-
@Berci your comment should be an answer in my opinion – supinf Dec 05 '19 at 11:43
-
1These two spaces are not linearly homeomorphic. There is no bi-continuous linear bijection between them. Reference: Lindenstrauss & Tzafriri I ... https://www.amazon.com/Classical-Banach-Spaces-Ergebnisse-Grenzgebiete/dp/3540080724/ – GEdgar Dec 05 '19 at 12:51
-
https://math.stackexchange.com/questions/218284/proof-of-pitts-theorem – daw Dec 05 '19 at 13:10