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23h
comment Is every Hilbert space a Banach algebra?
@user51514, yes that's true. This is essentially Parseval's identity.
1d
answered Is every Hilbert space a Banach algebra?
1d
comment Is every Hilbert space a Banach algebra?
@user51514 separability doesn't matter here.
1d
comment Is every Hilbert space a Banach algebra?
An infinite-dimensional C*-algebra is never reflexive so it cannot be Banach-space isomorphic to a Hilbert space! The quickest way to see this is to take a self-adjoint element $t$ with infinite spectrum and apply the spectral theorem to it getting a subalgebra isometric to $C_0(S)$.
Jul
24
answered What is the $w^{*}$-closure of the finite rank operators in $B(H)$?
Jul
24
revised Algebraically flavoured functional analysis book
edited body
Jul
24
answered Algebraically flavoured functional analysis book
Jul
24
revised One-sided version of the Nakayama lemma?
added 26 characters in body
Jul
24
revised One-sided version of the Nakayama lemma?
deleted 54 characters in body
Jul
24
revised One-sided version of the Nakayama lemma?
added 153 characters in body
Jul
23
revised Continuity of multiplication of operators in the strong operator topology - find an error
added 8 characters in body; edited title
Jul
17
revised Consequences of the Carathéodory conjecture
edited title
Jul
11
revised The Mazur–Ulam Theorem: A generalization to arbitrary topological vector spaces
edited title
Jul
11
answered The Mazur–Ulam Theorem: A generalization to arbitrary topological vector spaces
Jul
11
revised Isomorphic image of a simple algebra is simple
added 40 characters in body
Jul
10
reviewed Approve Cancellation problem: $R\not\cong S$ but $R[t]\cong S[t]$ (Danielewski surfaces)
Jul
9
revised Are there spaces “smaller” than $c_0$ whose dual is $\ell^1$?
added 328 characters in body
Jul
9
comment Banach spaces containing copies of $\ell^1$
I believe it is a legitimate question.
Jul
9
answered Banach spaces containing copies of $\ell^1$
Jul
9
comment Is there a continuous $f(x,y)$ which is not of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$
That's a great answer.