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2d
comment Showing that $f$ continuous
@sammath I added an open set proof.
2d
revised Showing that $f$ continuous
added 502 characters in body
2d
comment Different norm on $\ell_p$-space and Hilbert space
I don't know the answer to your second comment but I'm interested in seeing answers in this thread here.
2d
comment Different norm on $\ell_p$-space and Hilbert space
I'm not sure whether it will be a Hilbert space but I think it will. E.g. if you take the set of all $1.5$-summable sequences $c_n$ I see no reason why $\sum_n c_n \overline{c_n}$ would not define an inner product.
Oct
29
comment An abelian Banach algebra without characters
I assume by $1 \in A^\ast$ you mean multiplication by $1$ (which is zero by how multiplication is defined). I also understand that you use $A^\ast$ to denote $\Omega(A)$ since they coincide because both are multiplication by complex numbers.
Oct
29
comment An abelian Banach algebra without characters
@TomekKania I am very sorry for being so dense but could you please elaborate further why my second displayed equation does not rule out non-zero characters? To me it seems to be a solid proof that any character must be zero.
Oct
29
comment An abelian Banach algebra without characters
@TomekKania I'm sorry could you try to rephrase your comment? I don't yet understand what is wrong with my answer.
Oct
29
comment Different norm on $\ell_p$-space and Hilbert space
Using this answer here you know that if $p \le q$ you can always endow $\ell^p$ with the $q$ norm.
Oct
29
comment Different norm on $\ell_p$-space and Hilbert space
That's an interesting question.
Oct
29
answered Find a basis of $B$
Oct
29
comment An abelian Banach algebra without characters
I wonder if there are any "easy" examples, like $L^p$ spaces or continuous functions. Of course these are all non-examples but something "common" would be nice. Alas, I don't know any "common" algebra that has empty character space : (
Oct
29
answered An abelian Banach algebra without characters
Oct
28
awarded  Popular Question
Oct
27
revised Showing that $f$ continuous
added 667 characters in body
Oct
27
comment Showing that $f$ continuous
@sammath Regarding your edit: I will edit my answer accordingly.
Oct
27
comment Showing that $f$ continuous
@sammath I thought about it but I don't see how to do it. I'm sorry to not be of more help.
Oct
27
comment Function that is uniformly continuous but not bounded?
Wonderful. The shortest answers are the best answers.
Oct
27
answered Showing that $f$ continuous
Oct
26
comment Diagonalisable linear operator on infinite-dimensional vector space: definition problem
I think I have seen a definition for Hilbert spaces: If $e_n$ is an orthonormal basis for a Hilbert space $H$ then $T$ is diagonal if $Te_n = \lambda_n e_n$ for some $\lambda_n \in \mathbb C$. Not sure this helps.