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Oct
29
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 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
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
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
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.
Oct
26
comment Let V be a vector space. If every subspace of V is T-invariant, prove that there exist a scalar multiple c such that T=c1v
Related: math.stackexchange.com/questions/982355/…
Oct
26
comment If every subspace of V is T-invariant, prove that T is a multiple of the identity map.
Hello? I think your answer contains a few typos.
Oct
21
comment Using Regular Value Theorem to Show that $S^2$ is a $2$-dimensional Manifold.
There is not much to say but would one of you post an answer? I would do it but I don't think I should get upvotes for other people's work.
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila One can never be suspicious enough though : ) I apologize, my bad!
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila Ok. (Edit: True, the two downvoters seem to have changed their minds...)
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila Hence the downvote or what?
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
Can you make up your mind whether you want to down vote or up vote?
Oct
20
comment Show that the directional derivative is linear by definition
Could you help me understand what you're doing? In the title you ask about showing that the directional derivative is linear but then in the question you don't actually use the expression for the directional derivative (which would include the direction vector).
Oct
20
comment Show that the directional derivative is linear by definition
Thank you but I still don't understand. I think I'm confused by the question. I'll post a comment there...
Oct
20
comment If every subspace of V is T-invariant, prove that T is a multiple of the identity map.
Shouldn't it be $Tw = a_v w$? And given that you use $c_v$ further down, maybe you want to change $a_v$ into $c_v$ at the beginning? Or am misunderstanding your answer?
Oct
20
comment Sequence of bounded linear functionals on $C^1[0,1]$ that shows Principle of Uniform Boundedness fails without completion.
Why "hint"? I consider this a full answer.
Oct
20
comment $X,Y$ are Banach spaces, $T$ is linear, $x_n\to 0$ and $Tx_n\to y$, then $y=0$ and $T$ is continuous.
I don't understand the stuff after "Then, ...": you seem to be showing that $y=0$ but that is one of the assumptions.