Reputation
21,635
Next tag badge:
77/100 score
25/20 answers
Badges
9 55 151
Impact
~607k people reached

Nov
30
comment Canonical $\mathbb{P}$-name
@AmitKumarGupta: Thanks!
Nov
30
comment Proving $\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1$ by induction
@Sosy: Thank you, my pleasure!
Nov
30
comment Proving $\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1$ by induction
@DavidMitra: Thanks, David. Fixed.
Nov
30
comment Proving $\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1$ by induction
@DavidMitra: Let's assume it's a typo. Let's see whether I can fix this.
Nov
30
comment Proving $\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1$ by induction
Darn, you're right!
Nov
29
comment $f\in C[0,\infty]$ and $\lim\limits_{x\to \infty}f(x)=L<\infty$. Compute $\lim\limits_{n\to \infty} \int_{0}^{2} f(nx)dx$
@S.D.: No, don't worry : )
Nov
29
comment $f\in C[0,\infty]$ and $\lim\limits_{x\to \infty}f(x)=L<\infty$. Compute $\lim\limits_{n\to \infty} \int_{0}^{2} f(nx)dx$
@S.D.: That's nice of you, thanks!
Nov
29
comment $f\in C[0,\infty]$ and $\lim\limits_{x\to \infty}f(x)=L<\infty$. Compute $\lim\limits_{n\to \infty} \int_{0}^{2} f(nx)dx$
@S.D.: I was wondering whether I didn't already cover the part "version 2" in my answer?
Nov
28
comment Canonical $\mathbb{P}$-name
Thanks for your nice answer. When you write $V$, does it stand for the von Neumann universe? I read the Wikipedia article on forcing and it says: "...forcing consists of expanding the set theoretical universe $V$ to a larger universe $V^\ast$..." so I was wondering if it's "the universe" because it's always the von Neumann universe.
Nov
28
comment Stokes' theorem application to vector field
@BillCook: Nice, thank you!
Nov
27
comment Stokes' theorem application to vector field
@ArturoMagidin: I wrote "Stokes'", not "Stoke's". And no, I did not know all that. I agree with your last sentence.
Nov
27
comment Stokes' theorem application to vector field
@ArturoMagidin: Or Stokes' theorem.
Nov
24
comment Properties of dual spaces of sequence spaces
@t.b.: Isomorphic Banach spaces have isomorphic duals: Let $\varphi : V \rightarrow V^\prime$ be a linear bijective map between two Banach spaces (prime is not meant to denote dual) with $\varphi^{-1}: V^\prime \rightarrow V$ bounded and linear. Let $\lambda \in {V^\prime}^\ast$ be a linear functional. Define $\varphi^\ast: {V^\prime}^\ast \rightarrow V^\ast$ as $\lambda \mapsto \lambda \circ \varphi^{-1}$. Then $\varphi^\ast$ is an isomorphism: (i) linear, injective and surjective are clear (ii) ${\varphi^\ast}^{-1}$ is bounded follows using the open mapping thm. and surjectivity from (i)
Nov
22
comment Change the values of a measurable function on a negligible set
Taking the obvious map $\varphi: f \mapsto f$, linear and surjective are "obvious". It's well-defined: Let $f \neq f^\prime$ on $U$ with $\mu(U) = 0$. Then $\tilde{\mu}(U) = 0$ and so $f = f^\prime$ $\tilde{\mu}$-a-e. Finally, $$ \| f \|_{L^1 (\mu)} = \int_X |f| d \mu = \sup \{ \int_X s d \mu \mid s \text{ step function }, s \leq |f| \} =$$ $$ \sup \{ \sum_{i=1}^n \alpha_i \mu(Y_i) \mid \text{ for some } \alpha_i \} = \sup \{ \sum_{i=1}^n \alpha_i \tilde{\mu} \mid \text{ for some } \alpha_i \} = \| f \|_{L^1(\tilde{\mu})}$$.
Nov
22
comment Computing winding number
@joriki: Unintended. Thanks, joriki.
Nov
21
comment Properties of dual spaces of sequence spaces
Yes, I understood that (after reading your comment).
Nov
21
comment Properties of dual spaces of sequence spaces
Two definitions? I think I'm getting confused. I'm only talking about the isomorphism between $c_0(\mathbb{N})^\ast$ and $l^1(\mathbb{N})$, which I called $\varphi$. And then my original confusion, which still remains, is why $\varphi$ has to be an isometry.
Nov
21
comment Properties of dual spaces of sequence spaces
@t.b.: Ah. I didn't realise I had to check that $\varphi$ is linear, too. Thanks for pointing it out! I think their comment is missing a "don't" or something, I couldn't make sense of what they were saying. Reading it again, I still can't make sense of it.
Nov
21
comment Properties of dual spaces of sequence spaces
"1) ...that $\varphi$ actually is an isomorphism of Banach spaces..." seems to imply exactly that it needs to be an isometry on top of being an isomorphism in order for it to be an isomorphism of Banach spaces. Am I misreading?
Nov
21
comment Properties of dual spaces of sequence spaces
This was another instance of me not knowing the definitions. Kreyszig p. 109: "Isomorphisms for normed spaces are vector space isomorphisms which also preserve norms." Sorry.