Rudy the Reindeer
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21,635
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 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.