Rudy the Reindeer
Reputation
20,723
98/100 score
 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. Nov 21 comment Properties of dual spaces of sequence spaces Why do I want to show that $\varphi$ is isometric in a)? Nov 20 comment Show that $R^3 - \{x,y\}$ is homotopy equivalent to $S^2\vee S^2$ You seem to be unsure about what "homeomorphic" and "homotopy equivalent" mean. I think you really meant to write "homotopy equivalent" everywhere. Nov 20 comment Properties of dual spaces of sequence spaces @DavidMitra: True. I don't like to obey. : ) Nov 20 comment Properties of dual spaces of sequence spaces @t.b.: I corrected the typos. Why 'of course'? It's not obvious to me. Thank you, I like my writing much better now than before you helped me change it. Nov 19 comment Does $f\circ g$ injective imply $f$ injective for functions $f,g:A\to A$? Sorry for my previous comment, I read $f \circ g$ as $g(f(x))$. Nov 18 comment Show that every upper semi-continuous real function is measurable Yes, true. But that doesn't mean he remembers it... anyway, will remove my comment. Nov 17 comment Proof: $X^\ast$ separable $\implies X$ separable @t.b.: I don't care so much for the contradiction, I wanted to thoroughly understand what I did wrong. Now that it's obvious, it's just embarrassing. Thank you for your patience! Nov 17 comment Proof: $X^\ast$ separable $\implies X$ separable Thank you. Now I wonder how I could make such a flawed argument : ( (the answer is because my head is messed up) Regarding the "norm dense": I'd never heard it before. It probably means w.r.t. the norm but then is there ever a normed space with a metric different from the metric induced by the norm? I guess the answer is yes, otherwise the word wouldn't exist. Nov 16 comment Proof: $X^\ast$ separable $\implies X$ separable And is it okay to say "dense" instead of "norm dense"? Nov 16 comment Proof: $X^\ast$ separable $\implies X$ separable Hi Brian. Thanks for your helpful answer! I edited the case where I try to show that $x_n^\ast (x_n)$ does not equal $0$. Is case (i) now correct? Nov 15 comment Proof: $X^\ast$ separable $\implies X$ separable @t.b.: But in the lecture notes it doesn't say continuous. How do I know you're right? Nov 15 comment Proof: $X^\ast$ separable $\implies X$ separable @AsafKaragila: Then I mean the algebraic dual, I think.