Rudy the Reindeer
Reputation
99/100 score
 Jan 5 revised Injective functions also surjective? edited body; edited tags; edited title Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ Oh, I wasn't trying to make a point -- I was merely asking a question (to which I don't know the answer). Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ +1 by the way, for posting your solution, regardless of whether it's correct or not. Jan 1 comment $f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$ Isn't $L^1$ the space of Lebesgue integrable functions? If it is, could you justify why you can use the Riemann integral in your first equation (the one with the $\infty$ on the left hand side)? Dec 30 awarded Great Question Dec 29 comment Is this contradiction faulty? @ReinhildVanRosenú It seems to be a very confused person who wrote that website. For example, if $S$ is all the values where $f$ is $\le 0$ the person makes a case distinction for values where $f$ is $>0$. Perhaps you can find better proofs in a book. Or on Wikipedia. Dec 28 answered Is this contradiction faulty? Dec 24 revised Your favourite application of the Baire Category Theorem spelling fixed Dec 22 answered Modulo of a negative number Dec 22 comment Modulo of a negative number But then your answer states that method 1 is correct while method 2 is not. Or is it not? Dec 20 comment Properties of dual spaces of sequence spaces I will have to read the whole thread and this answer as it's not even clear to me what pairing here means exactly. Sounds like either $f$ or $g$ is fixed because we probably want to show that these angle brackets (in one argument) define an isomorphism. But I won't have time to do this before tomorrow night. Dec 20 comment Properties of dual spaces of sequence spaces @AnthonyPeter That's an excellent question. I will have to think about this as I do not know the answer off the top of my head. Dec 16 comment Contractible vs. Deformation retract to a point. This I explain in the first paragraph of my answer. Pick a point in $Y$ and an $\varepsilon$ ball around it. Can you find a neighbourhood inside the ball that is contractible? I don't think so -- it's not even connected! Dec 8 comment Let $S$ be a non-empty set with an associative, cancellative operation and for each $a\in S$, $\{a^n\}$ is finite, must S be a group? Dec 3 comment Examples of absolutely continuous functions that are not Lipschitz. @FardadPouran Yes, as I point out in my comment above. Nov 28 comment If an abelian group has more than 3 elements of order 2 then it must have at least 7 elements with order 2. Alan's answer has been undeleted. Nov 23 comment Orders of the elements in $\mathbb{Z}/8\mathbb{Z}$ You're confusing additive and multiplicative notation I suspect. The neutral element here is $0$ not $1$. Nov 23 comment Cauchy Residue Theorem Application Craig, I left a comment in response to your comment to my answer in another thread. Since you probably won't be notified (as you seem to have deleted your comment) I am notifying you here. Nov 23 comment Examples of absolutely continuous functions that are not Lipschitz. @Craig You are right, it's not clear at all, I only hint at how to show uniform continuity. It would have been much better to use Lebesgue integrability of the square root function as in definition (2) here‌​. I will edit the answer when I have time. Thanks a lot for pointing out this shortcoming. Nov 21 comment Continuous function on a compact metric space is uniformly continuous @shilov You should look at the last line of the proof. The inequality should explain it.