35,082 reputation
53199
bio website
location
age
visits member for 2 years, 8 months
seen 8 hours ago

2d
revised Completely continuous implies compact under separability assumption
added 76 characters in body
Jul
17
reviewed Close $i^{-1} F$ a sheaf if and only if $\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y$ is an isomorphism
Jul
17
reviewed Close Nonhomogeneous equations
Jul
17
reviewed Close A question of real analysis.
Jul
17
reviewed Close Finding first few terms in power series expansion of general solution
Jul
17
reviewed Close Advantage of Bootstrapping Confidence Intervals over Standard Error
Jul
17
reviewed Leave Open Compute $\int_1^e \frac{dx}{x(x+(\ln x)^2)}$
Jul
17
reviewed Close Homework - Proof of a property by inductive reasoning
Jul
17
reviewed Close Surgery results in a cylinder
Jul
15
comment Distance to a closed subspace. Prove $d(x, Ker(T))=max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}$.
possible duplicate of Distance between point and linear Space
Jul
14
revised Proof of $\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1$
added 53 characters in body
Jul
14
comment Show that $\sum_{\{a_1, a_2, \dots, a_k\}\subseteq\{1, 2, \dots, n\}}\frac{1}{a_1*a_2*\dots*a_k} = n$
Another approach. Denote this sum by $S$. Consider $p(x)=\prod_{k=1}^n\left(x+\frac{1}{k}\right)$. If you expand $p(x)$ and substitute $x=1$ you'll get after carefull gazing $S+1$. So the desired sum is $S=p(1)-1=\prod_{k=1}^n\frac{k+1}{k}-1$. By telescoping that product equals to $n+1$, hence $S=n$
Jul
14
revised $L^1(μ)$ is finite dimensional if it is reflexive
deleted 27 characters in body
Jul
14
revised $L^1(μ)$ is finite dimensional if it is reflexive
added 34 characters in body
Jul
14
revised $C([0,1])$ is not weakly sequentially complete.
added 2 characters in body; edited title
Jul
14
revised $L^1(μ)$ is finite dimensional if it is reflexive
added 1 character in body
Jul
14
comment $C([0,1])$ is not weakly sequentially complete.
Why is $L_\infty(K)$ is weakly sequentially complete? I think it is not.
Jul
14
answered $L^1(μ)$ is finite dimensional if it is reflexive
Jul
14
answered $C([0,1])$ is not weakly sequentially complete.
Jul
13
comment $L^1(μ)$ is finite dimensional if it is reflexive
Also you need to normalize $\chi_n$. Once you did it, you can show it is an isometry. Note: since $\chi_n$ have disjoint supports $|\sum a_n\chi_n|=\sum|a_n||\chi_n|$.