Jul
19
revised Completely continuous implies compact under separability assumption
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Jul
19
asked On $C_0(\Omega)$-module maps from $L_\infty(\Omega,\mu)$ to $L_q(\Omega,\nu)$
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
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 $\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1$
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Jul
14
comment Clever downvoting
@O.L. The problem is solved. Downvoters made to much downvotes and the system finally classified them as serial downvotes.
Jul
14
answered Show that $\sum_{\{a_1, a_2, \dots, a_k\}\subseteq\{1, 2, \dots, n\}}\frac{1}{a_1*a_2*\dots*a_k} = n$
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
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Jul
14
revised $L^1(μ)$ is finite dimensional if it is reflexive
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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
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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
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