1d
awarded Constituent
1d
comment How prove $\frac{2}{3}<\frac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\frac{3}{2}$
This super-duper technique is called trial and error
Dec
14
revised Why are logarithms not defined for 0 and negatives?
deleted 4 characters in body
Dec
13
comment Distance between a point and a m-dimensional space in n-dimensional space ($m<n$)
@Orient, clearly, you can simultaneously shift this subspace and vector p by some vector of this subspace. Thus you reduce this problem to the one discussed above.
Dec
12
revised A generalized version of the Riemann-Lebesgue lemma
deleted 6 characters in body
Dec
12
comment A generalized version of the Riemann-Lebesgue lemma
@N3buchadnezzar, of course!
Dec
9
awarded Supporter
Dec
8
awarded Caucus
Dec
8
comment Riemann integrals of abstract functions into Banach spaces
Riemann integral is not well behaved even for scalar valued case. I see no reason to waste time on its generalizations. Better take a look at Bochner and Pettis integrals. If you are brave enough look at Vector Measures. Joseph Diestel, John Jerry Uhl
Dec
8
comment Prove that a complex-valued homomorphism on a Banach algebra which is not identically 0, is a bounded linear functional of norm $1$
See theorem 2.7 here
Dec
6
revised When $C(X)$ is an injective $C(X)$-module?
edited tags
Dec
6
comment When $C(X)$ is an injective $C(X)$-module?
That is correct. Excuse me if I was too unclear. By the way, if you suspect that this question is deadly difficult I would like to see your arguments.
Dec
6
comment When $C(X)$ is an injective $C(X)$-module?
The stricter version, where embeddings have closed range.
Dec
6
asked When $C(X)$ is an injective $C(X)$-module?
Dec
3
answered Hahn–Banach theorem?
Dec
2
revised Relations between p norms
added 5 characters in body
Nov
30
comment Reference for a nice formula
you should answer your question then
Nov
26
awarded Revival
Nov
18
revised Contractively complemented subspaces of $c_0(I)$
added 1 character in body
Nov
18
comment How to show this space is NOT reflexive
Since $X^*=\ell_1$, then $X^{**}=\ell_\infty$. Note that $X$ is separable, while $\ell_\infty$ is not separable. Therefore $X$ and $\ell_\infty$ are not isomorphic.
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