Apr
29
comment Norm vs A-norm in non-Archimedean Functional Analysis
I think there are two reasons: 1) evaluation of the underlying field satisfy a stronger axiom 3; 2) most of the Banach spaces are constructed by standard operations over copies of underlying field.
Apr
22
comment Convolution of $L^1(G)$ functions with elements of $M(G)$.
My poor intuition says $A(G)$ separates compact sets but I don't know the proof. I'm not a specialist on this subject.
Apr
22
comment Convolution of $L^1(G)$ functions with elements of $M(G)$.
@Mambo it would be better if you ask your question via standard tools of the site. Do not use comments for this purpose.
Apr
16
comment Tensor product of algebras of operators on $SQ_p$ spaces
You should look at paper on p-operator spaces by Matthew Daws
Apr
16
awarded Notable Question
Apr
12
comment Should we have tags (maximal-ideals) and (prime-ideals)?
With the same reasoning we can merge tags for banach and normed spaces, but we must not do this. Soooo....
Apr
11
comment A net in $L^1(\mu)$ which converges vaguely in $M(X)$ to the point mass at $x$.
At the verry begining choose an open set $U$ of finite measure containing $x$. Then intersect all $V_n$'s with this $U$.
Apr
7
answered Lower bound of coefficients in an orthogonal decomposition
Apr
3
comment Does the function $\log(1+\exp(x))$ have a conventional name?
@Glen_b sorry, my bad
Apr
3
comment Does the function $\log(1+\exp(x))$ have a conventional name?
This is the sigmoid function en.wikipedia.org/wiki/Sigmoid_function
Apr
3
awarded Enlightened
Apr
3
awarded Nice Answer
Apr
2
comment Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$
@Nicolas fixed.
Apr
2
revised Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$
deleted 66 characters in body
Apr
2
answered Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$
Mar
31
awarded Popular Question
Mar
17
revised Why isometric isomorphic between Banach spaces means we can identify them?
added 11 characters in body
Mar
15
awarded Revival
Mar
15
comment Algebra of compact operators on $\ell_p$
Very nice answer, as usually.
Mar
15
comment Show that the subspace $Y = \{ x \in \mathcal{C}[a,b] \mid x(a) = x(b) \} \subset \mathcal{C}[a,b]$ is complete
This is the kernel of bounded linear functional $f:\mathcal{C}[a,b]\to\mathbb{C}:x\mapsto x(a)-x(b)$, hence closed, and even of cdimension 1.
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