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comment If a unit ball is compact then why a ball of radius 5 has to be compact too?
Note that in infinite-dimensional Banach spaces (which are the most interesting ones), the closed unit ball is not compact, nor is any other ball.
1d
revised What did I do wrong?
latex, tags
2d
answered prove that $u$ is equal a.e. to an absolutely continuous function
Jun
23
awarded  Enlightened
Jun
23
awarded  Nice Answer
Jun
21
answered Dense subsets in tensor products of Banach spaces
Jun
21
comment Dense subsets in tensor products of Banach spaces
No, certainly not for any choice of the norm $\|\cdot\|_B$: it could be totally unrelated to the $B_1, B_2$ norms. The usual assumption for the norm on a tensor product of Banach spaces is that it should satisfy $\|f \otimes g\|_B = \|f\|_{B_1} \|g\|_{B_2}$, see ncatlab.org/nlab/show/tensor+product+of+Banach+spaces, and in that case your claim is true.
Jun
20
comment Proving a differential inequality without performing iteration
If you set $f = \sqrt{g}$ you immediately get $f' \le \frac{1}{2}$. I guess you do have to account for points where $f=g=0$ but that should not be hard.
Jun
16
comment To understand some terminology of metric spaces
Maybe it would be more helpful if you could quote one of the explanations you found, and point out exactly where you are confused. Otherwise you are just going to get more explanations, with no guarantee that they will be any less confusing to you than the ones you've already seen. That is not very efficient for anybody.
Jun
16
answered Is the unit sphere in an infinite dimensional Hilbert space closed?
Jun
16
comment Is the unit sphere in an infinite dimensional Hilbert space closed?
Of course, you also have to prove that the norm is a continuous function. This isn't necessarily obvious from the definition.
Jun
16
comment A construction of sigma-algebras - surely not new, right?
@hot_queen: This is basically the notion of a Borel code, is it not?
Jun
16
comment Measure Theory by Halmos, theorem B, page 22
Maybe the word "covered" is confusing you? Here it means "subset". In other words, a set $F$ is covered by a finite union of sets in $\mathbf{E}$ if and only if there exist sets $A_1, \dots, A_n \in \mathbf{E}$ such that $F \subseteq A_1 \cup \dots \cup A_n$.
Jun
16
comment Equivalence of reflexive and weakly compact
@user91360: Yes, almost everywhere would be fine. The key point is that the set is bounded in $L^p$ norm for some $p > 1$ (here we can take $p=\infty$). More generally, you can look at the Dunford-Pettis theorem which says that the essential condition for weak compactness is uniform integrability; when working with a finite measure, an $L^p$ bounded set ($p > 1$) is always uniformly integrable.
Jun
16
comment Equivalence of reflexive and weakly compact
@user91360: In that question, it is shown that a particular closed convex subset $S$ of the unit ball (defined in the question) is weakly compact. It is not claimed, and not true, that every closed convex subset of the unit ball of $L^1$ is weakly compact. In particular, the unit ball itself is not weakly compact in $L^1$. There is no contradiction here.
Jun
15
comment Is a Banach space also a metric space?
A norm always induces a metric, but not every metric is induced by a norm...
Jun
15
answered For an infinite set, prove that the collection of subsets A such that A is finite, or A complement is finite is a sigma algebra
Jun
15
answered Convergence in probability, in the sense of weak convergence of measures
Jun
15
answered Condition for the existence of a Markov process from the properties of a semigroup $\{T_t\}$
Jun
15
revised a problem from durret's essentials of stochastic processes
align