All Questions
9 questions
0
votes
0
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51
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Trying to understand the proof for the criterion of compactness in $l_p$ space
I have the following theorem about the criterion of compactness in $l_p$ space
For the set $K\subset (l_p,||.||_p), p\geq 1$, following conditions
are equivalent:
i) $K$-totally bounded in $(l_p,||.||...
- 328
1
vote
0
answers
46
views
Prove that $A$ is relatively compact (or totally bounded) in $L^r(\Omega,\mu)$.
I encountered an exercise in my functional analysis course:
Let $(\Omega, \Sigma, \mu)$ be a measure space, and
$A\subseteq L^p (\Omega, \mu) \cap L^r (\Omega ,\mu)$, where $1\leq p <r <+\infty$...
- 716
2
votes
1
answer
270
views
Show that the closed unit ball in $L^2[a,b]$ is not sequentially compact.
I'm using Royden & Fitzpatrick's Real Analysis text to prep for an exam. This is one of the questions regarding the general properties of metric spaces.
Let $B=\{ f \in L^2[a,b] | \space ||f||_2 ...
- 257
1
vote
2
answers
220
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Why isn't $B = \{x \in l_{\infty} : \| x \|_{\infty} \leq 1 \}$ a compact metric space? [duplicate]
I have thought about an example of Cauchy sequence for which every subsequence has no limit in $B$, but I wasn't able to find one.
- 651
1
vote
2
answers
151
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What set $Y$ that $C_{c}(U) \subset Y$ and $Y$ separable w.r.t. $\vert \vert \cdot \vert \vert_{\infty}$
Let $U \in \mathbb R^{d}$ open and further $C_{c}(U)$ be the space of functions with compact support. Show that $C_{c}(U)$ is separable w.r.t. $\vert\vert \cdot \vert \vert_{\infty}$, and I have been ...
- 1,838
2
votes
2
answers
425
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Subset of metric space is bounded and closed but not compact.
We consider the set $l^1({\mathbb{N}})$ of the sequences whose associated series are absolute convergent: $$ l^{1} (\mathbb{N}) = \left\{ (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}} \mid \...
- 839
3
votes
3
answers
3k
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Unit ball in $l^{2}$ is bounded and closed but not compact.
I was told that the answer I gave to this question didn't actually show that the unit ball wasn't closed. Could someone please help me figure out how to do this? this is the original answer I gave ($S$...
- 523
3
votes
2
answers
682
views
Closed and bounded but not compact subset of $\ell^1$
I need to prove that
$$A=\{a\in \ell^1:\sum_{i=1}^{\infty}|a_n| \le 1\}$$
is closed, bounded and not a compact subset in $\ell^1$. Boundedness is trivial, but I get stuck in the other two. Proving ...
- 263
2
votes
1
answer
291
views
Prove that $ A\subset \ell_1 $ is compact iff $A$ satisfies the following property
$A$ is compact iff $ A $ is bounded and, given $\epsilon > 0$, there exists $ n_0$ such that $ \sum_ {k=n}^\infty |x_k|\le\epsilon $ for all $n \geq n_0 $ and for all $ x\in A $.
To prove ...
- 2,316