1
vote
1answer
49 views

Prove that a relatively compact subset of $L^p$ is bounded.

Let $p\in [1,\infty)$, $A\subset L^p(\mathbb R^m)$ relatively compact and $\lambda^m$ be the Lebesgue measure on $\mathbb R^m$. Prove: a) $A$ is bounded. b) $\lim_{y \to 0}\sup_{f \in A} ...
1
vote
1answer
24 views

Is a set of jointly bounded functions over a compact domain compact under p-norm?

Let $X$ be a metric space and a measurable space. Let $K$ be a compact set of nonzero measure and $r> 0$. Is a set $\{ f: K\rightarrow \mathbb R| |f|\leq r$ almost everywhere$\}$ compact with ...
1
vote
1answer
28 views

A compactness argument for small high frequencies

I would like to prove the following statement: Let $N\geq 1$, $1\leq q<\infty$ and let be $E$ a relatively compact subset of $L^q(\mathbb{R}^N)$. Then \begin{equation*} \sup_{u\in ...
1
vote
1answer
38 views

Is there any Banach space $X$ that $L^2(\Omega)$ is compactly embedded into?

Let $\Omega \subset \mathbb{R}^n$. Is there a good (*) Banach space $X$ that $L^2(\Omega)$ is compactly embedded into: $$L^2(\Omega) \subset\!\subset X$$? If not compactly embedding, I at least would ...
1
vote
1answer
53 views

Is this a totally bounded set in the space of continuous functions?

If $A=\{f\in C[0, 1]: \int^1_0|f(x)|^2\,dx\leq1\}$ and metric $d(f, g)$ is $(\int^1_0|f(x)-g(x)|^2dx)^\frac{1}{2}$. Is $A$ totally bounded? I know $A$ is clearly bounded since $d(f, 0)\leq 1$ under ...
0
votes
2answers
150 views

A relative compactness criterion in $\ell^p$

There is a relative compactness criterion for subset of $\ell^p$ that seems to me to be almost unheard-of (I say that because a google search provided no proofs nor references) but that is very ...
1
vote
1answer
91 views

Closed set in $l^1$ space

Let $$ X := \left \{ (a_n) : \sum_{n=0}^\infty |a_n| < \infty \right\}$$ with the metric $d(a_n,b_n) := \sum_n |a_n-b_n|$. Let $\delta_j^{(n)} := 1$ if $n = j$ and $0$ otherwise. Denote ...
3
votes
1answer
157 views

When is a subset of $\ell^2$ compact?

I have been looking on the internet for hours now and even asking in chat without an answer. When is a set $M\subseteq\ell^2$ compact? For $L^p$, there is the Arzelà–Ascoli theorem that provides a ...
2
votes
2answers
181 views

Compact inclusion in $L^p$

Is it true that there is a compact inclusion from $L^p$ to $L^q$ whith $q<p$? What is the counterexample if what I said is wrong? Thank you.
1
vote
1answer
48 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 ...
3
votes
1answer
140 views

$W^{1,p}$ compact in $L^\infty$?

Is $W^{1,p}(0,1)$ compactly contained in $L^\infty(0,1)$? Can I use this to show that I can select a sequence $(u_{n_k})$ from every bounded sequence $(u_n)$ in $W^{1,p}(0,1)$ such that $\lVert ...
5
votes
2answers
189 views

Closed set in $\ell^1$

Show that the set $$ B = \left\lbrace(x_n) \in \ell^1 : \sum_{n\geq 1} n|x_n|\leq 1\right\rbrace$$ is compact in $\ell^1$. Hint: You can use without proof the diagonalization process to ...
3
votes
1answer
151 views

Two Real Analysis Questions

If I have $ A = \{a \in \ell_2 : |a(n)| \leqslant c(n)\}$ for $c(n)\geqslant 0$ where $ n \in N $, and I want to show that is $A$ compact in $\ell_2$ iff $\sum{c(n)^2}<\infty$. How do I go about ...
2
votes
1answer
252 views

Bounded sequences that form compact sets or not

a) Give an example of a bounded closed subset of $$ A = \{(x_n) \in \ell^1: \sum_{n\geq1} x_n = 1\}$$ which is not compact. The metric we consider on A is induced by the normal norm on ...