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Assume without loss of generality that $a_1\ne0$ and consider, for every $n\geqslant1$, $$b_n=\frac{a_n}{A_n},\qquad A_n=\sum_{k=1}^na_k^2.$$ Then $$\sum_na_nb_n=\sum_n\frac{a_n^2}{A_n},$$ which diverges by a standard result. On the other hand, $A_n^2\geqslant A_nA_{n-1}$ and $a_n^2=A_n-A_{n-1}$ hence, for every $n\geqslant2$, $$... 6 The answer is: Not in general. For example, \Omega=(0,2\pi), u_n(x)=\sin nx. Then u_n\rightharpoonup 0, since$$ \int_0^{2\pi} f(x)\,\sin nx\,dx\to 0=u, $$for all f\in L^2[0,2\pi]. Meanwhile$$ v_n(x)=u_n^2(x)=\sin^2 nx=\frac{1}{2}-\frac{\cos (2nx)}{2}\rightharpoonup \frac{1}{2}=v, $$and u^2\ne v. Note. If \{u_n\} is bounded and ... 3 A look at the proof of Jensen's inequality is all you really need; there is no need for (more sophisticated) Hölder's inequality. For simplicity, scale f so that \|f\|_r=1. Let g=|f|^r. Jensen's inequality says \int_X g^{p}\ge 1 (for p=s/r>1), which is just the result of integrating the pointwise inequality$$g^{p} \ge 1+p(g-1) \tag{1}$$... 3 A solution, by my professor: Consider the closed unit ball in \ell^{\infty}(I). The extreme points of this are functions f: I \to \{-1, 1\}. Notice that if \phi: \ell^{\infty}(I) \to \ell^{\infty}(J) is an isometric isomorphism, then \phi takes extreme points to extreme points. We may assume without loss of generality that \phi(\chi_I) = \chi_J, ... 2 Note that \ell^{\infty}(I)=BC(I) (set of bounded continuous functions) when I is endowed with the discrete topology. In turn, BC(I)\cong BC(\beta(I)), where \beta(\cdot) denotes Stone–Čech compactification. Therefore, if \ell^{\infty}(I)\cong\ell^{\infty}(J), then BC(\beta(I))\cong BC(\beta(J)). The Banach–Stone theorem, in turn, implies that ... 2 The solution is correct. Just to beef up this post, I'll sketch a slightly different proof: the complement of l_0 is open. If x\notin l_0, let r=\frac12\limsup_{k\to\infty} |x(k)|. If \|x-y\|\le r, then$$\limsup|y(k)| \ge \limsup_{k\to\infty} |x(k)|-r =r$$hence y\notin l_0. By the way, this is the first time I see notation l_0 used for ... 2 Let w=\dfrac{\partial v}{\partial x}. Since v\in W^{1,\infty}, we have w\in L^\infty. The space L^\infty is dual to L^1. Thus, pairing a fixed L^\infty function with a convergent sequence of L^1 functions produces a convergent sequence of numbers. This is why$$\int b(u_k) w\to \int b(u)w$$For the other integral, note that a(u_k)w ... 2 If we define S as S=\operatorname{Supp}f, since f=f\cdot\mathbb{1}_S,$$\widehat{f}=\widehat{f}*\widehat{\mathbb{1}}_S,\tag{1}$$so the Fourier transform of your identity gives:$$\left(\widehat{f}\cdot(1-\widehat{g})\right)*\widehat{\mathbb{1}}_S=0.\tag{2}$$By the Riemann-Lebesgue theorem we know that g\in L^p implies \widehat{g}=o(1), so, if ... 2 No. For simplicity, consider the interval \Omega=[0,1] and construct a sequence of sets A_n such that the measures \lambda(A_n) tend to 0 but every point belongs to infinitely many A_n. For example A_1=[0,1/2], A_2=[1/2,1], A_3=[0,1/4],\ldots,A_6=[3/4,1], A_7=[0,1/8],\ldots. If f_n is the indicator function of A_n, that is f_n(x)=1 if ... 2 Let me offer a simpler proof. First, let c_{00} denote the subset of \ell_2 that consists of sequences with finite support, that is, sequences with finitely many non zero terms. Then your set, let's call it r, with rational entries, is a subset of this. I claim that r is dense in c_{00} and c_{00} is in turn dense in \ell_2. Proof Given ... 2 As the OP corrected guessed the right subset to study is the set A\subset\ell^\infty consisting of all the sequences with zeros and ones, i.e., \{a_n\}\in A if and only if a_n\in\{0,1\}, for all n. Clearly if \{a_n\},\{b_n\}\in A and \{a_n\}\ne\{b_n\}, then$$ \|\{a_n\}-\{b_n\}\|_\infty=1, $$since for some n |a_n-b_n|=1. Also ... 2 Yes, it's true and follows from the Holder's inequality: For any for a,b>0 such that \frac1a+\frac1b=1 we have$$\int|fg|\leq(\int |f|^a)^\frac1a(\int|g|^b)^\frac1b$$Now set f=|X|^p, g=1, a=\frac pq and b=\frac{q-p}q, this is possible for p<q:$$\int||X|^p\cdot1|\leq\left(\int\left(|X|^p\right)^\frac qp\right)^\frac pq\left(\int ...

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For the first part you have to use $\liminf$, as you don't still know that the limit exists. For the second part, you are thinking as if $X$ was $\mathbb R^n$, which it might not be. One way of attacking the problem along your line of thought would be to assume $\|f\|_\infty=1$ (i.e., work with $f/\|f\|_\infty$). Then, for $p>r$, $$... 1 If V is not separable, then L^p(X,\mu,V) is not either. Take \{v_j\} an uncountable set in V without a limit point and f_j\in L^p(X,\mu,V), such that f_j(x)=\varphi_j(x)v_j, where \varphi_j\ne 0 scalar. Clearly, there is no limit point in \{f_j\}. If X is separable, then again it is not certain that L^p(X,\mu,\mathbb R) is separable. It ... 1 Your ideas are essentially correct. However, the devil is in the details. You have to be careful about the values of q_{i,j}, and choose them correctly. For example, if in your sequence, |q_{i,i} - x_i|> 1 \quad \forall i, then you will never converge to the sequence (since we are always at least distance 1 away), even though you have pointwise ... 1 The general fact is, if f\in L^p\cap L^q with 1\leq p<q\leq \infty, then f\in L^r for any r\in (p,q). This is entirely elementary: write f=f_1+f_2, where f_1 is the restriction of f to S=\{x\mid \lvert f(x)\rvert\geq 1\}, then estimate L^r norm of f in terms of the norms of f_1 and f_2, and possibly the measure of S if ... 1 First Proof. Let f be a bounded Borel function and \varepsilon>0. Let s=\sup_{x\in K} \lvert f\rvert<\infty, M=\lfloor M\rfloor+1 and n\in\mathbb N, such that 1/n<\varepsilon. Let$$ E_{k,n}=\left\{x\in K: \frac{k}{n}\le f(x)<\frac{k+1}{n}\right\}, \quad -nM\le k\le nM. $$Define$$ f_n=\sum_{k=-nM}^{nM} \frac{k}{n}\chi_{E_{k,n}}, $$... 1 The commutator is compact if and only if f is in VMO. Source: On the compactness of operators of Hankel type by A. Uchiyama (free access; note that VMO is called CMO there). Since C_0(\mathbb R)\subset VMO, the answer to your question is positive. 1 In general, if you have a topological space X with a dense subset E, and if \phi,\psi:X\to\mathbb R are continuous functions, then the validity of \phi\le \psi on E implies that it holds on all of X. Passing to the limit preserves nonstrict inequalities. Here, the space X is W_0^{1,p}. The righthand side, \int |\nabla f|^p, is continuous ... 1 Following Jimmy R, we have$$ ||y_i-(x_n)||_{\infty}=\sup_{k \in N}|Pr_k(y_i)-Pr_k((x_n))|=\sup_{k >i}|Pr_k(y_i)-Pr_k((x_n))| \ge \sup_{k >i}|x_k|, $$where Pr_k denotes k-th projection. Since \lim_{i\to \infty}|x_i|=|x| > 0, we can choose such n_{0} that |x_k| >\frac{|x|}{2} for k \ge n_0. Hence for i\ge n_0 we get$$ ...

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Given $\epsilon>0$, there are $N$, $M$ so that for all $j$, $\ \ \ \ \ \ \int_{|x|\ge M} |f_j|<\epsilon$ and $\ \ \ \ \ \ \int_{|f_j|>N} |f_j|<\epsilon$. Let $A_j=\{\,x\mid |x|\ge M\,\}\cup \{ \,x\mid |f_j|>N\,\}$ and set $f_j^1=f_j\cdot\chi_{A_j}$. The $L_p$-norm of the other component, $f_j-f_j^1$, will be bounded, since it's a ...

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The function $$f(x) = \begin{cases} 1, &\text{if x is rational}, \\ 0, &\text{if x is irrational} \end{cases}$$ is measurable and bounded, hence integrable on any bounded interval. But $\sup_{x\in[0,1]} |f(x)-P(x)| \ge \frac12$ for any polynomial $P$ (indeed, any continuous function $P$).

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Here's a solution from first principles. As in lyj's answer, we use $\ell^\infty(I)$ to detect the number of pairwise disjoint subsets of $I$. Let $S$ be the unit sphere of $\ell^\infty(I)$. For a set $E \subset S$, let me say that $E$ is dispersed if for every distinct $f,g \in E$, $\|f + g\| \le 1$ and $\|f - g\| \le 1$. Lemma. If $E \subset S$ is ...

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Every $C^1$ domain, and more generally a Lipschitz domain, is a Sobolev extension domain, meaning that Sobolev functions on it can be extended to Sobolev functions on $\mathbb R^n$. In particular, all embedding theorems for Sobolev spaces hold on such domains. When $p>n$, Morrey's inequality gives a continuous embedding of $W^{1,p}$ into $C^\beta$ ...

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