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In fact, we can derive that $\,\dim S \le c^2$. Let me describe the proof as it is very elegant. Assume that $v_1,\ldots, v_n\in S$ are orthonormal functions, i.e., $\int_0^1 v_iv_j\,dx=\delta_{ij}$, and for a fixed $a=(a_1,\ldots,a_n)\in\mathbb R^n$ define $\varPhi_a :\mathbb R^n\to \mathbb R$ as $$\varPhi_a(x)=\sum_{j=1}^n a_jv_j(x).$$ Then $$... 3 From the inclusions of sets$$\ell_p\subset c_0\subset \ell_\infty,$$where c_0 denotes the set of convergent sequences and the separability of (c_0,\lVert \cdot\rVert_\infty) is separable we conclude that (\ell_p,\lVert\cdot\rVert_\infty) is separable. \ell_p endowed with the supremum norm is not a closed subspace of \ell_\infty because its ... 3 This question is much easier if you write the weak L^p norm in terms of decreasing rearrangements http://en.wikipedia.org/wiki/Lorentz_space#Decreasing_rearrangements Then \int_E f \, d\mu \le \int_0^{[0,\mu(E)]} f^*(s) \, ds \le \|f\|_{p,\infty} \int_0^{[0,\mu(E)]} s^{-1/p} \, ds . 2 Write s = tp+(1-t)q. Consider 1/p'+1/q'=1.$$ \int |f|^{s} = \int |f|^{tp+(1-t)q} \le \left\{\int |f|^{tpp'} \right\}^{1/p'} \left\{\int |f|^{(1-t)qq'} \right\}^{1/q'} $$Now you can choose$$ tpp'=p, (1-t)qq'=q\iff p'=\frac 1t, q'=\frac 1{1-t} $$because$$ \frac 1{p'}+\frac 1{q'} = t+(1-t)=1 $$Then$$ \int |f|^{s} \le ||f||_p^{p/p'} ||f||_q^{q/q'} ...

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$A$ is not closed. Take any function $f\in L^1$ such that $f$ is not in $L^2$, then approximate it by functions in $A$. I.e. $f(x) = \chi_{(0,1)}(x) x^{-1/2}$, then $f\in L^1$, $f\not\in L^2$. Define $f_n(x) = \min(n,f(x))$. Then $f_n\to f$ in $L^1$, $\|f_n\|_{L^2}\to\infty$. As to your second question: sets like $\{f\in L^1: \ f\ge 0 \}$ are closed as ...

2

Here is a counterexample. Let $\phi\colon\mathbb{R}\to\mathbb{R}$ be $C^\infty$ supported on $[0,1]$, positive with $\int_0^1\phi(x)=\int_0^1\phi(x)^2\,dx=1$. Let $$f_n(x)=\sum_{k=1}^n2^{k}\phi(2^{2k}(x-k)).$$ Then $$\int_{\mathbb{R}}f_n(x)\,dx=\sum_{k=1}^n2^{k}2^{-2k}<1.$$ Moreover $f_n(x)\le f_{n+1}(x)$ for all $x\in\mathbb{R}$. It follows that ...

2

The case $q = 1$ is done by the premises, so let's suppose $1 < q < p$. The idea is to use Hölder's inequality to get an estimate $$\int_0^1 \lvert u_n(t) - u_m(t)\rvert^q\,dt \leqslant \lVert u_n - u_m\rVert_{L^1}^\alpha \cdot \lVert u_n-u_m\rVert_{L^p}^\beta,$$ which then shows that $(u_n)$ is a Cauchy sequence in $L^q(0,1)$, since $\lVert u_n - ... 1 Do the normalization. Remember that$\|f\|_{L^{p,\infty}}^p=p\int_0^\infty \lambda_f(t)t^{p-1}dt$. We have that$\#\{|f(x)|\geq t\} \leq t^{-p}$but just inputting that isn't good enough so let's throw away some values. Note that the trivial bound$|X|$is lower that$t^{-p}$over some small values of$t$, and that if$t>1$,$\#\{|f(x)|\geq t\} <1$but ... 1 EDIT: the fact that the sequence is unbounded for$y = 0$is not a problem since we don't care about sets of measure$0$. Clearly $$\lim_{n \to \infty} \frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) = \frac{1}{1 + x^2}f(0).$$ Moreover we can find$N$such that if$n \ge N$then$\Big|f\Big(\frac{x}{\sqrt{n}}\Big) - f(0)\Big| \le 1$, then in particular ... 1 Since $$2^{np}\mu\{2^n\leqslant |f|^p\lt 2^{n+1}\}\leqslant\int |f|^p\chi_{\{2^n\leqslant |f|\lt 2^{n+1}\}}$$ we obtain after a summation that $$\int |f|^p\geqslant \sum_{n\geqslant 1}2^{np}\mu\{2^n\leqslant |f|^p\lt 2^{n+1}\}=\sum_{n\geqslant 1}2^{np}(\lambda_f(2^n)-\lambda_f(2^{n+1})).$$ The result follows from a summation by parts. 1$A$is closed, but not open. The sequence$x=(1,1,1,...)$is not an interior point. If$(x_n)$is a convergent sequence in$A$with the limit, then it also converges pointwise. But since$0\le x_{n,m}\le 1$, we also have$0\le \lim \limits_{n\to \infty} x_{n,m}\le1$, so$\lim \limits_{n\to \infty}x_n\in A.B$is open, but not closed. If$f\in B$, not that ... 1 It looks like a good candidate for Hölder inequality, with$\frac{1}{p}$and$p'$such that$p + \frac{1}{p'}=1$, so$p'=\frac{1}{1-p}$Let first assume that$f$is positive and real-valued :$\Gamma(E) = \int_X |f|^p 1_E d\mu \leq (\int_X(|f|^p)^{\frac{1}{p}} d\mu)^p(\int_X 1_E^{p'} d\mu)^{\frac{1}{p'}} = (\int_X|f| d\mu)^p(\int_X 1_E)^{1-p} $With the ... 1 I hope I didnt miss something, but this should work, For$p=1$the inequality trivialy holds and notice that$\forall p>0$we have, $$|a-b|^p\leq|a+b|^p$$ We have two cases to consider, (1)$p>1$: Note that$h(x)=x^p$is convex and monotone for$p>1$. Hence, ... 1 Let$f:X\to \mathbb{R}$. Assume that$f$is measurable, and that$\|f\|_p<\infty$for all large$p$. Suppose for convenience that$f\geq 0$. (If not, just work with$f^*:=|f|$.) We define $$\|f\|_{\infty}:=\sup \{r\in \mathbb{R}: \mu\left( \{x:|f(x)|\geq r\} \right)>0\}.$$ The case where$\|f\|_{\infty}=0$is trivial. If$\|f\|_{\infty}\neq 0$, let ... 1 First of all,$f_n\to f$in$L^2$implies$f_n\to f$in measure, so the second assumption is redundant. There is a standard counterexample to show that convergence in$L^p$,$1\le p<\infty$, does not imply convergence a.e., much less "almost uniformly". Namely, enumerate dyadic subintervals of$[0,1]$as$I_1,I_2,\dots$(order does not matter), and let ... 1 If$I$is an interval where$f'$is positive, we have, by integration by parts, $$\int_I |f'(x)|^pdx=\int_If'(x)f'(x)^{p-1}dx=-(p-1)\int_If(x)f''(x)f'(x) ^{p-2}dx,$$ and similarly if$f'$is non-negative. By summing over intervals$I$where$f'$has constant sign, we obtain $$\Vert f'\Vert_{L^p}^p=\int_a^b|f'(x)|^pdx\leq ... 1 Note that$$f_x = \frac{(1 + x^2) \frac 1 2 x^{-1/2} - x^{1/2} \cdot 2x}{(1 + x^2)^2} = \frac{\frac 1 2 x^{-1/2} - \frac 3 2 x^{3/2}}{(1 + x^2)^2}$$For large values of x, we have the estimate (where \lesssim means "up to some constants")$$|f_x(x)| \lesssim \frac{x^{3/2}}{x^4} = x^{-5/2}$$For small values of x, we have the similar estimate ... 1 Consider the unit interval [0,1] and$$f_\epsilon(x) = \frac 1 {\sqrt x} \chi_E = g_{\epsilon}(x)$$where E = [\epsilon, 1] and \chi_E is a characteristic function. Then f_{\epsilon}g_{\epsilon} \in L^1, and we have$$\|f_{\epsilon}\|_1 = \int_{\epsilon}^1 \frac{1}{\sqrt x} dm(x) \le \int_0^1 \frac{1}{\sqrt x} dm(x) = 1$$and likewise for ... 1 Here's an idea that might be helpful: that L^p measures spikiness or broadness of a function in some sense. In particular, L^{\infty} consists of functions with no spikes whatsoever, while L^1 functions can't be too broad; the higher p is, the more control you have over spikes, and less control over broadness. Since f \in L^p \cap L^q, we've ... 1 Hint: It follows as a corollary of the following statement: Let \{f_i:i\in I\} be a family of functions with f_i\in L^{p_i}(\Omega) and \dfrac{1}{p}=\sum \dfrac{1}{p_i}\leq1. Then \prod f_i\in L^p(\Omega) and$$\|\prod f_i\|_{L^p(\Omega)}\leq\prod\|f_i\|_{L^{p_i}(\Omega)}$$This is the interpolation inequality (H. Brezis "Functional Analysis, ... 1 If a function u is actually a radially symmetric function, then we have the integration formula$$ \int_{B(0,R)} u(x)\, dx = |\mathbb{S}^{n-1}| \int_0^R u(r)r^{n-1}\, dr, $$where n is the dimension of the space where u is defined. In your case, u(x)=|x|^{-\lambda}, and it is straightforward to apply the formula and check the integrability of u. 1 The inclusion L^2\subset L^1_w does not hold: take f(x):=\frac 1{\sqrt x\log x}\chi_{(1,\infty)}(x). The converse reduces to ask whether g\in L^1 implies x\mapsto xg(x)^2\in L^1. Define$$g(x):=\sum_{j=1}^\infty c_j\cdot\chi(j^2-a_j,j^2+a_j)(x),$$with the a_j small enough. Then g is integrable if and only if \sum_{j=1}^\infty c_ja_j is ... 1 Let B be any bounded set that contains the support of f. By Jensen's Inequality,$$ \frac1{|B|}\int_B|f(x)|^p\mathrm{d}x\le\left(\frac1{|B|}\int_B|f(x)|\mathrm{d}x\right)^{\large p} $$since |x|^p is concave for 0\le p\le1. 1 Unfortunately, a sequence can be weakly convergent without being pointwise convergent (f_n(x))=(\sin(n\pi x)) on the unit interval (which is not convergent except at 0). However we can show that for each simple function g (linear combination of characteristic functions of measurable sets),$$\lim_{n\to +\infty}\int_{(a,b)} f_ngdx= \int_{(a,b)} ... 1 Yes, your proof is good. In terms of writing, you start using a fixed$x\in\ell^1$without saying so. Also, the sentence that says "For that we will show..." doesn't really make sense; I understand it because I know how to prove the density, but otherwise it looks hard to understand. 1 I'm not sure if this is what you're looking for, but if$B$is bounded away from zero (i.e., if there exists$M$such that$\frac{1}{|B(x)|} \leq M$for all$x$, or almost all$x$), then we could say the following:$\$ \left| \int A - \int \frac{C}{B} \right| = \left| \int \left( A - \frac{C}{B} \right) \right| \leq \int \left| A - \frac{C}{B} \right| = \int ...

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