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4

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'} ...


2

$A$ 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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