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## Hot answers tagged lp-spaces

8

We know that, $L_\infty$ is not separable, so neither does its dual $L_\infty^*$. It is remains to recall that $L_1$ is separable.

4

Use an approximation argument: it is true when $f$ is a simple function (linear combination of characteristic function), and for any $s$ simple and each positive $x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$

3

If $1\le p<q<\infty$, then the embedding $$L^q(0, 1)\subset L^p(0, 1)$$ trivially is weakly compact. Indeed, an $L^q$-bounded sequence is $L^p$-bounded too, and by Banach-Alaoglu's theorem (which applies since $q\ne 1, \infty$) it has weakly convergent subsequences.

3

Take $x(n):=n^{-1/p}\log(n)^{-2/p}$, $n\geqslant 2$. Since $|x(n)|^p=n^{-1}\log(n)^{-2}$, $x$ belongs to $\ell^p$. For $q\lt p$, we can show that $|x(n)|^q\geqslant n^{-\beta}$ for some $\beta\lt 1$ and each $n$ large enough. Indeed, $$|x(n)|^q=\frac 1{n^{q/p}(\log n)^{2q/p}}.$$ Now take $\beta\in (q/p,1)$. Then $$|x(n)|^q=\frac 1{n^\beta}\cdot ... 3 For 1<p<\infty, strict convexity of the norm implies that the only shortest path is the line segment between these points. For p=1, the length of a path (x(t),y(t)) is just the sum of the lengths of one-dimensional paths x(t) and y(t). Both of those must go from 0 to 1. They will have length 1 if and only if the function is ... 3 Let s integrable and \varepsilon such that s\leqslant|f| on X and \displaystyle\int_X|f|\leqslant\varepsilon+\int_Xs. Then, for every measurable E\subseteq X, |f|-s\geqslant0 on X\setminus E hence \displaystyle\varepsilon\geqslant\int_X|f|-s=\int_E|f|-s+\int_{X\setminus E}|f|-s\geqslant\int_E|f|-s , which implies ... 3 Forming the convolution of the (scaled) bump function with indicator functions you get "qausi-indicator functions" in \mathscr D, in particular, there are \psi_n\in\mathscr D(\mathbb R) which are positive and equl to 1 on [-n,n]. It is then easy to see that the elements of E^{loc} are those distributions which, on every compact set, have the same ... 2 Is false. For each n\in{\Bbb N},$$f_n(x)=\cases{ x^{-n/(n+1)} & for $x\in(0,1)$,\cr 0 & for $x\not\in(0,1)$.\cr }$$Is easy to see that f_n\in L^p for 1\le p < (n+1)/n and f_n\not\in L^p for  p\ge(n+1)/n. The function$$f(x) = \sum_{n=1}^\infty{f_n(x-n)\over 2^n||f_n||_{L^1}}$$is in L^1\setminus L^p for all ... 2 In example 2, page 52, Weidmann does not consider multiplication operators, but the functional T \colon L^2(\mathbb R) \to \mathbb C$$ Tf = \int_{\mathbb R} \psi(x)f(x) \, dx which of course is defined on the whole of L^2(\mathbb R) for \psi \in L^2(\mathbb R). It's just the scalar product with \bar\psi. The multiplication operator is defined on ... 2 Instead of (0,T)\times\Omega work on I=[0,1]. Let \begin{align*} f_n(x)=\begin{cases} x^{-2} & x\in[1/n;1]\\ 0 & x\in [0;1/n] \end{cases} \end{align*} Then |\{|f_n|>k\}|\leq |\{x^{-2}>k\}|=k^{-1/2} but \lim_{n\to\infty}\int_0^1f_n(x)dx=\infty. 2 And another proof: Let \epsilon>0 and choose a large enough so that \|f 1_{[a,\infty)} \|_p < {\epsilon \over 2}. Now choose L \ge a large enough so that |\frac{1}{x^{1-\frac1{p}}}\int_0^a f(t)\,dt | < {\epsilon \over 2} whenever x \ge L. Then we have |\frac{1}{x^{1-\frac1{p}}}\int_0^x f(t)\,dt | = |\frac{1}{x^{1-\frac1{p}}}\int_0^a ... 2 Your counterexample is correct. You have in addition to assume that \nu is also \sigma-finite (and apply Radon-Nidokym-Lebesgue, as you say correctly). Folland refers to the example of conditional expectation from probability theory, where \mu is a (probability and hence) finite[!] measure. In this case of course, \nu is also finite (and hence, ... 2 Define f:[0,2\pi]\to\mathbb{C} as:f(\theta)=\sum_{n=1}^{+\infty}x_n e^{ni\theta}.$$Then f is a L_2-function with zero mean over (0,2\pi), and we have:$$\|f\|_1 \leq \sqrt{2\pi}\,\|f\|_2 \tag{1}$$due to the Cauchy-Schwarz inequality and$$\|f\|_2 \leq \|f'\|_2\tag{2}$$due to Wirtinger's inequality, or just Parseval's identity. (1) and (2) ... 1 First, let a\in\mathbb{R} be arbitrary. As you noted, this implies$$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\lim_{n\rightarrow\infty}\left\Vert f-a\right\Vert _{L^{p}\left(\Omega_{n}\right)}=\left\Vert f-a\right\Vert _{L^{p}\left(\bigcup\Omega_{n}\right)}, $$and thus$$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert ...

1

My answer: only 2 is true. It is not closed as $\lim e_n$ does not exist. It is bounded as $||e_n||=1$ It is not compact as one can not find a finite sub-cover containing S. Neither does it contain a convergent subsequence as $\lim e_{n_k}$ does not exist.

1

Every pair of orthonormal vectors $e_i\neq e_j$ has distance precisely $d(e_i,e_j)=\sqrt{2}$. So $S$ is closed as it contains only isolated points, it is bounded as all its elements have norm one, it is not compact as it contains infinitely many isolated points, it contains a convergent subsequence the constant ones?!

1

Hint: Use the Uniform boundedness Principle to show that $T$ is bounded, and the use a)

1

This is the kernel of a continuous linear functional, hence closed. We have $|I(f_n)-I(f)|\leq\|f_n-f\|$, if $I(f_n)$ are all $0$ so is $I(f)$.

1

Promoting comment to answer: Both parts (a) and (b) are about $\|(x_n)\|$. In the first, you should calculate this norm and find when it's bounded. In the second, when it goes to zero. Keep in mind that these are sequences, not series. The sequence $n^p$ is bounded iff $p\le 0$, and goes to zero iff $p<0$. Your "$<-1$" comes from ...

1

Suggestions: Recall that every weakly convergent sequence is bounded in the norm. Think of what it means for $n_{k}u_{k}$ to be bounded in the norm for every sequence $n_k$ Conclude that your sequence $u_k$ is very special.

1

Because $L^{p}$ spaces expose the subtle nature of arguments. You have reflexive, non-reflexive, separable, non-separable, algebra, Hilbert, Banach, etc.. And, interpolation works between such spaces because of the exponent. They're good spaces for testing conjectures. They're the original spaces that firmly established the need to separate a space from its ...

1

To partially address 2.: In $L_1(X,\mu)$ with $\mu(X)<\infty$, a weakly convergent sequence is uniformly integrable. This follows, for instance, from the Dunford-Pettis theorem. See Theorem 3 from this paper (which is certainly worthy of careful study) for the statement and proof of the Dunford-Pettis theorem. So, if $(f_n)$ is weakly convergent and ...

1

Yes, it is correct. Maybe some steps could be justified: for example, write: "Using Cauchy-Schwarz inequality, we have for each integer $n$, $$\tag{*}\left(\int_A f_nf\right)^2\leqslant \left(\int_Af_n^2\right)\left(\int_Af^2\right).$$ On one hand, since $f_n\to f$ weakly in $L^2$ and $\chi_Af\in L^2$, the LHS of (*) converges to $\left(\int_A ... 1 By one hand, if$\Omega\subset \mathbb{R}^N$is any open set then,$C_0^\infty(\Omega)$is dense in$L^p(\Omega)$, as you can see, for example, in Brezis chapter 4. Once $$C_0^\infty(\Omega)\subset C^\infty(\Omega)\cap L^p(\Omega)\subset L^p(\Omega),$$ the result follows. On the other hand, if$-\infty<a<b<\infty$, you can see in chapter 8 of ... 1 You can make things easy and approximate $$\Vert f_n- f_m\Vert_2^2\le 4\cdot |1/(2n)-1/(2m)|$$ Note$f_n-f_m$is$0$of the interval with endpoints$1/(2n)+1/2$and$1/(2m)+1/2$. -- David Mitra 1 An example which can be adapted as required by fiddling a little with the exponents occurring below. Let$I_0=[0,1)$,$I_n=[\sum_{k=1}^n \frac{1}{k^2}, \sum_{k=1}^{n+1} \frac{1}{k^2})$, which is of length$1/n^2$. Define$f(x)=1/n^{1/2}$on the$I_n$'s ($n\ge 1$) and$0$elsewhere.$f$is clearly$L^1$but not$L^p$if$p\ge 2$. Also,$f\equiv 0$on$[0,1). ... 1 Since everything is non-negative and measurable, we can write \begin{align} \int_\mathbb{R} \lvert f(x)\rvert^p\,dx &= \int_\mathbb{R} \int_0^{\lvert f(x)\rvert^p}\,dt\,dx\\ &= \iint_{0 \leqslant t \leqslant \lvert f(x)\rvert^p}\,dt\,dx\\ &= \int_0^\infty m\left(\{ x : \lvert f(x)\rvert \geqslant t^{1/p}\}\right)\,dt \end{align} for anyp ...

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