# Tag Info

0

Sure, for $\delta$ small enough, we have $|\xi_\delta| \leq \delta$ and thus $$\bigg|\delta \int_{-\xi_\delta}^{\xi_\delta} |v_\delta '|^2 \bigg| \leq \delta \int_{-\delta}^{\delta} |v_\delta '|^2 ,$$ which easily implies what you want.

0

You totally can, since $C$ is independent of $n$. Sometimes not doing so is illustratively useful, but you can.

1

Put $E_N = \{x \in \Omega: |h(x)|^p \geq 1/N\}$ for $N \in \mathbb{N}$. Observe that $E_N \subset E_{N+1}$, and $\Omega = \cup E_N$. For any $g \in L^p(\Omega)$, apply the Lebesgue's monotone convergence theorem to $\{|g|^p\chi_{E_N}\}$: $$\lim_{N\rightarrow \infty} \int_{E_N}|g|^pd\mu = \|g\|_p^p.$$ Given $\epsilon > 0$, pick $N$ such that ...

0

One approach is to show that every continuous function on $[0,1]$ is a uniform limit of linear combination of Haar functions. Suppose $\phi$ is continuous, $n$ is a positive integer, and let $$\psi = \sum_{j\le n, k}\langle \phi, f_{j,k}\rangle f_{j,k}$$ By construction, $\psi$ is a piecewise constant function; it is constant on subintervals of the form ...

0

It turns out the set $T_p$ goes by the name (superellipse) or Lamé curve. The parametrization is: $$x(\theta) = \pm \cos^{2/n} \theta \\ y(\theta) = \pm \sin^{2/n} \theta \\ \text{ for } 0 \le \theta < \frac{\pi}{2}$$

1

For $0 \le s \le t$, the selfadjointness of $\Delta$ gives \begin{align} \frac{\partial}{\partial s}(u(t-s),v(s)) & =-(u'(t-s),v(s))+(u(t-s),v'(s)) \\ & =-(\Delta u(t-s),v(s))+(u(t-s),\Delta v(s)) =0. \end{align} Therefore, $$(u(t-s),v(s)|_{s=0} = (u(t-s),v(s))|_{s=t} \\ (u(t),v_0) ... 1 The answer is "no", even for f measurable and bounded (or indeed, a characteristic function), and even just for convergence almost everywhere (or indeed, anywhere). This is the main result in: Walter Rudin, "An Arithmetic Property of Riemann Sums", Proc. Amer. Math. Soc. 15 (1964), 321–324. 1$$ X_{n+1} = X_n + T_{n+1} $$where T_{n+1} is independent of X_1,\ldots,X_n and$$ \Pr(T_{n+1}\in A) = \int_A \frac{dt}{(2+t^2)^{3/2}} $$for every measurable set A. (Thus T_{n+1} has a t-distribution with 2 degrees of freedom.) 2 Consider first the functions$$h_1(x)=\frac{|\pi-x|}{2},\,\,h_2(x)=-2\ln\left|\sin\frac{x}{2}\right|.$$It is shown here that the Fourier series of h_1 on (-\pi,\pi) is \sum_{j=1}^{\infty}\frac{\sin jx}{j}, and the Fourier series for h_2 on (-\pi,\pi) is \sum_{j=1}^{\infty}\frac{\cos jx}{j}. Therefore, the Fourier series of the function \sin ... 1 It is true that \|\psi_{n}\star f-f\|_{L^2}\rightarrow 0 and n\rightarrow \infty for any f \in L^2. If you start with the wedge f as described, then \|\psi_{n}\star f - f\|\rightarrow 0 and$$ \|(\psi_{n}\star f)'-f'\|=\|\psi_{n}'\star f-f'\|=\|\psi_{n}\star f'-f'\|\rightarrow 0. $$This is because, even though f has only a piecewise ... 0 Let C = \sup_{n\in \mathbb N} \|u_n\|_\infty. Let \epsilon >0 and \Omega' \subset\subset \Omega so that L^n(\Omega\setminus \Omega') <\epsilon. Then$$\|u_n\|^2_{L^2(\Omega)} \le \|u_n\|^2_{L^2(\Omega')} + C^2\epsilon.$$Given that u_n \to u strongly in \Omega', we have$$ \begin{split} \limsup_{n\to \infty} \|u_n\|^2_{L^2(\Omega)} ...

1

We will use the following inequality: $$\forall a,b\ge 0, s\ge 1: \, (a+b)^s\leq 2^{s-1}(a^s+b^s)$$ This is easily proved by noting that the function $f(t)=t^s$ is convex for $t\ge 0$ and therefore you have $$f(\frac{1}{2}a+\frac{1}{2}b)\leq \frac{1}{2}f(a)+\frac{1}{2}f(b)\Leftrightarrow \big(\frac{a+b}{2}\big)^s\leq \frac{1}{2}a^s+\frac{1}{2}b^s$$ ...

0

By definition weak* convergence of $f_k$ to $f$ in $X=L^{\infty}(\mathbb R)$ means: $$lim_{k \to \infty}\langle f^*,f_k\rangle=\langle f^*,f \rangle \space (\forall f^*\in X^*)$$ This is a convergent sequence of reals, hence we have $$sup_{k \in \mathbb N}\Vert \langle f^*,f_k \rangle \Vert = sup_{k \in \mathbb N} \Vert \iota(f_k)(f^*)\Vert < \infty ... 0 The statement in the title is true, since by the second condition the only sequence that applies is the zero sequence. However, the title seems unrelated to the question.$$ \left|\sum^∞_{k=1}(x_n)_k(x_m)_k\right|\le\|x_n\|·\|x_m\| $$per the Cauchy-Schwarz inequality. For any scalar product and its norm, the CSI follows from$$ 0\le\Bigl\|\;\|v\|·u ± ...

3

We have to assume $\mathbb{E}(|X|^r)<\infty$; otherwise the expession $\|X_n-X\|_{L^r}$ might not even be finite. Note that $$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$ for any $\epsilon>0$ and $n \in \mathbb{N}$. Obviously, $$\int_{|X_n-X| \leq ... 1 (i) We can even find some f \in L^1[0,\infty) that is continuous and does not satisfy \lim\limits_{x\rightarrow\infty} f(x)=0 Consider \phi a positive bump function and still continuous, with support in [-1/2,1/2] and \int_\mathbb{R} \phi(x) dx=1. We can also assume \phi(0)=1. Then \int_\mathbb{R} \phi(nx) dx=n for 1/n \in \mathbb{N} and ... 1 For (i): Take an integrable bump function \phi and form$$f(x) = \sum_{k=0}^\infty \phi(2^k(x-k)).$$1 Counterexamples: (i)$$f(x)=\begin{cases}e^{-x},& x\notin \mathbb N\\1,&\text{otherwise}\end{cases}$$(ii)$$f(x)=\begin{cases}{1\over x},& x\neq 0\\1,&\text{otherwise}\end{cases}$$1 Let M=\|u\|_{\infty}. Since G' is continuous, there exists a constant K such that |G'(s)|\leq K for all s\in [-M,M]. Thus |G'\circ u|\leq K; it follows that |G'\circ u|^p is bounded by K^p. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ... 1 The best intuition I can get for your result comes from one dimension. Think of u as being 1 on an interval (a,b) and then diminishing down to 0 outside a bigger interval (c,d)\supset (a,b). As you exponentiate u to higher and higher powers, it becomes dominated by what happens on (a,b) and goes to zero outside (a,b). Also, no matter how ... 1 This seems to a question about Maple programming rather than math, and as such is better suited to stackoverflow.com or mapleprimes.com and not here. Some things you could try: restart: N := 10: P := [seq(plots:-implicitplot( abs(x)^p+abs(y)^p=1, x=-1.1..1.1, y=-1.1..1.1, gridrefine=2, ... 1 If f=\infty on a set of positive measure, say E, then$$\int |f|^p \ge \int |f|^p 1_E =\infty.$$|\int f|\le \int|f| =0 because |f|=0 \mu-a.e. and for g\in L^+ this can be shown using the definition of the integral (i.e., \sup\{\int \phi : \phi \text{ is simple s.t. }0\le\phi\le g\}). Also the converse is not true in general (only for ... 1 Let E_n=\{x:|f|>n, \:n\in \Bbb{N}\}. Then by Chebyshev's inequality$$ \mu(E_n)\leqslant \frac1{n^p}\int_{X}|f|^pd\mu\quad\text{and }\quad\lim_{n\to\infty}\mu(E_n)=0 $$Let E=\bigcap_{n=1}^{\infty}E_n. Then E=\{x:|f|=\infty\}. Since E_1\supset\cdots\supset E_n\supset\cdots\supset E, by Monotone class theorem$$ \mu(E)=\lim_{n\to\infty}\mu(E_n)=0 ...

1

Expanding on Did's comment, let $\epsilon > 0$ and let $\delta > 0$ be such that if $1 - \delta < x < 1$ then $|f(x) - a| < \epsilon$. Notice that $\int_0^1kx^k\,dx = 1$. Then \begin{align} \Big|\int_0^1kx^kf(x)\,dx - a\Big| = &\ \Big|\int_0^1kx^k(f(x) - a)\,dx\Big|\\ \le &\ \int_0^{1 - \delta}kx^k|f(x) - a|\,dx + \int_{1 - ...

0

As pointed out in the comments, if $\left|f(x)\right|\leqslant C$ for almost every $x$, then using increasingness of the map $t\mapsto t^p$ for non-negative $t$, we have $\left|f(x)\right|^p\leqslant C^p$ hence $\left|f\right|^p$ satisfies the definition of $L^\infty(\Omega)$.

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As pointed out in the comments, the attempt is good. Furthermore, the argument works for $\mathbb L^p$ where $1\leqslant p<+\infty$.

1

What we need is the following: if $(f_\delta)_{\delta\in (0,1]}$ is a family of function indexed by a real number $\delta$ such that the function $t\mapsto \sup_{\delta\in(0,1]}\left|f_\delta(t)\right|$ is integrable and for each $t$, $\lim_{\delta\to 0}f_\delta(t)=0$, then $\lim_{\delta\to 0}\int_{\mathbb R}f_\delta(t)\mathrm d\lambda(t)=0$. To prove ...

1

We have to prove finiteness of $$I:=\int_{[0,+\infty)}\int_{[0,+\infty)}|f(t)|e^{x-t}\mathbf 1\{t\geqslant x\}\mathrm dm(t)\mathrm dm(x).$$ Using Fubini-Tonnelli's theorem (that is, switching the integral when the integrand is a non-negative function), we infer that $$I=\int_{[0,+\infty)}e^{-t}|f(t)|\int_{[0,t)}e^x\mathrm dm(x)\mathrm dm(t).$$

1

For (a): Note that $$\int_I|f_n-f|^2\le \|f_n-f\|_\infty^2 \lambda(I),$$ where $\lambda(I)$ is the length of the interval. For (b): Note that $$\int_I|f_n-f|=\int_I(|f_n-f|\cdot 1)\le \|f_n-f\|_2 \lambda(I)^{1/2}.$$

1

The answer is no. Take $\Omega_1 = \Omega_2 = \mathbb{R}$ with the usual Lebesgue measure and set $f(x_1,x_2) = e^{-(x_1-x_2)^2}$. Then $\|f(\cdot, x_2)\|_{L^1(\Omega_1)} = \|f(x_1, \cdot)\|_{L^1(\Omega_2)} = const. < \infty$ for all $x_1, x_2$ but $f \notin L^1(\mathbb{R}^2)$.

0

The left hand side is the essential supremum of $|f_n-f|$ on $A,$ i.e., the lowest essential upper bound of $|f_n-f|$ on $A.$ The right hand side is the (ordinary) supremum of $|f_n-f|$ on $A,$ i.e., the lowest upper bound of $|f_n-f|$ on $A.$ The inequality holds because every upper bound is an essential upper bound: the set where it does not hold is the ...

2

I wouldn't know about the proof in the book, but here's a proof. It could probably be streamlined some - you should see what it looked like a few days ago. Going to change some of the notation; this is going to be enough typing as it is. Going to assume we're talking about real-valued functions, so that for every $f$ there exists $E$ with $\left|\int_E ... 2 By assumption$p\neq q$. Case 1:$p,q\in(1,+\infty)$. Assume we have a surjection$T:\ell_p\to\ell_q$, then$T^*:\ell_{q'}\to \ell_{p'}$is an embedding. Since$p',q'\in(1,+\infty)$, we get a contradiction because by corollary of Pitt's theorem theses spaces are totally incomparable. Thus for this case a desired surjection doesn't exists. Case 2:$q=1, ...

0

In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence. If each component of $u$ is weakly ...

1

I would rather say that the opposite is true, and that it comes from Jensen's inequality: since $x \mapsto x^{1/p}$ is concave for $p>1$, one has: $$\left( \int g \right)^{1/p} \geq \int \left(g^{1/p}\right).$$ Then you apply this result to the function $g = \sum_k \vert f \vert^p$.

2

If $(\Omega,\mathcal{T},\mu)$ is a measured space such that $\mu(\Omega)<\infty$ then you have $$\forall (p,q)\in[1,\infty], p\leq q\Rightarrow L^q\subset L^p \textrm{ (with continuous injections)}$$ This doesn't apply here as $\mu(\mathbb{N})=\infty$ if $\mu$ is the usual counting measure. However, denoting ...

3

Your conclusions about the function $f$ are correct. The inclusions are generally false (they are true for finite measure spaces).

0

In physics vector-valued $L^{2}$-functions occur frequently. As an example let $f$ be a function from $\mathbb{R}^{n}$ into the vector space $V$ with $% |.|_{V}$ the norm on $V$. Then we denote the $L^{p}$-space as $L^{p}(\mathbb{% R}^{n},V,d^{n}x)$ or something similar. Now \begin{equation*} \parallel f\parallel _{p}^{p}=\int d^{n}x|f(x)|_{V}^{p} ...

1

I guess that by "$u$ is a vector", you mean "$u$ is a vector-valued function", right? In this case, indeed the definition would be to use the definition of $l^p$ norm inside the integral term.

2

Since $f_n\xrightarrow{L^1(\Omega)} f$, there is a subsequence $(f_{n_k})$ of $(f_n)$ such that $f_{n_k} \rightarrow f$ pointwise a.e. on $\Omega$. By Fatou's lemma, $$\|f\|_{L^2}^2 \le \varliminf_{k\to \infty} \|f_{n_k}\|_{L^2}^2\le \varliminf_{k\to \infty} M = M.$$

1

In general, this is false (see below), but if $(f_n)_n$ is bounded,it is true, since we have for every $f \in L^2$ that $$\langle f_n , f\rangle = \langle f_n g_n , f\rangle + \langle (f_n (1-g_n), f\rangle ,$$ where the first summand on the right converges to $\langle h,f\rangle$ and where $$|\langle f_n (1-g_n), f\rangle \leq \|f_n\|_2 \|(1-g_n)f\|_2 \to ... 1 Since on \mathbb{R}^n any two norms are equivalent there is a constant C such that$$||v||_2\le C\sum_{i=1}^n|v_i|$$for any v=(v_1\dots,v_n). (so, yes). 0 It's true for any open \Omega: you have to replace u by the function$$ \tilde{u}(x) = \begin{cases} u(x) & x \in \Omega \\ 0 & \text{else} \end{cases} $$Then you get bounds on the L^p(\Omega)-norms from the L^p(\mathbb{R}^n)-norms, and exactly the same proof applies. See, for example, p.64ff. of Lieb and Loss, Analysis, and particularly ... 1 You can use spherical coordinates. In this way, the terms in \cos \theta_i and \sin\theta_i only appear in the numerator (the denominator depends only on r). Consequently, the considered integral is convergent if and only if I:=\int_0^{+\infty}r^{n-1+2|\alpha|}/(1+r^2)^{2k}\mathrm dr converges. Since the only problem is at infinity, I is convergent ... 0 Define for each n \in \mathbb{N}$$I_n(x) := \int_a^b x(t) y(t) 1_{\{|y| \leq n\}}(t) \, dt.$$Then I_n:L^1 \to \mathbb{C} is a linear operator and, since$$|I_n(x)| \leq n \|x\|_{L^1},$$the operator is also continuous. Moreover, since by assumption x \cdot y \in L^1 for any x \in L^1, it follows directly from the dominated convergence theorem ... 1 Rewrite the inequality that you want to prove as$$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$Set v:=|u|^{\frac{p+m-1}{2}}. The LHS above becomes$$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}$$with q=\frac{2p}{p+m-1} and \alpha=\frac{p}{p+m-1}. Apply the NGN ... 1 By decomposing f into real part and imaginary part and notice that$$|\int_X \mathrm{Re}(f)1_E\,d\mu|\leq |\int_Xf1_E\,d\mu|,$$we may assume that f is real. By decomposing f into positive part and negative part and notice that$$|\int_X f^+1_E\,d\mu|=|\int_X f1_{f\geq 0}1_E|\leq C\mu(\{f\geq 0\}\cap E)^{1-1/p}\leq C\mu(E)^{1-1/p},$$we may assume ... 1 Note that$$\int_X|f|=\int_X\chi_E|f|+\int_X\chi_{E^c}|f|.$$1 Consider that if a_n<p for each n with \lim_{n\to \infty}a_n=0, and x=(x_n)_{n\in N}=(n^{-1/(p-a_n)})_{n\in N} then x\notin l_q for any q<p. I tried putting (x_n)^p=n^{-1}(\log n)^{-2} for all sufficienly large n, to get x\in lp. For n>1, I got a_n=p y_n/(1+y_n) where y_n=2(\log (\log n))/\log n. And n\geq 4\implies ... 1 Hint: Think about$$\sum_{n=2}^{\infty}\frac{1}{n^a (\ln)^b} for $a, b> 0.$ This diverges for all $a, 0< a <1,$ but converges for $a= 1,b>1.$

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