# Tag Info

1

divide the interval [0,1] to N subintervals and take the function $F_N$ as below: in odd subintervals equal 1 in even subintervals equal 0

2

First, $A\subset l^2$ is the collection of sequences that only take finitely many values, their entries are all zero after some point (you said constant, but the constant must be zero), and $A$ is dense in $l^2$. Now let us show $B$ is dense in $A$ (which would mean $B$ is also dense in $l^2$), let $a\in A$ and $\epsilon$ be given, we show there exists a ...

1

The span $S$ of the functions $\{ \chi_{[0,x]} : 0 \le x \le 1\}$ is a dense linear subspace of the Hilbert space $L^{2}[0,1]$. To see that the span is dense, it is enough to show that $(f,\chi_{[0,x]})=0$ for all $x$ and some $f \in L^{2}[0,1]$ implies $f=0$. By the Lebesgue differentiation theorem, the following holds a.e.: $$... 1 Alternatively and directly: Assume by contradiction that (e_n)_{n} does not converge weakly to zero. Then there exists a \epsilon >0 and a functional f\in l_\infty^\ast s.t. |f(e_n)|\geq \epsilon for infinitely many n\in\mathbb{N}. By passing to that subsequence, we have that |f(e_{n_k})|\geq \epsilon for all k\in\mathbb{N}. Let \lambda_k ... 0 From here we know that (e_n)_n converges weakly to 0 iff it is bounded and every subsequence converges quasi-uniformly to 0. Clearly, (e_n)_n is bounded and pointwise converging to 0. Given a subsequence (e_{n_k})_k, \epsilon>0 and n_0\in \mathbb{N}. Set \alpha_1 = {n_0+1} and \alpha_2={n_0+2}. Then for any m\in\mathbb N we have ... 2 Think of a special case: f and g are zero except on a small interval, like "two copies of the box-function, which is 1 on [-1/2, 1/2] and 0 elsewhere". Then f \star g(u) is, roughly, "how much f looks like g, flipped over and shifted by u. In the case where g is symmetric (i.e., an even function), the "flip over" can be ignored, and ... 5 Because there is a maximum (well, supremum) in its definition. Answered here: I want to show that |f(x)|\le(Mf)(x)| at every Lebesgue point of f if f\in L^1(R^k) The important properties are: M is a bounded (nonlinear) operator on L^p for 1<p<\infty, and also from L^1 to weak L^1. This fact offers more control over the local behavior ... 1 Let f_{\alpha,\beta}(x) = \frac{1}{x^\alpha (\log x)^\beta}. You have shown that f_{\alpha,\beta} \in L^p([2, \infty)) whenever: p > 1/\alpha, or p = 1/\alpha and \beta > \alpha. So, in particular, \{p : f_{1/2,1} \in L^p([2,\infty))\}=[2, \infty]. You have also shown that f_{\alpha,\beta} \in L^p([0,1/2]) whenever: p < ... 1 L^p (\mu) is always reflexive for 1<p<\infty. EDIT: For the cases p=1 or p=\infty, this is almost never true. But what is still true in the case p=1 (if the measure is sigma finite) is that the dual space of L^1 is L^\infty. In the non sigma finite case, this can fail. If I recall correctly, Rudin even proves it for non sigma-finite ... 1 Hint: For the first case, you can actually show that$$||f||_p \leq ||f||_\infty \mu(X)^{1/p}$$if \mu(X) < \infty. For the second one, consider X = \mathbb R with the Lebesque measure. Can you found a bounded positive function f on \mathbb R so that$$\int_\mathbb R f dx = + \infty?$$(Don't think too hard) 0 Hint Can you bound |(x_n)_j - x_j| \le \|x_n - x\|_p? Take the canonical basis as a counter example for b). (what is it's point-wise limit? What is \|i^p e_i - e\|_p?) 0 Your statement in the comments that an operator T is continuous on a Banach space X if, for a sequence f_n \rightarrow f in X then we have Tf_n \rightarrow Tf. That is about the operator T being continuous. The functions f_n and f are simply an arbitrary collection of elements in the space that form a convergent sequence and its limit. ... 0 First of all, in the proof you need to assume something else about the function g. Usually one takes g\in C_c(\mathbb{R}^n), the space of continuous functions with compact support. This implies that g is uniformly continuous, allowing one to prove that \|g(x-a)-g(x)\|_p\to0 as a\to0. The proof shows that$$ ...

2

A square-integrable function on $\mathbb{R}^{n}$ is not necessarily integrable, but it is integrable on any bounded set $S$ because of the Cauchy-Schwarz inequality: \begin{align} \int_{S} |e^{-2\pi i(x,y)}f(x)|\,dx & = \int_{S}|f(x)|\,dx \\ & \le \left(\int_{S}1\,dx\right)^{1/2}\left(\int_{S}|f(x)|^{2}\,dx\right)^{1/2} \\ & \le ... 1 The sub. u=\tan(x) gives\int_{0}^{\infty}{\frac{u^{p}}{1+u^{2}}du}=\int_{0}^{1}{\frac{u^{p}+u^{-p}}{u^{2}+1}du}\leq \int_{0}^{1}{(u^{p}+u^{-p})du}=\frac{2}{1-p^{2}}$$whenever p\in (0,1) 3 Hint:$$\mu(|f| > \varrho) = \int_{|f|> \varrho} \, d\mu \leq \int_{|f|>\varrho} \frac{|f|}{\varrho} \, d\mu.$$This inequality is known as Markov's inequality. Remark: For f \in L^p(\mu), p \geq 1,$$\mu(|f|>\varrho) \leq \frac{1}{\varrho^p} \int |f|^p \, d\mu.$$2 For the first part, we can actually write$$\sum_{j=1}^N|x_j|^p=\lim_{n\to \infty}\sum_{j=1}^N|x_j^n|^p\leqslant \sup_l\lVert x^l\rVert_p^p.$$As N is arbitrary, we get that x belongs to \ell^p. For the second part, we approximate the element z by the sequence whose N first terms are the corresponding to z, and the other ones are 0. Call ... 0 These inequalities do not seem to be the best ones to me. Why don't you try this: Note that the E_n simply divide the domain of E_n according to the integer part of |f|. In E_n, f is "sandwiched" between n-1 and n. Writing this with functions:$$(n-1)\chi_{E_n}\leq|f|\chi_{E_n}\leq n\chi_{E_n}.$$We can take the power p on the inequalities ... 2 Hint 1: If you do not have sets of arbitrarily small measure, one of the inclusions L^r \subset L^s holds. For instance, consider \sum_{n=1}^\infty \frac{1}{n} and \sum_{n=1}^\infty \frac{1}{n^2}. How does this relate to \left(\int_X |f(x)|^p d\mu(x)\right)^{1/p}? Hint 2: If you do not have sets of arbitrarily large measure, the other inclusion ... 2 If \mu(X)<\infty, then$$ p<q \quad\Longrightarrow\quad L^q(X)\subset L^p(X). $$This is due to the fact that (Holder inequality)$$ \int_X \lvert\, f\rvert^p\,d\mu=\int_X \lvert\, f\rvert^p\cdot 1\,d\mu =\left(\int_X \lvert\, f\rvert^q\,d\mu\right)^{p/q} \left(\int_X 1^{q/(q-p)}\,d\mu\right)^{(q-p)/q}. $$Hence \|\,f\|_p\le ... 1$$ \sup_{|y|>|x|} \frac{1}{(1+|y|)^{n}} = \frac{1}{(1+|x|)^{n}}$$and$$\int_{\mathbb R} \frac{1}{(1+|x|)^{n}} dx = 2\int_0^\infty\frac{1}{(1+|x|)^{n}} dx \leq 2\left(\int_0^1 1 dx +\int_1^\infty\dfrac{1}{x^n}dx\right) $$1 Obviously,$$\int_{\mathbb R} \sup_{|y|>|x|} \frac{1}{(1+|y|)^{n}} dx = \int_{\mathbb R} \frac{1}{(1+|x|)^{n}}dx=\cdots$$0 Let$$ f_n(x)= \begin{cases} n^2&\text{for $0\leq x\leq\frac1n$},\\ 0&\text{for $\frac1n< x\leq1$}. \end{cases} $$The sequence is unbounded, hence not weakly convergent, but convergent in the sense of 1 and 2. 1 Let f(x)=1/x^2 for |x|>1 and 1 for |x|\leq1. In the sense of your definition g(x)\equiv1 is a contraction of f, but is not in L^2(\mathbb{R}). 1 For the case in which 1 \leq p < \infty: Let \displaystyle f = \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A and \displaystyle g = \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B where A, B both have nonzero finite measure, are disjoint and \chi is the indicator function. Notice that$$\|f\|_p = \bigg( \int \vert f \vert^p \, d\mu ...

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For the first inequality, please use Lemma 4.1 in ref [X. X. Huang and X. Q. Yang. A unified augmented Lagrangian approach to duality and exact penalization. Mathematics of Operations Research, 28(3):533–552, 2003.] Note that $0<r\,/q<1$.

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1) No, every such operator is bounded. For simplicity, I assume that all functions are real-valued. If $T$ is not bounded, there is a sequence $f_n$ with $\Vert f_n \Vert_p \leq 2^{-n}$ and $\Vert Tf_n\Vert_p \geq 2^n$ (why?). Then Fatou's Lemma (or monotone convergence) shows $f :=\sum_n |f_n|\in L^p$ with $\Vert f\Vert_p \leq 1$. But $|f_n|\leq ... 2 How to find these... By linearity of the integral one has: $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|^p\mathrm{d}\lambda=\int|f|^p\mathrm{d}\lambda+\int|g|^p\mathrm{d}\lambda$$ So take a block and shift it slightly: $$f:=\chi_{[0,1]}:\quad f_\varepsilon(x):=f(x-\varepsilon)$$ Then for an appropriate choice: ... 1 I'll give you a hint. Let $$f(x)=\begin{cases}2,x\in[0,3/4],\\0,x>3/4,\end{cases}\quad g(x)=f(1-x).$$ Then the parallelogram law says that in hilbert spaces we have $$2\|f\|^2+2\|g\|^2=\|f-g\|^2+\|f+g\|^2.$$ Can you calculate the norms above? 1 Suggestion:$E$is a set where$f$is "large". When$f > 1$,$|f|^p$increases and decreases in direct relation to$p$. So if you know$f$integrates at the$q$th power, what other types of powers will be integrable on this set? Likewise when$|f| < 1$we are talking about the "tail" of the integral if$\Omega$has infinite measure, or simply the ... 0 There are many norms. This is but an example. Let$W=\{w_i\}_{i=1}^\infty$,$w_i>0$, be a weight-sequence. Then $$\|x\|_{p,W}=\sum_{í=1}^\infty|x_i|^pw_i$$ defines a norm. If$W$is bounded, then it is a norm on$\ell^p$. If moreover$\inf_{i} w_i>0$, then it is equivalent to the$\ell^p$norm. But if$\inf_{i}w_i=0$, the norms are not equivalent. 4 If$\,\mu(X)<\infty$, then the answer is YES. In such case $$\left|\int_X f_n\,d\mu-\int_X f\,d\mu\,\right|\le\int_X\lvert\,f_n-f\rvert\,d\mu\le\mu(X)\cdot\sup_{x\in X}\lvert\,f_n(x)-f(x)\rvert,$$ and as$\,f_n\to f$uniformly, then$\,\sup_{x\in X}\lvert\,f_n(x)-f(x)\rvert\to 0\$, and hence $$\int_X f_n\,d\mu\longrightarrow\int_X f\,d\mu.$$

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