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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 ...


-1

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 $0<r\,/q<1$.


1

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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