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8

It is certainly not dense. The linear functional $$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n} $$ Is continuous on $l^2$. Thus, the set $T^{-1}(0)$ must be a closed subspace of $l^2$. If it were dense, we'd have $l^2 = T^{-1}(0)$. But the sequence ${\frac{1}{n}}$ obviously isn't in this space.


6

First note that the sequence is bounded: $$ |\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty. $$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it ...


4

As felipeh noted, the problem reduces to $p=1$ by replacing $f$ with $g = |f|^p$. (The case $p=\infty$ should be treated separately.) Also, the annulus can be replaced by the sphere $S^{n-1}$. Indeed, define a function $h$ on the unit sphere by $h(\xi) = \int_r^1 g(t\xi)\,dt$. Then $h\in L^1(S^{n-1})$, with a norm comparable to $\|g\|_{L^1(A)}$ because ...


4

Do what you did. Except don't plug in the $L^2$ norm: $$\int_0^x|f'(t)|\,dt\le\sqrt x\left(\int_0^x|f'(t)|^2\,dt\right)^{1/2}.$$ Now (prove and) use the fact that $\int_0^x|f'|^2\to0$ as $x\to0$.


3

Normal Human already gave a good answer, but I came up with another one that I would like to share. It is based on the coarea formula and I tried to make it feel natural, with few arbitrary choices. As has been noted, it suffices to consider $L^1$ functions, so let $f\in L^1$. (For $f\in L^p$ we have $|f|^p\in L^1$.) This argument works for any ...


3

This is false. Consider $$\mu = \sum_{n=-\infty}^\infty \infty \delta_n + \lambda$$ where $\infty \delta_n$ is the measure that has infinite point mass at $n$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$. Let $\phi(n) = n,$ for $n\in \mathbb{Z}$ and $\phi(x)=1$ for $x$ not an integer. Then $f \in L^1(\mu)$ implies $f(n) = 0$ for all $n\in ...


3

Edit Here is a proof that works if $\mu$ is semifinite. Suppose $\phi \notin L_\infty(X, \mu).$ Then given any $R>0$ there is a set of positive measure $E \subset X$ such that $\vert \phi \vert > R$ on $E.$ Let $f \in L_1(X,\mu)$ such that $f$ vanishes outside of $E$. Taking absolute value, we can assume $f$ is real and non-negative. Then $$\Vert Mf ...


3

Here are some examples of relations between the spaces: If $p,q\in[1,\infty)$, there is a bijection $L^p\to L^q$, namely $L^p\ni f\mapsto |f|^{p/q-1}f\in L^q$. If $A$ has finite measure, then $p\geq q$ implies $L^p(A)\subset L^q$. If $f\in L^p$ and $g\in L^q$ so that $1/p+1/q=1/s$ (assuming $p,q,s\in[1,\infty]$), then the pointwise product $fg$ is in ...


3

Corollary 3 of Chapter 7 of Royden: If $E$ is measurable with finite measure and $1\leq p_1< p_2\leq \infty$, then $L^{p_2}(E)\subseteq L^{p_1}(E)$. Furthermore, $||f||_{p_1}\leq c||f||_{p_2}$ for all $f\in L^{p_2}(E)$ where $c=[m(E)]^{\frac{p_2-p_1}{p_1p_2}}$ if $p_2<\infty$ and $c=[m(E)] ^{\frac{1}{p_1}}$ if $p_2=\infty$. Royden remarks in an ...


3

To derive Nash's inequality from $$\|f\|_2^2\leq 2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2\tag{1}$$ you should choose the value of $\lambda$ that minimizes the right hand side. To this end, consider the function $$g(\lambda)=2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2$$ of a real variable $\lambda$. Then ...


3

In fact $\sin (n_kx)$ diverges a.e. for every subsequence $n_k.$ Proof: Fix a subsequence $n_k$ and let $E= \{x\in \mathbb {R}: \lim_{k\to\infty}\sin (n_kx)\,\,\text {exists}\}.$ Let $f$ be the pointwise limit function on $E.$ Let $a> 0$ and put $E_a=E\cap [-a,a].$ On $E_a$ we have $f\in L^1\cap L^2.$ By the DCT, $$\int_{E_a}f^2 = \lim_{k\to\infty} ...


2

A different proof. If $\{\sin(n_k\,x)\}$ converges a.e., then $$ (\sin(n_{k+1}\,x)-\sin(n_k\,x))^2 $$ converges to $0$ a.e. By the dominated convergence theorem $$ \lim_{k\to\infty}\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=0, $$ but $$ \int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=1. $$


2

For convenience, scale $\phi$ so that $\|M_{\phi}\|_{\mathcal{L}(X)}=1$. For $\delta >0$, the $\chi_{\delta}$ be the characteristic function of the set where $|\phi| \ge 1+\delta$. Then, for any $f \in L^{1}$, $$ (1+\delta)\int |f\chi_{\delta}|d\mu \le \int |f\chi_{\delta}| |\phi|d\mu \le \int |f\chi_{\delta}|d\mu. $$ Thus $\|f\chi_{\delta}\|=0$ ...


2

The ordinary heat equation is $$ \frac{\partial F}{\partial t}=\frac{\partial^{2}F}{\partial x^{2}},\\ F(0,x)=f(x). $$ The initial heat distribution is $f$. The ordinary heat kernel is the Guassian, and the resulting time evolution operator $T(t)=e^{tL}$ is a constractive $C_0$ semigroup on every ...


2

The functions $g_m(t) = H_m(t)\,e^{-t^2/2} $ just give an orthogonal base of $L^2(\mathbb{R})$, since: $$ \int_{-\infty}^{+\infty} f_m(t)^2\,dt = 2^m m! \sqrt{\pi}. $$ An orthonormal base is given by: $$ f_m(x) = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}}\,H_m(x)\, e^{-x^2/2}. $$ We may notice that if $m$ is odd then $\int_\mathbb{R}f_m(x)\,dx = 0$, while: $$ ...


2

Here are some interesting facts about the relations between different $L^p$-spaces over the same measure space $(X,\Sigma,\mu)$ (based on Section 6.1 of Folland, 1999): If $0<p<q<r\leq\infty$ and if $f\in L^q$, then there exist $g\in L^p$ and $h\in L^r$ such that $f=g+h$. If $0<p<q<r\leq\infty$, then $L^p\cap L^r\subseteq L^q$. If ...


2

HINT: The Cauchy-Schwarz Inequality reveals that $$\begin{align} \left|Af(x)\right|^2 &= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\ &\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy \end{align}$$ And thus, the square of the operator norm is $$\begin{align} ||A||_2^2&=\sup_{f\in \mathscr{L}^2} ...


2

You know (why?) that $L^2 \neq L^{3/2}$, so choose an $f \in L^{3/2}\setminus L^2$, and consider the map $$ L^3 \mapsto \mathbb{C} \text{ given by } g \mapsto \int fgd\mu $$ This is a bounded linear map. Suppose it is the restriction of a bounded linear map on $L^2$, then $\exists h \in L^2$ such that $$ \int fgd\mu = \int hg d\mu \quad\forall g\in L^3 $$ ...


2

Given a measure space $(X,\mathcal M,\mu)$, we define $$L^1(X) = \left\{f : X\to\mathbb C \;\mid\; \int_X |f|\mathsf d\mu < \infty\right\}$$ and $$\mathcal L^1(X) = L^1(X) / \sim $$ where $\sim$ is the equivalence relation on the bracketed set defined by $f\sim g$ iff $f=g$ a.e. That is, $$\mu(\{x\in X : f(x)\ne g(x)\}) = 0. $$ In other words, we identify ...


2

If by “die down,” you mean “converge to $0$”, then the answer is no. Consider the following function: \begin{align*} f(x)\equiv \begin{cases} 0&\text{if $x\in[0,1)$,}\\ 1&\text{if $x\in[1,1+1/1^2)$,}\\ 1&\text{if $x\in[2,2+1/2^2)$,}\\ 0&\text{if $x\in[2+1/2^2,3)$,}\\ 1&\text{if $x\in[3,3+1/3^2)$,}\\ 0&\text{if $x\in[3+1/3^2,4)$,}\\ ...


1

Hint: try the Radon-Nikodym theorem.


1

This is true for any $p\in[1,\infty)$. First, since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)$. Say $\epsilon>0$. Choose $K\subset\Omega$ with $K$ compact, so that if $g=f\chi_K$ then $$||f-g||_p<\epsilon.$$ Now if $\phi_n$ is a smooth approximate identity with compact support then the convolution $g*\phi_n$ is smooth, ...


1

you just need to show that characteristics of intervals can be approximated in $L^p$ norm, and then extend this result arbitrary measurable sets (using the regularity of the Lebesgue measure). Finally extend this result to simple functions. This is also Lusin Theorem. Whose main idea is to use Urysohn's functions. So the answer of your why question is ...


1

For any $s>1/2$ the norm of $H^s$ controls the $L^\infty$ norm. Indeed, when the series of Fourier coefficients is absolutely convergent, their sum bounds $\sup|u|$. By the Cauchy-Schwarz inequality, $$ \|u\|_{L^\infty}\le \sum_{n \in \mathbb{Z}} |\hat u_n| \le C(s,P) \left( \sum_{n \in \mathbb{Z}} \bigg(1 + \frac{4 \pi^2 n^2}{P^2}\bigg)^{s} ...


1

Actually, I made a mistake: The question is not an exercise in the book, but rather a complement and a rather famous result in functional analysis, known as the Dunford-Pettis theorem (see Uniform Integrability Wiki). The proof can be found in several textbooks and in a short research note here.


1

This is not quite what was asked for, but I thought it worth mentioning: Below is (the non-trivial) Proposition 11.1.9 in Kalton and Albiac's Topics in Banach Space Theory: $\ \ \ $(i) For $1\le p\le2$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $p\le q\le2$. $\ \ \ $(ii) For $2< p<\infty$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $q=2$ ...


1

By the inequality $|Af| \leq A|f|$, we may assume that $f$ is non-negative. Then from the Tonelli's theorem (a.k.a. Fubini's theorem for non-negative functions), $$\| Af \|_2^2 = \int_0^1 \int_0^1 f(y)f(z) \left( \int_0^1 \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}} \right) \, dydz $$ and we may try to estimate the following function $$ k(y, z) = \int_0^1 ...


1

After you use Holder, it is $\|f'\|_{L_2([0,x])}$ and it goes to $0$ when $x$ goes to $0$, because $f'$ is in $L_2([0,1])$.


1

We need to prove that $$ \sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty. $$ Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives $$ x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1 $$ Do $z$-transform of the equation, i.e. multiply by $z^{k+1}$ and ...



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