1
vote
0answers
35 views

Prove that $l^{\infty}(\mathbb{Z^+})$ is not separable.

Let $l^{\infty}(\mathbb{Z^+})$ be the set of all bounded complex functions on $\mathbb{Z^+}$. Then prove that $l^{\infty}(\mathbb{Z^+})$ is not separable. My attempt: Suppose $E\subset ...
1
vote
1answer
43 views

Uniform Boundedness Principle for $L^p( \mathbb{R})$

Suppose $\{f_n\}$ is a sequence in $L^p$ such that for each $g\in > L^q$, the sequence $\{\int f_n g\}$ is bounded. Then $\{f_n\}$ is bounded in $L^p$. $(1\leq p<\infty)$ Proof: Argue by ...
0
votes
0answers
23 views

weak convergent sequence in $L^p(\mathbb{R})$ with $(1\leq p < \infty)$ implies norm is bounded

$f_n \rightharpoonup f$ in $L^p(\mathbb{R})$ with $(1\leq p <\infty)$ implies $||f_n||_p$ are bounded. And for $p = \infty$, if $f_n \xrightarrow{w^*} f$, then $||f_n||_\infty$ are ...
1
vote
0answers
9 views

Most general type of $L^p(X,\ V)$ space where compactly-supported continuous functions are dense

Let $(X,\ \tau)$ be a topological (locally compact?) space, $(X,\ \mathcal{F},\ \mu)$ a measure space, $(V,\ \|\cdot\|)$ a Banach space, $1\leq p < \infty$ and $\|\cdot\|_p$ a function defined for ...
3
votes
0answers
32 views

$\sin(nx)$ does not contain Cauchy subsequence in $L^p([0,2\pi]) $ for $1\leq p < \infty$

$\sin(nx)$ does not contain Cauchy subsequence in $L^p([0,2\pi]) $ for $1\leq p < \infty$ My attempt: Set $f_n(x) = \sin(nx)$. Argue by contradiction, suppose there exists a Cauchy ...
9
votes
1answer
56 views

There is no continuous mapping from $L^1([0,1])$ onto $L^\infty([0,1])$

There is no continuous mapping from $L^1([0,1])$ onto $L^\infty([0,1])$. Proof: suppose $T:L^1 \rightarrow L^\infty$ continuous and onto. $L^1$ is separable, let $\{f_n\}$ be a countable dense ...
0
votes
2answers
39 views

$L^{\infty} (X, \mu)$ is not separable? [on hold]

Show that $L^{\infty} (X,\mu)$ is not separable if $X$ contains sequences of disjoint sets of strictly positive measure?
6
votes
1answer
46 views

Which normed space have Fatou's property?

There is tool in mathematics, more specifically in Analysis, which has immense applications in handling delicates estimates, limiting arguments, etc... ; namely Fatou property. Let $E$ be a normed ...
0
votes
0answers
23 views

About convergence in L^1 [duplicate]

Let $(f_n)_{n \in \mathbb{N}} \subset L^1$ and $f \in L^1$ such $f_n \longrightarrow f \ a.e. $ and $ ||f_n||_1 \longrightarrow ||f||_1 $; then $ ||f_n - f||_1 \longrightarrow 0$. Why?
2
votes
1answer
37 views

Compactness of $L^p$ inclusion into $L^q$

Let $ i \colon L^p (0, 1) \longrightarrow L^q (0, 1) $, when $ p \ge q $ the canonical inclusion. It is clearly continuos but never compact: I cannot succeed in showing the last point. Thanks for the ...
3
votes
1answer
28 views

The isomorphism of the Morrey space $L^{2, n} (\Omega)$ and $L^{\infty} \cap L^2 (\Omega)$

The Morrey space $L^{2, \nu} (\Omega)$ for an open set $\Omega \subset \mathbb{R}^n$ and for $1 \leq \nu \leq n$ is defined as $$L^{2, n} (\Omega) = \{ f \in L^2(\Omega); \hspace{2mm} [f]^2_{L^{2, ...
3
votes
1answer
56 views

Is $C^\infty_0$ dense in $C^\infty$ w.r.t. $\|\cdot\|_{L^p}$ and $\|\cdot\|_{W^{1,p}}$?

Is the space $C^\infty_0(\Omega)$ of smooth functions with compact support, dense in the set of smooth functions $C^\infty(\Omega)$ with respect to the norms $\|\cdot\|_{L^p}$ and the Sobolev-Norm ...
0
votes
1answer
64 views

Weak convergence of scaled elements implies norm convergence

Let $u_{k}\in l^2{\mathbb{(Z)}}$ be a sequence such that for every sequence $n_{k} \in \mathbb{Z}$ the sequence $n_{k}u_{k}\rightharpoonup 0$. Prove that $ u_{k} \rightarrow 0$ in $l^{q}(\mathbb{Z}) , ...
2
votes
1answer
24 views

Find the values of $p$ for which the given sequence converges in $l^p$ norm or weakly

We consider $E_p=(c_{00}, ||\cdot||_{p})$ (where $c_{00}$ are the sequences that are zero except in a finite number of values and $1\leq p \leq \infty$) and the sequence:$$(x_n)_{n\in ...
0
votes
1answer
32 views

Dense subset of $L^{2}$ such that $x^{-1/2}f \in L^{1}$ and $\int_{[0, 1]}x^{-1/2}f\, dx = 0$

Does there exist a dense set of functions $f \in L^{2}([0, 1])$ such that $x^{-1/2}f(x) \in L^{1}([0, 1])$ and $\int_{0}^{1}x^{-1/2}f(x)\, dx = 0$? I've noticed that $\int_{0}^{1}x^{-1/2}f(x)\, dx = ...
0
votes
2answers
42 views

Integrable function with given condition is in $L^p$

Suppose $f:\Bbb R \to \Bbb R$ is integrable and there exist constant $c\gt 0$ and $\alpha \in (0,1)$ such $$\int_A |f(x)|dx\le cm(A)^\alpha$$ for every Borel measurable set $A\subset \Bbb R,$ where ...
3
votes
1answer
37 views

What does local space of a given Banach space says intuitively?

We put, $\mathcal{D}(\mathbb R)=$ The space of $C^{\infty}-$ functions on $\mathbb R$ with compact support Example: For instance bump function is in $\mathcal{D}(\mathbb R)$ Let $E$ is a Banach ...
1
vote
2answers
49 views

Showing a sequence is in $\ell^2$ [duplicate]

I am working on the following problem. Suppose that $\{a_j\}_{j=1}^{\infty}$ is a sequence with the property that, whenever $\{b_j\}_{j=1}^{\infty} \in \ell^2$, one has ...
1
vote
0answers
47 views

When does an integral operator belong to the Schatten - von Neumann class in terms of its kernel?

It is well known that an integral operator $X: L^2(\mu)\to L^2(\nu)$ with kernel $k(x, y)$ belongs to the Schatten -- von Neumann class $\mathfrak S_2$ if and only if $\int |k(x, y)|^2\, d\mu(x)\, ...
4
votes
1answer
81 views

Proof that $(L^1)\neq(L^\infty)^\ast$

I have seen a "proof" that $L^1\neq(L^\infty)^\ast$ which goes as follows: show that there is an element of $(L^\infty)^\ast$ which is not in the image of the canonical map ...
2
votes
1answer
38 views

Limit of the p-norm of a function on subdomains equals the p-norm of the function on the union domain

Let $\Omega$ be an open subset in $\mathbb{R}^n$. Given a measurable function $f$, define $$ ||f||_{p,\Omega}=\inf_{a\in\mathbb{R}}||f-a||_{L^p{(\Omega})}. $$ Let $\{\Omega_n\}$ be a sequence of open ...
1
vote
1answer
32 views

Prove this inclusion: $\bigcup_{k<p}\ell^k\subsetneq\ell^p$

Let $1<p<\infty$. I have to prove that $$ \bigcup_{k<p}\ell^k\subsetneq\ell^p. $$ I am not able to find a counterexample to prove the inequality.
2
votes
2answers
67 views

Properties of a set in $\ell^2$ space

Let $\ell^2 = \{x= (x_1,x_2,x_3,\ldots): x_n\in \mathbb C\text{ and } \sum_{n=1}^\infty |x_n|^2 < \infty\}$ and $e_n \in \ell^2 $ be the sequence whose $n$-th element is 1 and all other elements ...
2
votes
1answer
54 views

Domain of multiplication operator

Edit: This question arose due to a misunderstanding, which has now been resolved. Let $\psi \in L^{2}(\mathbb{R})$ be a continuous function. Let $M_{\psi}$ be the multiplication operator on ...
3
votes
1answer
61 views

Weak convergence $f_n \rightharpoonup f$ in $L^2(\mathbb{R})$ and $f_n^2 \rightharpoonup g$ in $L^1(\mathbb{R})$ implies $f^2\leq g$ a.e.

$f_n \rightharpoonup f$ in $L^2(\mathbb{R})$ and $f_n^2 \rightharpoonup g$ in $L^1(\mathbb{R})$, then $f^2\leq g$ a.e. Could you guys help me check the proof please, thanks! Proof: to show $f^2 ...
0
votes
1answer
34 views

Canonical inclusion $L^q(0,1) \to L^p(0,1)$ is compact?

Does there exist $q>p$ such that the canonical inclusion $L^q(0,1) \to L^p(0,1)$ is compact? My answer is no. Since we know that $L^\infty (0,1) \to L^p(0,1)$ is not compact, take $\{\sin(nx)\}$ ...
0
votes
1answer
36 views

$\ell^p$ spaces' inclusion

$$ \ell^s\subsetneq \bigcup_{k<p}\ell^k\subsetneq \ell^p\subsetneq\bigcap_{k>p}\ell^k\subseteq \ell^q $$ for any $1\le s<p<q$. Any idea to prove these inclusions? Counterexamples for the ...
2
votes
1answer
57 views

Is a linear operator on $\ell^2$ defined by the inner product necessarily bounded? [duplicate]

If $a=\{a_n\}\in \ell^\infty(\mathbb{R})$ and $\langle a,x \rangle<\infty$ for all $x\in \ell^2(\mathbb{R})$, (where $\langle a, x\rangle=\displaystyle \sum_{k=1}^\infty a_kx_k$), then is $a\in ...
1
vote
0answers
40 views

Proof of separability of Lp spaces

The following is a proof in Brezis book. It shows the separability of $L^{p}$ spaces: I have a few questions regarding the proof. Questions: It says 'it is easy to construct a function $f_{2} ...
1
vote
1answer
20 views

Showing that a function is bounded in $L^1$ given a bound on its distribution function

Let $f \in L^2((0,T)\times\Omega)$ where $\Omega$ is a compact manifold. Suppose I know that for every $k > 0$, $$\mu(\{|f| > k\}) \leq Mk^{-\frac 12}$$ for some constant $M$ (which is ...
2
votes
1answer
43 views

Need help with application of Hardy-Littlewood inequality (Marcinkiewicz space and distribution functions)

I am going over this work here. I couldn't understand the equality where the Hardy-Littlewood inequality is used. I think $\delta$ here is a weight so we can take it to be $1$ for simplicity. Would ...
0
votes
0answers
18 views

Duality set for $L~p$ spaces, $1<p<\infty$.

I need to show that, given $f \in L^p$, $1<p<\infty$, the duality set $F(f)$ is equal to the point $$\|f\|_p^{2-p}|f|^{p-2}\overline{f}.$$ I have a hint: this is a consequence of convexity of ...
1
vote
3answers
47 views

Dense subsets of $(L^p(\Omega),\|\cdot\|_p)$

The following results hold. Theorem Let $\Omega\subset\mathbb{R}^n$ be an open set. Then $C^0_c(\Omega)$ is dense in $(L^p(\Omega),\|\cdot\|_p)$, if $1\le p<\infty$. Theorem Let ...
1
vote
1answer
53 views

$C^0$ is a closed subspace of $L^{\infty}$

Let $\Omega\subset\mathbb{R}^n$ be an open bounded set. Let $f\in C^0(\bar\Omega)$. I have to prove that $\|f\|_{\infty}=\|f\|_{L^{\infty}}$. One implication is trivial. Let's consider the other one. ...
1
vote
1answer
42 views

Banach valued sequence spaces $\ell^p(X)$

Let $X$ be a Banach space and $\ell^p(X)$ denote the space of sequences $x_i\in X$ for which the norm $\big(\sum_{i=1}^\infty\|x_i\|^p\big)^\frac1p$ is finite, when $X=\mathbb{R}$ we get the usual ...
1
vote
1answer
29 views

Given $u \in L^1$, is there approximating sequence $u_n \in L^\infty$ uniformly bounded in $L^p$?

Let $u \in L^1(U)$ where $U$ is a bounded domain. Is it possible to find a sequence $u_n \in L^\infty $ converging to $u$ in $L^1$ such that the $u_n$ are uniformly bounded for all $n$ in some $L^p$ ...
0
votes
1answer
32 views

$L^\infty(\Omega)$ space

Consider Lebesgue spaces $L^p(\Omega)$, $\Omega$ is a bounded domain. Let $f \in L^p(\Omega)$ for all $p$. Is it true that $f \in L^\infty(\Omega)$?
0
votes
0answers
26 views

Weak star convergent sequence and $L^\infty(0,T;L^\infty(\Omega))$

Suppose $u_n$ is uniformly bounded in $L^\infty(0,T;L^\infty(\Omega))$ where $\Omega$ is a bounded domain. From this answer, we know that the dual space of $(L^1(0,T;L^1))^* = ...
2
votes
1answer
34 views

Weak-star lower semicontinuity in $L^\infty$

Let $u_n \rightharpoonup^* u$ in $L^\infty(\Omega)$. Do we get something like $$\lVert u \rVert_{L^\infty} \leq \liminf_{n \to \infty} \lVert u_n \rVert_{L^\infty}$$ i.e. a weak-star lower ...
4
votes
2answers
53 views

Reflexivity of $\ell^p$

I'm having bad difficulties in understanding how to prove that $\ell^p$ with $1<p<\infty$ are reflexive spaces. Every text I have consulted give that as a trivial result because "observing that ...
4
votes
2answers
48 views

Can we expect, $S(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R), (1<p<\infty) $?

It is well-known that $L^{1}(\mathbb R)$ is a closed with respect to convolution(product), that is, $L^{1}(\mathbb R)\ast L^{1}(\mathbb R)\subset L^{1}(\mathbb R),$ more specifically, if $f, g\in ...
2
votes
0answers
41 views

Convergence of Step Function Defined by Averages

For a function $f \in L^2[0,T]$, and a uniform partition $P = \{0=t_0, t_1, \ldots, t_n = T\}$ of the domain, we can define a step function approximation as the average value over each interval in the ...
1
vote
1answer
40 views

Strong Convergence in L1 Implies Weak Convergence in L2?

If I have $f_n \to f$ in $L^1(D)$, where $D \subset \mathbb{R}$ is compact, is it accurate to say $f_n \rightharpoonup f$ in $L^2(D)$? The argument is as follows: consider a simple function $\phi = ...
0
votes
0answers
23 views

Pointwise (in time) convergence in $H^{-1}$ implies pointwise weak convergence in $L^q$, why?

Let $u_n \to u$ in $C^0([0,T];H^{-1}(\Omega))$ and suppose $\lVert u_n \rVert_{L^\infty(0,T;L^\infty(\Omega))} \leq C$ for all $n$. It follows that for almost all $t$, $u_n(t)$ is bounded in ...
5
votes
1answer
78 views

$\widehat{f\ast g}= \hat{f} \cdot \hat{g}$ for $f, \hat{f} \in L^{p}(\mathbb R)\cap C(\mathbb R) (1<p<\infty, p\neq 2), g\in \mathcal{S}(\mathbb R)$?

It is well-known that, for $f,g \in L^{1}(\mathbb R).$ Then, by Fubini's theorem, one can derive, $\widehat{f\ast g} = \hat{f} \cdot \hat{g},$ (that is, Fourier transform takes, convolution to point ...
7
votes
1answer
54 views

How to obtain the inequality $\int |f|^2\log(|f|) \leq \frac{n}{4}\log\lVert f \rVert^2_{L^p} $ from Jensen's inequality?

Let $f$ be a positive function with $\lVert f \rVert_{L^2}=1$. Let $p= 2n/(n-2)$. How to obtain $\int |f|^2\log(|f|) \leq \frac{n}{4}\log\lVert f \rVert^2_{L^p}$ from Jensen's inequality? Here all ...
0
votes
1answer
26 views

Formula for $L^{q}$ norm using $C_{c}^{\infty}$ functions

We put, $L^{p}=L^{p}(\mathbb R), L^{q}=L^{q}(\mathbb R);$ $\frac{1}{p}+\frac{1}{q}=1;$ ($p$ and $q$ are conjugate exponents); and $<f,g> =\int_{\mathbb R} f(x)g(x) dx.$ Fix $g\in L^{q}, ...
4
votes
2answers
46 views

p-norm of a function

Let $f\in L^1(\mu)\cap L^\infty(\mu)$. I have proved for any $1<p<\infty$, $f\in L^p(\mu)$, $w(p)=||f||_p$ is continuous w.r.t. $p$, and $\lim_{p\to \infty}||f||_p=||f||_\infty$. Is $w(p)$ ...
0
votes
1answer
27 views

An application of Holder's inequality to show one norm is smaller than another

Let $p(s) = r(s) + m-1$ where $r:[0,T) \to [q,\infty)$ where $q \geq 2$ and $m > 1$ is fixed. Let $\text{Vol}(\Omega) = 1$. Then can we show that $$\lVert u \rVert_{L^{r(s)}(\Omega)} \leq ...
1
vote
1answer
36 views

Proving that weak limit in $L^p$ and strong limit in $H^{-1}$ are the same

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. Let $p \geq 1$ and suppose that $u_n \rightharpoonup u$ in $L^p(\Omega)$ and $u_n \to v$ in $H^{-1}(\Omega)$. How to show that $u=v$? I can do ...