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4 votes
1 answer
81 views

weak convergence and pointwise implies $L_p$ convergence

Suppose $f_i \to f$ weakly in $L^p(X, M, \mu)$, $1 < p < \infty$, and that $f_i \to f$ pointwise $\mu$-a.e. Prove that $f_i^+ \to f^+$ and $f_i^- \to f^-$ weakly in $L^p$. My proof: Since $f^\pm ...
Mr. Proof's user avatar
  • 1,575
3 votes
1 answer
75 views

What is wrong with this proof that a linear, bounded, time invariant operator on $L_p$ must be a convolution?

I'm trying to understand if this is true and how to prove it, "If $T$ is a bounded, time invariant operator on $L_p(\mathbb{R})$, then $T$ is a convolution operator.'' Here's an attempt at a ...
travelingbones's user avatar
1 vote
1 answer
51 views

Weakly sequentially closed set in $L^p$

Let $\Omega\subset \mathbb{R}^n$ be bounded and Lebesgue measurable, $p \in [1,\infty]$, and $a,b \in L^p(\Omega)$. Consider the set $$ K = \big\{ u \in L^p(\Omega):\, a(x) \leq u(x) \leq b(x) \, \...
Andymt's user avatar
  • 23
1 vote
1 answer
45 views

Weak Star Convergence of Integral Averages

Suppose $U$ is a Banach space and let $V\subseteq U$ be bounded, convex and (norm-)closed. Consider the Bochner-Lebesgue space $L^r(0,T;U)$ with $T>0$ and $r\in[1,\infty]$ consisting of strongly ...
Joe S's user avatar
  • 23
2 votes
0 answers
78 views

Convergent subsequences in $L^p(\mathbb R^n)$

This is a particular fact that I ended up proving in the process of attempting one of my recent homeworks, but I don't think I've seen this particular fact online even though it feels like a fairly ...
person's user avatar
  • 1,503
1 vote
1 answer
44 views

Does the weak limit of a sequence in $L^2([0,1])$ vanish on the limit set of vanishing sets?

Suppose $h_n$ is a sequence of non-negative functions in $L^2([0,1])$ converging weakly to $h$ (i.e., for every $g\in L^2([0,1])$ it holds $\int g\cdot h_n \,d\lambda \to \int g\cdot h\, d\lambda$). ...
Michael's user avatar
  • 365
1 vote
0 answers
39 views

Approximating a weakly convergent sequence "uniformly" by a dense subspace

For a fixed $p \in (1,\infty)$, consider $L^p(\Omega)$ for a bounded domain $\Omega$ in the Euclidean space and a sequence $\{ f_{n} \} \subset L^p(\Omega)$ converging weakly. That is, for each $g \in ...
Keith's user avatar
  • 7,829
2 votes
1 answer
66 views

Sequence with bounded $L^p$ norm which converges in measure also converges weakly

Let $\{f_n \}_{\mathbb{N}}$ be a bounded sequence in $L^p(X, \Sigma, \mu)$ (i.e, there exists $M > 0$ such that $\|f_n \|_{L^p} \leq M$ for all $n \in \mathbb{N}$), where $(X, \Sigma, \mu)$ is a ...
Matheus Andrade's user avatar
1 vote
1 answer
51 views

Weak convergence argue

If we take a sequence $\{u_n\}$ that converges weak to u in $L^2(\Omega)$, where $\Omega$ is bounded and $g_n\to g$ weak-* in $L^{\infty}(\Omega)$ then how can I obtain this limit, for all $\varphi \...
Luiza Camile's user avatar
1 vote
1 answer
52 views

Convergence of a sequence in $L^2(\mathbb{R}^N)$ that is also bounded in $H^1(\mathbb{R}^N)$

Let $(u_n)_n\subset H^1(\mathbb{R}^N)$ be a sequence and $u\in L^2(\mathbb{R}^N)$ verifying: $(u_n)_n$ is bounded in $H^1(\mathbb{R}^N)$ $(u_n)_n$ converges to $u$ in $L^2(\mathbb{R}^N)$ Then, does $...
UCL's user avatar
  • 381
3 votes
1 answer
137 views

Lower Semicontinuity of $L^p$ norms with varying exponents

In a previous post (see continuity of $L^p$ norms with respect to $p$) it is shown that in a measure space $(\Omega,\Sigma,\mu)$, if $1\leq p_0\leq p\leq p_1\leq+\infty$, then the function $\Phi\...
Tytiro's user avatar
  • 114
0 votes
1 answer
53 views

Weak star convergence in $L^{\infty}$ and strong convergence in $L^2$ imply strong convergence in $L^p$

I can't see why the following would hold. Can someone please point out why? If $f_n$ convergence weak-* in $L^{\infty}(\Omega)$, and converges strongly in $L^2(\Omega)$, then it must also converge ...
Morcus's user avatar
  • 585
1 vote
0 answers
99 views

Analogue of Skorohod representation theorem for Wasserstein metric and convergence in $L^p$?

Background. The Skorohod representation theorem says that if $X_n \to_{\mathrm{d}} X$ (i.e. $X_n$ converges to $X$ in distribution as $n \to \infty$), then we can construct a new probability space ...
Ziv's user avatar
  • 346
2 votes
2 answers
302 views

Weak convergence in Sobolev space $W^{1,p}$ implies strong convergence in $L^p$ [duplicate]

Let $p\neq \infty$ and let $\Omega\subseteq \mathbb{R}^N$ be a regular domain (a bounded open set with $C^1$ boundary). I want to prove that If $\{u_n\}$ weakly converges in $W^{1,p}(\Omega)$ then it ...
Kandinskij's user avatar
  • 3,536
0 votes
1 answer
126 views

Example of a sequence in the closed unit ball of $\ell^1$ which does not have any weakly convergent subsequence.

I am looking forward to having an example of a sequence in the closed unit ball of $\ell^1$ which does not have any weakly convergent subsequence. I need this to prove that the closed unit ball in $\...
RKC's user avatar
  • 835

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