Questions tagged [lp-spaces]
For questions about $L^p$ spaces. That is, given a measure space $(X,\mathcal F,\mu)$, the vector space of equivalence classes of measurable functions such that $|f|^p$ is $\mu$-integrable. Questions can be about properties of functions in these spaces, or when the ambient space in a problem is an $L^p$ space.
672
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Limit of $L^p$ norm
Could someone help me prove that given a finite measure space $(X, \mathcal{M}, \sigma)$ and a measurable function $f:X\to\mathbb{R}$ in $L^\infty$ and some $L^q$, $\displaystyle\lim_{p\to\infty}\|f\|...
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7
answers
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$L^p$ and $L^q$ space inclusion
Let $(X, \mathcal B, m)$ be a measure space. For $1 \leq p < q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample ...
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If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$
Let $1\leq p < \infty$. Suppose that
$\{f_k, f\} \subset L^p$ (the domain here does not necessarily have to be finite),
$f_k \to f$ almost everywhere, and
$\|f_k\|_{L^p} \to \|f\|_{L^p}$.
Why is ...
- 8,633
39
votes
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answer
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The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. [duplicate]
If we are in a sequence space, then the $ l^{p} $-norm of the sequence $ \mathbf{x} = (x_{i})_{i \in \mathbb{N}} $ is $ \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} $.
The $ l^{\...
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votes
6
answers
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How do you show monotonicity of the $\ell^p$ norms?
I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).
- 8,633
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votes
2
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The Duals of $l^\infty$ and $L^{\infty}$
Can we identify the dual space of $l^\infty$ with another "natural space"? If the answer is yes, what can we say about $L^\infty$?
By the dual space I mean the space of all continuous linear ...
- 2,475
43
votes
3
answers
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"Scaled $L^p$ norm" and geometric mean
The $L^p$ norm in $\mathbb{R}^n$ is
\begin{align}
\|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}.
\end{align}
Playing around with WolframAlpha, I noticed that, if we define the "scaled" $L^p$ ...
- 3,774
37
votes
1
answer
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Strong and weak convergence in $\ell^1$
Let $\ell^1$ be the space of absolutely summable real or complex sequences. Let us say that a sequence $(x_1, x_2, \ldots)$ of vectors in $\ell^1$ converges weakly to $x \in \ell^1$ if for every ...
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Show that $\lim _{r \to 0} \|T_rf−f\|_{L_p} =0.$
I am having a hard time with the following real analysis qual problem. Any help would be awesome.
Suppose that $f \in L^p(\mathbb{R})$, where $1\leq p< + \infty$. Let $T_r(f)(t)=f(t−r)$. Show ...
- 1,723
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votes
3
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On the equality case of the Hölder and Minkowski inequalities
I'm following the book Measure and Integral of Richard L. Wheeden and Antoni Zygmund. This is problem 4 of chapter 8.
Consider $E\subseteq \mathbb{R}^n$ a measurable set. In the following all the ...
- 10.5k
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votes
1
answer
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A Hamel basis for $\ell^p$?
I am looking for an explicit example for a Hamel basis for $\ell^{p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
- 307
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4
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Convergence of integrals in $L^p$
Stuck with this problem from Zgymund's book.
Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1<p<\infty$. Assume that $\|f_{n}\|_{p} \leq M < \...
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votes
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Why is $L^{\infty}$ not separable?
$l^p (1≤p<{\infty})$ and $L^p (1≤p<∞)$ are separable spaces.
What on earth has changed when the value of $p$ turns from a finite number to ${\infty}$?
Our teacher gave us some hints that ...
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answer
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If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic
If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$.
Maybe I would have to use the Rademacher's functions.
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2
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Translation operator and continuity
I came across a text that proves that translation operator $T_a(f):=f(x-a)$ where $a\in\mathbb{R}^n$ and $f\in L^p(\mathbb{R}^n)$ is continuous. The proof follows:
$$||f(x-a)-f(x)||_p=||f(x-a)-g(x-a)+...
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