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.
5,707
questions
8
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Convergence of functions in $L^p$
Let $\{f_k\} \subset L^2(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is a bounded domain and suppose that $f_k \to f$ in $L^2(\Omega)$. Now if $a \geq 1$ is some constant, is it possible to say ...
5
votes
1
answer
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Sharp interpolation inequality for Lebesgue spaces
Suppose $f\in L^{p}(\mathbb{R}^{n}) \cap L^{q}(\mathbb{R}^{n})$. How can I prove that for any $p \lt r \lt q$,
$$
\lVert f \rVert_{r} \leq
(\lVert f \rVert_{p})^{(1/r-1/q)/(1/p-1/q)}
(\lVert f \...
1
vote
0
answers
213
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Convergent in $L^1(0,1)$ but not in $L^2(0,1)$ help understanding a paper from arxiv
http://arxiv.org/pdf/math/0205003v1
In around equation (1.1) the author says
"By necessity all authors have been led in one way or another to the natural approximation
$$F(n) := \sum_{a=1}^n \mu(a) ...
16
votes
4
answers
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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 < \...
5
votes
3
answers
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When $L_p = L_q$?
As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!
3
votes
1
answer
333
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Convergence of integrals in $L^p$ and $L^{p/(p-1)}$
Let $X$ be a measure space and let $f_{n}$ be a sequence of functions which converge pointwise to a function $f$ in $L^{p}(X)$ where $p>1$ and suppose $g_{n}$ is a sequence of functions which ...
66
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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).