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.

$L^p$ spaces are defined for $p\in(0,\infty]$ as follows:

Let $(X,\mathcal F,\mu)$ be a measure space. For $p$ with $0 < p<\infty$ we write $L^p(X,\mu)$ ($L^p(X)$ or $L^p$ when there is no ambiguity), for the vector space of equivalence classes (for equality almost everywhere) of measurable functions $f$ such that $\int_X|f|^p\mathrm{d}\mu$ is finite.

When $1\leq p<\infty$ we endow $L^p$ with the norm $$\lVert f\rVert_p:=\left(\int_X|f(x)|^p\mathrm{d}\mu(x)\right)^{1/p}$$ When $0 < p < 1$ we write $$\lVert f\rVert_p:=\int_X|f(x)|^p\mathrm{d}\mu(x)$$ which induces a metric on $L^p$.

For $p=+\infty$, $L^\infty$ is the space of equivalence classes of functions $f$ such that we can find a constant $C$ with $|f(x)|\leq C$ almost everywhere. Then $\lVert f\rVert_{\infty}$ is the infimum of constants $C$ satisfying the latter property, and is called the essential supremum of $|f|$.

These spaces are sometimes called Lebesgue spaces. If we work with counting measure on, for example, the set $\mathbb N$, we get sequence spaces $\ell^p$ and $\ell^\infty$ as special cases.

5636 questions
3
votes
3 answers

What does mean : element of $L^p$ are equivalence class rather than function.

What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $f\in L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent…
3
votes
2 answers

Real Analysis Folland, Proposition 6.13, The Dual of $L^p$

Proposition 6.13 - Suppose $p$ and $q$ are conjugate exponents and $1\leq q < \infty$. If $g\in L^q$, then $$\|g\|_{q} = \|\phi_g\| = \sup\{\left|\int fg\right|: \|f\|_{p} = 1 \}$$ If $\mu$ is semifinite, this result also holds for $q =…
Wolfy
  • 6,495
3
votes
1 answer

Factorization of $L^{p}$ spaces

Is the following theorem true and if so does it have a name? Theorem: Let $p$, $q$, $r\in\left(0, \infty\right]$ with $\frac{1}{q}=\frac{1}{p}+\frac{1}{r}$. Every $L^{q}$-function is the product of a $L^p$-function and a $L^{r}$-function. I believe…
3
votes
3 answers

$\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^\infty(\Omega)}$

is true that $\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^{\color{blue}\infty}(\Omega)}$ ? I can't see a proof for this :/ ( of course, $\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^{\color{blue}2}(\Omega)}$ ) I'm using the usual…
yemino
  • 814
3
votes
0 answers

Simple $L^p$ space question

Let $X$ be a measure space with positive measure $\mu$. If $\|f_n-f\|_p\to 0$ (i.e. $f_n$ converges to $f$ in the $L^p(\mu)$ norm) does it follow that $f_n(x)\to f(x)$ almost everywhere?
Robert
  • 39
2
votes
1 answer

Properties of $L^p$ spaces

Let $\Omega$ $\subseteq$ $\Bbb R^{n}$ be bounded and $u$ be a measurable function on $\Omega$ Such that $|u|^{p} \in {L^{1}_{\mathrm{loc}}(\Omega)}$ for some $p \in \Bbb R$. Then how to show the following? $$\lim_{p\rightarrow0}…
2
votes
1 answer

a function that is in $L^2$ (the right version)

Sorry I made a mistake when posting the last question. Actually my question is: can you give a $f(x) $ such that $ f \in L^2 ( \mathbb R)$ but $ x^{-\frac{1}{2}} f \notin L^1 ( \mathbb R ) $. Thanks!
user112564
  • 3,552
  • 1
  • 23
  • 51
2
votes
1 answer

a function that is in $L^2$

Can anyone give me an example of $f(x) $ such that $ f \in L^2 ( \mathbb R)$ but $ x^{\frac{1}{2}} f \notin L^1 ( \mathbb R ) $. Thanks! It seems that $f(x) = x^\alpha$ doesn't work...
user112564
  • 3,552
  • 1
  • 23
  • 51
2
votes
1 answer

Bounded function in $L_1([0,\infty))$ but not in $L_2([0,\infty))$

Is it possible to find a continuous bounded function $f$ such that the integral \begin{equation} \int_{0}^{\infty}|f(x)|dx \end{equation} is finite and the integral \begin{equation} \int_{0}^{\infty}|f(x)|^2dx \end{equation} diverges? Edit What…
2
votes
1 answer

Checking a property of a closed vector subspace of $L^1$.

Based on this question, I asked myself the following: Let $\Omega=(0,1)$ and let $X=\overline{\text{span}\{x^{1/n-1}:n\in\mathbb{N}\}},$ where the closure is taken in $L^1(0,1)$. Does $X$ satisfy the conditions for the claim in the linked…
UserA
  • 1,650
2
votes
2 answers

Show that there exist $f_1 \in L^{p_1}$ and $f_2 \in L^{p_2}$ such that $f = f_1 +f_2$.

Let $1 \le p_1 < p \le p_2 \le \infty$ and let $f\in L^p$. Show that there exist $f_1 \in L^{p_1}$ and $f_2 \in L^{p_2}$ such that $f = f_1 +f_2$. Just for clarify, we consider the $L^p$-spaces on $\mathbb{R}^n$. How can I start this exercise?
Gabrielek
  • 1,898
2
votes
1 answer

If $u_n\to u$ in $L^p$ and $u_n\to v$ in $L^q$ do we have that $u=v$?

If $u_n\to u$ in $L^p$ and $u_n\to v$ in $L^q$ do we have that $u=v$ ? Attempt I know that we have a subsequence $u_{n_k}$ that converge to $u$ and $v$ a.e. and thus that $u=v$ a.e., but do we have that $u=v$ everywhere ?
user386627
  • 1,828
2
votes
1 answer

$L^2$ function with compact support is in $L^1$?

I'm reading Real analysis(Stein) and confused by a statement: If $f\in L^2$, then $f\chi_R$ is in both $L^1$ and $L^2$, where $\chi_R$ is the characteristic function of $[-R, R]$. But $f$ in $L^2$ does not mean that $f\in L^1$. Such an example…
2
votes
0 answers

What motivated mathematicians to realize that $L^p \simeq (L^q)^*$?

Here $p, q$ denote conjugate exponents. Essentially, what problems were (I assume Riesz) and other's working on that made them realize this connection? Were there example functions they found that hinted this may be true? The following question…
2
votes
2 answers

Examples of functions that are in $L^p(X,\mu)$ but not in $L^q(X,\mu)$, where $1< p < q <\infty$ for each $p$ and $q$?

It is known that if $\mu(X) < \infty$, we have the inclusion $L^q(X,\mu) \subset L^p(X,\mu)$ when $1