Linked Questions

1
vote
2answers
4k views

Proving that $L^p \subset L^q$ when $1 \le q \le p$ [duplicate]

Let $(E,\mathcal{F},\mu)$ be a measure space such that $\mu(E)=1$ and let $L^p=L^p(E, \mathcal{F},\mu)$. Prove that $L^p \subset L^q\text{ if } 1 \le q \le p$. I let $f \in L^p$. Then $(\int_E |f|^pd\...
0
votes
1answer
1k views

How to prove that $L^q$ space is a subset of $L^p$ space if $1 \leq p < r < q < \infty$? [duplicate]

How to prove that $L^p$ space is a subset of $L^q$ space whenever $1 \leq p \leq q < \infty$? I tried to solve it using the $L^p$ space definition. Tell me how to proceed further?
2
votes
1answer
921 views

Embedded Lp spaces [duplicate]

Let $L^\infty(Ω,F,P)$ be the vector space of bounded random variables $(X ∈ L^\infty (Ω,F,P)$ means that there exists a constant C such that $|X(ω)|≤C$, a.s.$)$. Show that $$L^\infty(Ω,F,P)⊂L^2(Ω,F,P)⊂...
1
vote
1answer
536 views

Show that $L^p(U) \subseteq L^q(U)$ for $p > q \geq1$. [duplicate]

Let $U \in \mathbb{R^n}$ be a closed space and $p > q \geq1$. Show that $L^p(U) \subseteq L^q(U)$. I need some hint to start with this!
0
votes
0answers
351 views

Proof of $L_q(\Omega) \subset L_p(\Omega)$ when $1\leq p \leq q \leq \infty$ [duplicate]

I need to prove that $L_q(\Omega) \subset L_p(\Omega)$ when $1\leq p \leq q \leq \infty$ when $\Omega$ is bounded. The hint given is that I should use Hölders inequality. How do I start my proof ...
1
vote
0answers
187 views

Problem # 25, page 95, from Stein and Rami [duplicate]

Let $(X,M,\mu)$ be a measure space with $\mu(X) < 1$. Show that for any $1\le p<q$, we have $$L^q (X,\mu)\subset L^p(X,\mu).$$ Let $\ell^p(Z)$ denote the $L^p$ space of the integers equipped ...
1
vote
0answers
177 views

Lp space contained in the intersection of Lq, where 1≤q≤p. [duplicate]

Let $(X, \Sigma, \mu)$ be a measure space with $\mu$(X) $< \infty$. Let $p \in [1,\infty)$. I would like to prove that $L^p \subseteq \underset{1 \leq q \leq p}\cap L^q$. I'm not sure how to go ...
1
vote
0answers
116 views

Finite meaure space with $f \in L^p$ [duplicate]

Given a finite measure space $(X,\Sigma,\mu)$, for $1<p<\infty$, if $f \in L^p(X)$, then $f \in L^1(X)$. Can anyone show me how to start the proof? Thanks.
1
vote
0answers
50 views

Set inclusion involving the set of locally integrable functions [duplicate]

Let $0 <p < q < \infty$. Why exactly does the inclusion $L_{\mathrm{loc}}^q(\mathbb{R}^n) \subseteq L_{\mathrm{loc}}^p(\mathbb{R}^n)$ hold? For clarification:
108
votes
4answers
12k views

Is it possible for a function to be in $L^p$ for only one $p$?

I'm wondering if it's possible for a function to be an $L^p$ space for only one value of $p \in [1,\infty)$ (on either a bounded domain or an unbounded domain). One can use interpolation to show that ...
7
votes
5answers
318 views

Does $\mathbb{E}\left[X\right]=\infty$ imply $\mathbb{E}\left[X^{2}\right]=\infty$?

I'm trying to prove that if $\mathbb{E}\left[X\right]=\infty$ then $\mathbb{E}\left[X^{2}\right]=\infty$ for every random variable $X$. I know that if $X(w)>1$ I'll get that $X^2(w)>X(w)$ so $\...
6
votes
2answers
5k views

Inclusion of $l^p$ space for sequences

Inclusion of $L^p$ spaces for functions has been discussed here. Does this apply to $l^p$ space of sequences similarly? I tried to show the following: For $1\leq p<q<\infty$, $l^q\subset l^p$ ...
7
votes
1answer
3k views

When $L^p \subset L^q$ for $p <q$.

Assume that $(X, \mathfrak{B}, m)$ is a measure space such that there exists a constant $\alpha>0$ such that for every $E \in \mathfrak{B}$ the following holds: $$ m(E)=0 \ \ or \ \ m(E)\geq \...
3
votes
4answers
848 views

Are there relations between elements of $L^p$ spaces?

I have read about dual spaces and the relation $1/p+1/q=1$ as mentioned in the Wikipedia page. Are there any more theorems or relations that connect elements between the $L^p$ spaces for different $p$'...
2
votes
2answers
787 views

Compact inclusion in $L^p$

Is it true that there is a compact inclusion from $L^p$ to $L^q$ whith $q<p$? What is the counterexample if what I said is wrong? Thank you.

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