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

Question

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 $p$'s?

Answer 1

Corollary 3 of Chapter 7 of Royden:

If $E$ is measurable with finite measure and $1\leq p_1< p_2\leq \infty$, then $L^{p_2}(E)\subseteq L^{p_1}(E)$. Furthermore, $||f||_{p_1}\leq c||f||_{p_2}$ for all $f\in L^{p_2}(E)$ where $c=[m(E)]^{\frac{p_2-p_1}{p_1p_2}}$ if $p_2<\infty$ and $c=[m(E)] ^{\frac{1}{p_1}}$ if $p_2=\infty$.

Royden remarks in an example however that for measurable set $E$ of infinite measure, there are no inclusion relationships among the $L^p$ spaces.

Answer 2

Here are some interesting facts about the relations between different $L^p$-spaces over the same measure space $(X,\Sigma,\mu)$ (based on Section 6.1 of Folland, 1999):

• If $0<p<q<r\leq\infty$ and if $f\in L^q$, then there exist $g\in L^p$ and $h\in L^r$ such that $f=g+h$.

• If $0<p<q<r\leq\infty$, then $L^p\cap L^r\subseteq L^q$.

• If $0<p<q\leq\infty$ and $\mu(X)<\infty$, then $L^q\subseteq L^p$.

• Suppose that $0<p<q<\infty$. Then, $$\exists f\in L^p\setminus L^q\Longleftrightarrow\forall\varepsilon>0,\exists S\in\Sigma:0<\mu(S)<\varepsilon,$$ and $$\exists f\in L^q\setminus L^p\Longleftrightarrow\forall K>0,\exists S\in\Sigma:K<\mu(S)<\infty.$$

• Perhaps the most surprising one: If $X=(0,\infty)$, $\Sigma$ is the Borel $\sigma$-algebra on $(0,\infty)$, and $\mu$ is the Lebesgue measure, then there exists for any given $p\in(0,\infty)$ some $f\in L^p$ such that $f$ is not a member of any other $L^q$-space for $q\in(0,\infty)$ and $q\neq p$!

Answer 3

Here are some examples of relations between the spaces:

• If $p,q\in[1,\infty)$, there is a bijection $L^p\to L^q$, namely $L^p\ni f\mapsto |f|^{p/q-1}f\in L^q$.

• If $A$ has finite measure, then $p\geq q$ implies $L^p(A)\subset L^q$.

• If $f\in L^p$ and $g\in L^q$ so that $1/p+1/q=1/s$ (assuming $p,q,s\in[1,\infty]$), then the pointwise product $fg$ is in $L^s$. This follows from Hölder's inequality.

Answer 4

This is not quite what was asked for, but I thought it worth mentioning:

Below is (the non-trivial) Proposition 11.1.9 in Kalton and Albiac's Topics in Banach Space Theory:

$\ \ \ $(i) For $1\le p\le2$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $p\le q\le2$.

$\ \ \ $(ii) For $2< p<\infty$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $q=2$ or $q=p$.

Moreover, if $L_q$ embeds in $L_p$, then it embeds isometrically.

Here, "$X$ embeds into $Y$" means there is a linear operator $T:X\rightarrow Y$ that is an isomorphism onto $T(X)$. Equivalently, $X$ embeds into $Y$ iff $Y$ contains a subspace isomorphic to $X$.