List of counter-examples to $\mathcal{L}_p(\mathbb{R})$ inclusions.

Question

Given $1 \leq p < q \leq \infty$, it is well-known that $$\ell_p \subseteq \ell_q$$ and that $\mathcal{L}_p(\mu) \supseteq \mathcal{L}_q(\mu)$ whenever $\mu$ is finite. However $\mathcal{L}_p(\mathbb{R})$ spaces with Lebesgue measure fall in no such categories. I was thus trying to find some counter-examples to the most common inclusion that one could think of. Here is my list so far and any help to check my solution and to extend it would be the most welcome :

• $\mathcal{L}_p(\mathbb{R}) \not\supseteq \mathcal{L}_q(\mathbb{R})$ for $1 \leq p < q < \infty$. Consider the function $f(x) := \frac{1}{x^{\frac{1}{p}}}\chi_{[1,\infty)}$.

• $\mathcal{L}_p(\mathbb{R}) \not\subseteq \mathcal{L}_q(\mathbb{R})$ for $1 \leq p < q < \infty$. Consider $f(x) := \sum\limits_{n=2}^\infty \left(\frac{1}{n\log^2(n)}\right)^{\frac{1}{p}}\chi_{[n,n+1)}$.

• $\mathcal{L}_p(\mathbb{R}) \not\subseteq \mathcal{L}_\infty(\mathbb{R})$ for $1 \leq p< \infty$. Consider $f(x) := \frac{1}{x^{\frac{2}{p}}}\chi_{[1, \mathbb{R})} \in \mathcal{L}_p(\mathbb{R})$.

• $\mathcal{L}_p(\mathbb{R}) \not\supseteq \mathcal{L}_\infty(\mathbb{R})$ for $1 \leq p < \infty$. Consider $f(x) :\equiv 1$.

• $\mathcal{C}(\mathbb{R}) \not\subseteq \mathcal{L}_p(\mathbb{R})$ for $1 \leq p< \infty$. Consider $f(x) := x^\frac{1}{p}$.

• $\mathcal{C}(\mathbb{R}) \not\subseteq \mathcal{L}_\infty(\mathbb{R})$ for $1 \leq p < \infty$. Consider $f(x) := x$.

• Here is a function that is in $\mathcal{L}_p(\mathbb{R})$ and no other $\mathcal{L}_q(\mathbb{R})$ spaces : $$\sum\limits_{n=1}^{\infty} \left(\frac{c_n}{x^{r_n}}\right)^{\frac{1}{p}}\chi_{[0,1]} + \sum\limits_{n=2}^\infty \left(\frac{1}{n\log^2(n)}\right)^{\frac{1}{p}}\chi_{[n,n+1)},$$ where $r_n := 1-\frac{1}{r^n}$ and $c_n := \frac{r_n}{\int_0^1 \frac{1}{x^{r_n}}}\; dx$.

Answer 1

Theorem 1: Let $1 \leq p<q \leq \infty$. If $X$ contains sets of arbitrarily small positive measure, then $L^p \setminus L^q$ is nonempty.

Theorem 2: Let $1 \leq p<q \leq \infty$. If $X$ contains sets of arbitrarily large finite measure, then $L^q \setminus L^p$ is nonempty.

Theorem 3: Let $1 \leq p<q<r\leq \infty$. If $f \in L^p \cap L^r$ then $f \in L^q$.

Sketch of proof of Thm 1: take $A_n$ disjoint with measure scaling like $n^a$. Define $f$ to be $n^b$ on $A_n$ and zero elsewhere. Choose $a,b$ appropriately.

Sketch of proof of Thm 2: take $A_n$ disjoint with measure in between two fixed positive numbers. Define $f$ to be $n^c$ on $A_n$ and zero elsewhere. Choose $c$ appropriately.

Sketch of proof of Thm 3: split $X$ into $A=\{ |f| \leq 1 \}$ and $B=\{ |f|>1 \}$. On $A$, $|f|^p \geq |f|^q$; on $B$, $|f|^r \geq |f|^q$. (When $r=\infty$ you have to do something different.)

Theorems 1 and 2 account for most of your examples. In view of Theorem 3, the only thing that can actually be done that we have not yet done is find a function which is $L^p$ for just one $p$. To do that, you can sum up functions which are in $L^q$ for $q \in [p,p+1/n)$ (constructed using the idea in Theorem 1 and 2).

Answer 2

In fact, given a measure $\mu$, let $$M=\{\mu(E):0<\mu(E)<\infty\}.$$

Then $L^p\subset L^q$ whenever $p\le q$ if and only if $\inf M>0$, and $L^p\subset L^q$ whenever $p\ge q$ if and only if $\sup M<\infty$.