Example and counterexamples on inclusions among $L^p$ spaces

Question

What are some examples of functions $f$ that satisfy the following properties?

1. $f \in L^1(\mathbb{R})$ but $f \not\in L^p(\mathbb{R})$ for $p > 1$
2. $f \in L^\infty(\mathbb{R})$ but $f \notin L^p(\mathbb{R})$ for $p \in [0,\infty)$.
3. $f \in L^p(\mathbb{R})$ for $p \in (1,\infty)$, but $f \not\in L^q(\mathbb{R})$ for $q \neq p$.

Answer 1

For 3, take $f(x)=1/\sqrt[q]{x}$ .Then $f \in L^p(\mathbb{R})$ but $f\notin L^q(\mathbb{R})$ if $p>q$ and $q>1$(because $p/q>1$ if $p>q$)

For 2, take $f=1$.This function is bounded but not integrable in any $L^p(\mathbb{R})$

Answer 2

You can expect two kinds of singularities that break integrability: Singularities coming from infinite volume supports and those coming from unbounded values of the function.

For example if $\epsilon>0$: $$\int_1^{\infty}\frac1{x^{1+\epsilon}}\,dx=-\frac1\epsilon(0-1)=\frac1\epsilon\qquad\int_0^1\frac1{x^{1+\epsilon}}\,dx = -\frac1{\epsilon}(1-\infty)=\infty$$ On the other hand: $$\int_1^{\infty}\frac1{x^{1-\epsilon}}\,dx=\frac1\epsilon(\infty-1)=\infty\qquad\int_0^1\frac1{x^{1-\epsilon}}\,dx = \frac1{\epsilon}(1-0)=\frac1\epsilon$$

So powers of $1/x$ that are larger than $1$ are integrable over infinite volume but not over the singularity, whereas powers of $1/x$ smaller than $1$ are integrable over their singularity but not over the infinite volume.

It follows that $\chi_{(0,1)}(x)\,\frac{1}{x^{\alpha}}$ is in $L^p$ iff $1/p>\alpha$, whereas $\chi_{(1,\infty)}(x)\,\frac{1}{x^{\alpha}}$ is in $L^p$ iff $1/p<\alpha$. So if you want a function $f$ to lie in $L^p$ but not in $L^q$ for any $q\neq p$ it is enough to consider $$f=\chi_{(0,1)}(x)\cdot\sum_n2^{-n}\frac1{x^{1/p-1/n}}+\chi_{(1,\infty)}(x)\cdot\sum_{n\in\mathbb N}2^{-n}\frac1{x^{1/p+1/n}}$$ (Here I am assuming you are interested also in non-integer $p$, otherwise the sums can be forgotten and you just take the $n=2$ value.)

Clearly all summands in the first term are in $L^p$, since $1/p-1/n<1/p$ for all $n$ and you then have a sum of terms that have norm $\sqrt[p]{n/p}$ multiplied with $2^{-n}$. This sum converges absolutely in $L^p$ norm and since $L^p$ is complete the limit exists and lies in $L^p$. You can do the same with the second sum.

Furthermore for every $q<p$ there exists an $n$ so that $p-1/n>q$ and then the first sum doesn't lie in $L^q$. Similarly the second sum doesn't lie in $L^q$ if $q>p$.

So this function $f$ is in $L^p$ but not in any $L^q$ for $q\neq p$.

$L^\infty$ is a bit special, but if you want a function in $L^\infty$ but not in $L^p$ for $p<\infty$ just take the constant function $1$.