# Lebesgue integrable p and q

Be $E=[0,\infty]$ and $1\leq p\leq \infty$, Exhibit a function $f$ that $f\in \mathcal{L}^p$ and $f\notin \mathcal{L}^q$ when $q \neq p$

I try with simple functions but i dont know if are lebesgue integrable or dont

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what does "if are leb" mean? – Arturo Magidin Jul 11 '12 at 19:11
I assume at least one of the $p$s should be a $q$. – Chris Eagle Jul 11 '12 at 19:16
By "simple functions", do you mean the technical meaning of simple functions (functions that take only finitely many values, and the inverse image of each value is Lebesgue measurable)? Or do you mean "easy functions", using 'simple' in an informal sense? – Arturo Magidin Jul 11 '12 at 20:09

## 2 Answers

Since your question is a bit ambiguous, I will give hints to get a stronger result that Arturo's. Let $p$ be in $[1, \infty]$. With Arturo's method, for any $q \neq p$, you can find a function $f$ such that $f \in \mathbb{L}^p$ but $f \notin \mathbb{L}^q$. Actually, you can do better, and find a function $f$ such that, for all $q \neq p$, one has $f \notin \mathbb{L}^q$.

Before starting, let us look at two examples.

• If we work on $[0,1]$, then $\mathbb{L}^p ([0,1]) \subset \mathbb{L}^q ([0,1])$ whenever $p \geq q$. This is a consequence of Jensen's or Hölder's inequality. For $p < q$, a function which belongs to $\mathbb{L}^p ([0,1])$ but not to $\mathbb{L}^q ([0,1])$ is, for instance, a function which has a pole of suitable order. A function such as $x^{-\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}\right)}$ does the trick. By taking function which are barely integrable, you can do better. For instance, the function $(x \ln (x/2)^2)^{-\frac{1}{p}}$ belongs to $\mathbb{L}^p ([0,1])$, but to no $\mathbb{L}^q ([0,1])$ for $q > p$.

• If we work on $\mathbb{N}$ with the counting measure, then $\ell^p \subset \ell^q$ whenever $p \leq q$. For $p > q$, a function which belongs to $\ell^p$ but not to $\ell^q$ is a function which has a suitable decay at infinity. A function which behaves like $n^{-\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}\right)}$ works (but recall that, this time, $p > q$, and not the other way around). By taking function which are barely integrable, you can again do better. For instance, the function $(x \ln (2x)^2)^{-\frac{1}{p}}$ belongs to $\ell^p$, but to no $\ell^q$ for $q < p$.

The spaces $\mathbb{L}^p ([0,+\infty))$ are kind of a mix between the $\mathbb{L}^p ([0,1])$ spaces and the $\ell^p$ spaces. That's why you have no inclusions between the $\mathbb{L}^p$ spaces. If $q > p$, a function with a suitable pole at $0$ will belong to $\mathbb{L}^p ([0,+\infty))$ but not to $\mathbb{L}^q ([0,+\infty))$; it is as if we were working with $\mathbb{L}^p ([0,1])$. If $q < p$, a function with a suitable decay at infinity will belong to $\mathbb{L}^p ([0,+\infty))$ but not to $\mathbb{L}^q ([0,+\infty))$; it is as if we were working with $\ell^p$. So, if we take, for instance...

$$f(x) = \frac{1}{(x \ln (x/2)^2)^{\frac{1}{p}}}, \ x \leq 1;$$

$$f(x) = \frac{1}{(x \ln (2x)^2)^{\frac{1}{p}}}, \ x \geq 1,$$

I'll let you see what happens...

NB: the factors $2$ in the logarithm are here only to ensure that the functions are not dumbly not-integrable at $1$. What interests us is their behavior at $0$ and at infinity, no some kind of accident that might happen in-between.

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Hint. $f(x) = \frac{1}{x^{\alpha}}$ is integrable on $[1,\infty]$ if and only if $\alpha\gt 1$, and on $(0,1)$ if and only if $\alpha\lt 1$. If $p\gt q$, Maybe you can find a function whose $p$th power has a large enough exponent but whose $q$th power does not; andif $q\lt p$, maybe you can find a function whose $p$th power is large enough but whose $q$th power is not?

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