The inclusion of $L^p(\mathbb R)$ and $L^q(\mathbb R)$

Question

Let $p > q$, and I am looking for counter examples of continuous functions which is in $L^p(\mathbb R)$ but not in $L^q(\mathbb R)$ and continuous functions in $L^q(\mathbb R)$ but not $L^p(\mathbb R)$.

While restricting attention to $(0,\infty)$ it suffices to manipulate with functions like $x^{-\alpha} |logx|^b$ using indicator functions, but this method does not work if I want continuous functions and $\mathbb R$.

Hint: look for functions that decay at different rates near infinity, e.g. Like x^-r for different r. — Zach Boyd, Sep 10 '17 at 04:27
you may not find any continuous function for your purpose here, as if $f\in C(\mathbb R)\cap L^p(\mathbb R)$, then $f$ is bounded, so $f\in L^q(\mathbb R)$ — Arun, Sep 10 '17 at 04:29
Arun, shouldn't an example such as this work, with suitably chosen rates of growth? https://math.stackexchange.com/questions/482934/continuous-unbounded-but-integrable-functions The boundedness should not be an issue. — Zach Boyd, Sep 10 '17 at 04:36
Do you have an idea on how to construct such continuous functions?@ZachBoyd — zach, Sep 10 '17 at 04:44

Eric Towers · Accepted Answer · 2017-09-10T15:11:02.070

2

Assuming $1 \leq q$.

Let $f_r(x) = \min\{1,|x^r|\}$. Note $f_r(x)$ is continuous at $x \in \mathbb{R}$. We see $f_r \in L^m$ for $m>1$ if $mr < -1$, which is to say, if $r < -1/m$. Since $q < p$, $-1/q < -1/p$, and if we pick $r \in (-1/q, -1/p)$, $f_r$ is a continuous function in $L^p$ but not $L^q$.

The train of triangles function at Continuous unbounded but integrable functions works for the opposite non-inclusion.

edited Sep 10 '17 at 15:11

answered Sep 10 '17 at 05:33

Eric Towers

67,037

Shouldn't the max be a min? – Zach Boyd Sep 10 '17 at 05:49
@ZachBoyd : Oops. Yes... – Eric Towers Sep 10 '17 at 15:10
Is there such a function which is independent of $q$? – zach Sep 10 '17 at 18:25
You could try the boundary case r=-1/p and see if the details work out. – Zach Boyd Sep 10 '17 at 19:27
I do not think this example works if I want it to be independent of $q$. – zach Sep 10 '17 at 19:48
@zach : Do you have a condition on a function in $L^p$ of the form $r \leq [\text{something}]$? If not, there is no way to put a function in $L^p$ and not in $L^q$ for $q \in (p-\delta, p)$ where $\delta > 0$ depends on the gap between $r$ and [something]. – Eric Towers Sep 11 '17 at 01:34

The inclusion of $L^p(\mathbb R)$ and $L^q(\mathbb R)$

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