There is no inclusion between $L^p[0,+\infty)$ and $L^q[0,+\infty)$

Question

I want to show that there is no inclusion between $L^p[0,+\infty)$ and $L^q[0,+\infty)$, for any $1\le p <q<+\infty$.
I know the intuition behind this:
I must find a function which decays very fast but must almost blow up somewhere: that function is in $L^p[0,+\infty)$ but not in $L^q[0,+\infty)$. I also want a fuction which does not blow up at all anywhere but which decays slowly, so that it is in $L^p[0,+\infty)$ but not in $L^q[0,+\infty)$.
However, I am having a hard time finding two such examples.

Did you try $x^\alpha$ for $x \neq 0$ and $0$ at $x = 0$? – William M. Mar 30 '17 at 18:44 — William M., Mar 30 '17 at 18:44
Treat $[0,1]$ and $[1,\infty)$ separately – zhw. Mar 30 '17 at 18:52 — zhw., Mar 30 '17 at 18:52

score 3 · Accepted Answer · edited Mar 31 '17 at 11:05

3

Hint to get you started: Suppose $p=1,q=2.$ Consider $f(x)=\dfrac{\chi_{(0,1)}(x)}{\sqrt x}, g(x) = \dfrac{\chi_{[1,\infty)}(x)}{x}.$

edited Mar 31 '17 at 11:05

Barbara

849

answered Mar 30 '17 at 18:58

zhw.

105,693

score 1 · Answer 2 · answered Mar 31 '17 at 07:25

1

For each $p>1$ consider

$$f_p(x) = \frac{1}{x^{1/p} (\ln(x)^2+1)}\mathbf{1}_{(0,\infty)}(x).$$

Then

$$f_p\in L_q(\mathbb{R})\iff p =q.$$

answered Mar 31 '17 at 07:25

Santiago

1,586

There is no inclusion between $L^p[0,+\infty)$ and $L^q[0,+\infty)$

2 Answers2