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Is it true that $L^p(\mathbb R) \subseteq L^q(\mathbb R)$ for $1 \le p <q <\infty$? I haven't been able to find a counterexample, so I'm startig to suspect it is true.

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    $\begingroup$ See here. $\endgroup$ Commented Apr 11, 2014 at 8:27
  • $\begingroup$ i think you mean that $q<p$ $\endgroup$ Commented Apr 14, 2014 at 23:36

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This is not true in general, since "increasing" the power can make a function non integrable near zero for example, and "decreasing" the power can make it non integrable near infinity. For example, take $f(x) = \frac{1}{ \sqrt{x}} $ in $]0;1]$ and zero otherwise, then f is integrable, but not $f^2$ since it's $\frac{1}{x}$ near zero.

Conversely, you can take $f(x) = \frac{1}{x}$ except in $[-1;1]$ where it is zero. Then $f^2$ is integrable but not $f$.

However, when your measure is finite (meaning $\int_{R} 1 d\lambda < \infty $) which is true in a probabilistic setting, then the inclusion is true, because you have, thanks to Hölder inequality (I use $r=\frac{p}{q}$ and $s$ such that $\frac{1}{r} + \frac{1}{s} = 1$) then

$\int f^q.1 \leq (\int f^{qr})^{\frac{1}{r}} (\int 1^s)^{\frac{1}{s}} = (\int f^{p})^{\frac{1}{r}} (\int 1)^{\frac{1}{s}} < \infty$

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  • $\begingroup$ How can one define $1/\sqrt{x}$ for negative $x$? $\endgroup$
    – Twnk
    Commented Apr 11, 2014 at 8:57
  • $\begingroup$ Sorry, corrected. $\endgroup$
    – yago
    Commented Apr 11, 2014 at 9:02
  • $\begingroup$ So the first example shows that $L^2([0,1])\nsubseteq L^1([0,1])$ and the second one that $L^1([-1,1])\nsubseteq L^2([-1,1])$. But what about $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$? (integrals $\int_{-\infty}^\infty$) $\endgroup$
    – Twnk
    Commented Apr 11, 2014 at 9:05
  • $\begingroup$ Well, you can just extend these functions by taking zero elsewhere. Actually the point is that interesting phenomenon here occurs near zero or infinity. The non-inclusion is somehow local. If it bothers you, you can take constant values such that these functions are continuous. I think you may even find examples with smoother functions $\endgroup$
    – yago
    Commented Apr 11, 2014 at 9:09
  • $\begingroup$ No you can't extend it with any constant value it wouldn't be integrable anymore. But what I wanted to say is that you can do whatever you want outside of the interval, as long as it doesn't make it non integrable. For the first example, you could use : $f(x) = \frac{1}{\sqrt{|x|}} 1_{]0;1]}(|x|) + \frac{1}{x^2}1_{[1;\infty[}(|x|)$ which is never null $\endgroup$
    – yago
    Commented Apr 11, 2014 at 9:25
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Actually, here's a counter-example, for $p=1$ and $q=2$ (but easy to adapt to any $p < q$).

Define $S_n = \sum_{i=1}^n {1 \over i^3}$ and $f_n$ the positive function $f_n(x) = \delta_{S_n < x < S_{n+1}}n$

Then define $f = \sum_{n=1}^\infty f_n$. It looks like stairs going infinitely high on the $y$ axis, but which steps grow very narrow, so fast they don't get beyond a point on the "x" axis (namely, the $n$-th step has height $n$ and width $1 \over n^3$).

It is easy to see that $f$ is in $L_1$ ($\int f = \sum {1 \over n²} = \pi^2/6$) but not in $L_2$ ($\int |f|^2 = \sum {1 \over n} = +\infty$).

Loosely speaking, it depends on where the weight is : in asymptotic behavior near zero, or in "peak" behavior around some point.

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