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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.

Twnk
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2 Answers2

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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

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

yago
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  • How can one define $1/\sqrt{x}$ for negative $x$? – Twnk Apr 11 '14 at 08:57
  • Sorry, corrected. – yago Apr 11 '14 at 09:02
  • 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 $\int_{-\infty}^\infty$) – Twnk Apr 11 '14 at 09:05
  • 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 – yago Apr 11 '14 at 09:09
  • 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 – yago Apr 11 '14 at 09:25
  • But if the extension is made by taking zero elsewhere, it would remain measurable, wouldn't it? – Twnk Apr 11 '14 at 09:29
  • Yes, this is the simplest and most efficient way to construct a working example on $L(R)$ from a one on $L(]0;1])$, since it doesn't affect measurability nor integrability, that is the extended function is integrable if and only if the original one is. I just wanted to point out that if you find these examples a little too "artificial", you could also extend it to a smoother, non null function. – yago Apr 11 '14 at 09:34
  • @twink $L^2([0,1]$ is in $L^1([0,1]$ because $[0,1]$ is finite with respect to the lebesgue measure the first example however shows that:$ L^1([0,1])\nsubseteq L^2([0,1])$ the function can be integrable but not square integrable – toufik_kh.17 Apr 14 '14 at 02:52
  • @YannHamdaoui it seems to me that your second example shows the opposite of what you stated. $f$ is integrable and $f^2$ is not. since the integral of $f$ will be just zero and the integral of $f^2$ is infinite .please correct me if i am wrong – toufik_kh.17 Apr 14 '14 at 02:56
  • @toufik_kh.17 In the original post, $f$ is indeed not integrable, since $\int f = \int_{- \infty}^{-1}f + \int_{1}^{+\infty} f$ but for example $\int_{1}^{+\infty}f=\int_{1}^{+\infty} \frac{1}{x} = [ln(x)]1^{+\infty} = +\infty$ so integral of $f$ is infinite. On the other hand, $\int{1}^{+\infty}f^2=\int_{1}^{+\infty} \frac{1}{x^2} = [-\frac{1}{x}]_1^{+\infty} = 1$ and the same for the other part, so I think my example work. (This is somehow informal since in Riemann's theory the first integral should not even be written down since it is not well defined, but I think you get my point) – yago Apr 14 '14 at 16:56
  • ok i see , you are restricting your space to $[1, \infty]$ – toufik_kh.17 Apr 14 '14 at 23:34
  • Yes perhaps my wording was misleading but it says $f(x)=\frac{1}{x}$ $except$ in $[-1,1]$ in the original answer – yago Apr 15 '14 at 09:26
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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.

Valentin Waeselynck
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