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I could show that $\|\cdot\|_p$ is decreasing in $p$ for $p\in (0,\infty)$ in $\mathbb{R}^n$. Following are the details.

Let $0<p<q$. We need to show that $\|x\|_p\ge \|x\|_q$, where $x\in \mathbb{R}^n$.

If $x=0$, then its obviously true. Otherwise let $y_k=|x_k|/\|x\|_q$. Then $y_k\le 1$ for all $k=1,\dots,n$. Therefore $y_k^p\ge y_k^q$, and hence $\|y\|_p\ge 1$ which implies $\|x\|_p\ge \|x\|_q$.

The same argument works even for $x\in \mathbb{R}^{\mathbb{N}}$.

I am wondering whether the result is true for functions $f$ in a general measure space $(\mathcal{X}, \mu)$. The same technique doesn't seem to work in general.

I know that its certainly not true in the case when $\mu(\mathcal{X})<\infty$, as in this case $\|f\|_p\le \|f\|_q\cdot \mu(\mathcal{X})^{(1/p)-(1/q)}$ for $p<q$, so in particular if $\mu$ is a probability measure then, in fact, $\|f\|_p$ is increasing in $p$.

The question is the following. If $f$ is a real valued function on a measure space $(\mathcal{X}, \mu)$ and if $\|f\|_p$ is defined for all $p>0$, is there any result like $\|f\|_p$ is decreasing in $p$?

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It is decreasing if $\mu$ is counting measure. Here's an argument that can easily be adapted. –  t.b. Oct 26 '11 at 6:59
    
Suppose $\mu$ is a measure such that $\mu(\mathcal X) < \infty$, but not necessarily $1$. Then what can you say about $\| f \|_p$ where $f$ is the constant function $1$? –  Srivatsan Oct 26 '11 at 7:01
    
Here is an other one math.stackexchange.com/questions/4094/… –  AD. Oct 26 '11 at 7:04
    
I would say this is a duplicate. –  AD. Oct 26 '11 at 7:05
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This is not a duplicate. The other question handles only sequence spaces. Here, the OP specifically says that he knows the argument for $\mathbb R^\mathbb N$, and that he wants the answer for a general measure space. –  Srivatsan Oct 26 '11 at 7:12

1 Answer 1

up vote 3 down vote accepted

Sometimes we can not say anything. Consider the case $L^p(\mathbb{R})\,$ for $p=1\,$ and $p=2\,$ and look at the characteristic function $\chi_{E}\,$ of a measurable set $E.$ Then $$\|\chi_E\|_{L^1}=\int_E dx =m(E)$$ while $$\|\chi_E\|_{L^2}=\left(\int_E dx\right)^{1/2}=\sqrt{m(E)}.$$ Now if $m(E)>1\,$ we have $m(E)>\sqrt{m(E)}\,$ while $m(E)<\sqrt{m(E)}\,$ in the case $0<m(E)<1$.

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I am not concerned about $\infty$. I am concerned only about $p\in (0,\infty)$. If I am not wrong, a counter example should be like the following. Find $f$ and $0<p<q<r<\infty$ such that $\|f\|_p <\|f\|_q$ and $\|f\|_q>\|f\|_r$ or $\|f\|_p >\|f\|_q$ and $\|f\|_q<\|f\|_r$. –  Ashok Oct 26 '11 at 11:00
    
@Ashok Changed accordingly. –  AD. Oct 26 '11 at 11:40
    
Yes, its correct now. –  Ashok Oct 26 '11 at 12:30
    
@Ashok Great :) –  AD. Oct 26 '11 at 12:49

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