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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

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 $p$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](http://math.stackexchange.com/questions/69125/inequality-between-ellp-norms).2011-10-26
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    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 $1$?2011-10-26
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    Here is an other one http://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q/4122#41222011-10-26
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    I would say this is a duplicate.2011-10-26
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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.2011-10-26
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    @SrivatsanNarayanan Sure.2011-10-26
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    @SrivatsanNarayanan When $f=1$ a.e., increasing if $\mu(\mathcal{X})\le 1$, and decreasing if $1<\mu(\mathcal{X})<\infty$. But I am concerned about the case when $\mu(\mathcal{X})=\infty$.2011-10-26
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    I don't see why $ {y}_{k} \leq 1 $. Could you explain?2015-10-27
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    @Drazick: Because, for any $1\le k\le n$, $\|x\|_q = (|x_1|^q+\dots + |x_n|^q)^{1/q}\ge (|x_k|^q)^{1/q} = |x_k|$.2015-11-02

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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.

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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 such that $\|f\|_p <\|f\|_q$ and $\|f\|_q>\|f\|_r$ or $\|f\|_p >\|f\|_q$ and $\|f\|_q<\|f\|_r$.2011-10-26
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    @Ashok Changed accordingly.2011-10-26
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    Yes, its correct now.2011-10-26
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    @Ashok Great :)2011-10-26