Is $p$-norm decreasing in $p$?

Question

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 $p$?

It is decreasing if $\mu$ is counting measure. Here's an argument that can easily be adapted. — t.b., Oct 26 '11 at 06:59
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 07:01
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#4122 — AD - Stop Putin -, Oct 26 '11 at 07:04
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 07:12
@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 $\mu(\mathcal{X})=\infty$. — Ashok, Oct 26 '11 at 09:07
@Drazick: Because, for any $1\le k\le n$, $|x|_q = (|x_1|^q+\dots + |x_n|^q)^{1/q}\ge (|x_k|^q)^{1/q} = |x_k|$. — Ashok, Nov 02 '15 at 10:25

AD - Stop Putin - · Accepted Answer · 2011-10-26T11:38:15.333

12

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 $0<m(E)<1$.

edited Oct 26 '11 at 11:38

answered Oct 26 '11 at 10:40

AD - Stop Putin -

10,970

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 - Stop Putin - Oct 26 '11 at 11:40
Yes, its correct now. – Ashok Oct 26 '11 at 12:30
@Ashok Great :) – AD - Stop Putin - Oct 26 '11 at 12:49

Is $p$-norm decreasing in $p$?

1 Answers1

Linked