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

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 $p$?

Solutions Collecting From Web of "Is $p$-norm decreasing in $p$?"

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 $0<m(E)<1$.