# $L^p$ with negative $p$ [closed]

In spaces $L^p$ for $p \in (0,1)$ the "norm", defined as for $1 \leq p <\infty$, does not satisfies triangle inequality, however satisfy reversed Hoelder and Minkowski inequalities. But $\|\cdot \|^p$ satisfies triangle inequality, condition $\|x\|^p=0$ iff $x=0$, and $\|\lambda x\|^p=|\lambda|^p \|x\|^p$ for scalar $\lambda$ and vectors $x$. Hence, in that case, $L^p$ is F-space (in general not locally convex).

Assume now that $p<0$. If we would define "norm" $\| \cdot\|$ in the standard way (such as for $1 \leq p < \infty$) then it hold reversed Hoelder inequality and reversed Minkowski inequality $\|f+g\|\geq \|f\|+\|g\|$ for positive $f,g$. (I don't know is it true for all $f,g$.)

Is it possible in some natural way to introduce in $L^p$, with negative $p$, norm, metric, topology etc.

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## closed as not a real question by Qiaochu YuanAug 30 '11 at 17:47

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with negative $p$, I don't see how possible to define a norm: $\| f\|= \frac{1}{(\int\frac{1}{\|f\|^{p}}dx)^{1/p}}$. It seems noway to find a $\|f\| = 0 \leftrightarrow f = 0$ –  newbie Aug 30 '11 at 10:44
I think we must assume $\|0\|:=0$. –  Richard Aug 30 '11 at 10:46
Richard: what about functions that vanish on a set of positive measure? I think that is what @newbie is implying. While you can define $\|0\| := 0$, it looks like for any function that vanish on a set of positive measure you have $\|f\| = 0$. (In fact, if you are working over a measure space with infinite total mass, then any bounded function will have $\|f\| = 0$.) –  Willie Wong Aug 30 '11 at 12:08
I think you have right. It seems my question is not suitable. –  Richard Aug 30 '11 at 12:58