On the space $L^0$ and $\lim_{p \to 0} \|f\|_p$

Question

For $0 < p < \infty$, the definitions of the spaces $L^p$ are very natural. Then, we of course want $L^\infty$ and $L^0$ to be some kind of limits of $L^p$ spaces.

What does the parameter $p$ tell about a function $f \in L^p$? If $f \in L^p$ for very large $p$, it means that $f$ behaves very well near its singularities. If $f \in L^p$ for very small $p$, it means that $f$ decreases fast at infinity.

The limiting case $p \to \infty$ then of course means that $f \in L^\infty$ if $f$ has no singularities in some sense (essentially bounded). Then we define $\|f\|_\infty = \operatorname{ess sup}_{x\in X} |f(x)|$, which is a norm in $L^\infty$. In some cases it even holds that $\|f\|_\infty = \lim_{p \to \infty} \|f\|_p$.

How about $p \to 0$? I've never really seen anything said about this space, so I guess it's not very interesting. But I still started to think about it. Now, $f \in L^0$ should mean that $f$ is extremely bounded at infinity, meaning that $f$ is compactly supported. Of course because we are talking about measure and integration theory, the notion of compact support has to be modified accordingly. I guess we want to define that $f \in L^0$ if the support of $f$ is compact modulo sets of measure $0$. Is this how we should think about $L^0$? If it is, then we probably want to define $\|f\|_0 = \mu(\operatorname{spt} f)$, or something like that.

So, what is the number $\lim_{p \to 0} \|f\|_p$? For measurable compactly supported bounded functions it is easy to see that \begin{align} \lim_{p \to 0} \int_X |f|^p = \mu(\operatorname{spt} f)\,. \end{align} But if we want to take the limit of $\|f\|_p$, the integral should be raised to the power $1/p$, which tends to infinity, which complicates things.

So my questions are: what should the space $L^0$ be, what should be its "norm" $\|f\|_0$ and when is $\|f\|_0$ the limit of $\|f\|_p$?

Also, why does the wikipedia page on $L^p$ spaces refer to $L^0$ as the space of measurable functions? (In the case $\mu(X) < \infty$ it would make sense since $L^0$ restricts the behavior of its elements at infinity, and in the $\mu(X)< \infty$ case there is no "behavior at infinity".)

Answer 1

If $\mu(X)=1$, there exists $p>0$ with $f\in L^p(\mu)$, and if we agree that $\log(0)=-\infty$ and $\exp(-\infty)=0$ then $$\lim_{p\to0}||f||_p=\exp\left(\int\log|f|\,d\mu\right), $$the geometric mean of $|f|$. Hence some people sometimes write $||f||_0$ for the geometric mean.

Assume $f\ge0$. First, Jensen's inequality with $\phi(t)=e^{pt}$ shows that $$||f||_p\ge||f||_0\quad(p>0).$$

Note that $$\log(t)\le c_pt^p\quad(t\ge0).$$So if $f\in L^p$ then $\int\log^+f<\infty$, and hence $\int\log f$ exists (as an element of $[-\infty,\infty)$). Fix $q>0$ such that $$\int f^q\log^2(f)<\infty.$$

Now $$e^t\le 1+t+\frac{t^2}{2}e^t\quad(t\ge0)$$and $$e^t\le 1+t+\frac{t^2}2\quad(t\le 0).$$

Suppose $0<p<q$. If $f=0$ on a set of positive measure then Holder shows that $||f||_p\to0$. Assume that $f>0$ almost everywhere. Let $f=e^g$. Let $E$ be the set where $g\ge0$ and $F$ the set where $g<0$. Now $$\int_Ef^p=\int_Ee^{pg}\le\int_E\left(1+pg+\frac{p^2g^2}{2}f^p\right)\le\int_E\left(1+pg+\frac{p^2g^2}{2}f^q\right),$$while$$\int_F f^p\le\int_F\left(1+pg+\frac{p^2g^2}{2}\right).$$ Hence $$\int f^p\le 1+p\int\log f+cp^2,$$where $c=\frac12\int(1+f^q)\log^2f$. Hence $\limsup_{p\to0}||f||_p\le||f||_0$.