# Why are $L^p$-spaces so ubiquitous?

It always baffled me why $L^p$-spaces are the spaces of choice in almost any area (sometimes with some added regularity (Sobolev/Besov/...)). I understand that the exponent allows for convenient algebraic manipulations, but is there more behind it than mathematical convenience?

What bugs me about $L^p$-spaces is that they don't build a scale (of inclusions) but still only allow for one parameter, so by making a choice of exponent you make a choice about two (to my current knowledge) unrelated properties of your function, a) its behavior at singularities (which get milder with high exponent) and b) its tail behavior (which gets less nice with high exponent). How can it still be a good idea to ask "does this operator map $L^p$ to $L^p$" rather than "what does this operator do with singularities and what does it do with tails"? Of course answers to the latter will be harder to formulate and prove, but is that all?

• Can you come up with a well-defined norm that "only captures singularities" and another that "only captures tails"? I think our inability to do this is the problem.
– Ian
Jun 22 '14 at 1:37
• I really think it is because of Holder's inequality. Holder is what we've got and somehow we have get our PDE results. Turns out we just do a lot of algebra from there. Jun 22 '14 at 1:41
• When our measure space doesn't have subsets of arbitrarily large measure, then the $L^p$ spaces do build a scale (a downwards directed one). And probability theory is no small area of study.
– user123641
Jun 22 '14 at 2:39
• See here (math.stackexchange.com/questions/810029/…) for a related question (short summary of my answer: reflexivity is awesome ;) ) Note that I do NOT see your question as a duplicate, as you raise a new point (local vs. global behaviour). One class of spaces that is build to make this distinction are the so-called Wiener amalgam spaces. But I don't know if there is any (textbook) literature on these except for research papers. Jun 22 '14 at 2:49
• Another point is the scaling argument. The $L^p$ norms on $\mathbb{R}^n$ are homogeneous with respect to the dilation $x\mapsto \lambda x$. This corresponds to the physical fact that those norms have a dimension. Therefore they are good suited to analyse dimensionally homogeneous phenomena. Oct 15 '14 at 14:58

This is indeed a very good and natural question, as one usually learns that there is a whole spectrum of $L^p$-spaces but then in practice only $L^2$ (and, to a less extent, $L^1$ and $L^\infty$) seems to pop up. Why should we care about $L^{\frac{3}{2}}$? Of course this question has many possible answers, and I find that a convincing one comes from the context of nonlinear analysis.

The basic observation is the fact that $$\lVert u^k\rVert_{L^p}=\lVert u\rVert_{L^{kp}}^k.$$ So, when dealing with nonlinear problems, we can expect that we will have to play some trick with the index $p$. We won't be able to stay in the comfortable $L^2$-space all the time.

For an example of this let us consider the following PDE: $$\tag{1} -\Delta u (x)= u^2(x), \qquad x\in \mathbb{R}^3.$$ The associated linear inhomogeneous problem $$\tag{2} -\Delta u= h$$ can be solved very satisfactorily in the functional setting of $L^2$-space via the Fourier transform: assuming that everything lies in $L^2(\mathbb{R}^3)$, we can Fourier transform termwise in (2) and write $\hat{u}(\xi)=\lvert\xi\rvert^{-2}\hat{h}$, which can then be anti-transformed back to $$u(x)= \left(\lvert 4\pi y\rvert^{-1} \ast h\right) (x)\stackrel{\text{def}}{=}(-\Delta)^{-1} h.$$ (Note that $\lvert 4\pi y\rvert^{-1}$ is exactly the fundamental solution of the Laplace operator). Setting $h=u^2$, we can now reformulate the nonlinear equation (1) as follows: $$\tag{3} u=(-\Delta)^{-1}\left( u^2\right),$$ which is now an equation of fixed-point type. We want to approach it via the contraction mapping principle, by showing that the mapping $$\Phi(u)=(-\Delta)^{-1}\left( u^2\right)$$ is contractive on some complete metric space to be specified later. To do so we need some estimates on $\Phi$ and those can be provided by the Hardy-Littlewood-Sobolev inequality, which in our case ($\alpha=2,\ n=3$) reads $$\lVert (-\Delta)^{-1} f\rVert_{L^q(\mathbb{R}^3)} \le C \lVert f\rVert_{L^p(\mathbb{R}^3)}, \qquad 2+\frac{3}{q}=\frac{3}{p}.$$ (The condition on $p$ and $q$ can be recovered via the scaling argument, by observing that both sides of this inequality are homogeneous with respect to the scaling $f(x)\mapsto f(\lambda x)$, and therefore the degrees of homogeneity must match). With $f=u^2$ this inequality reads $$\tag{4} \lVert \Phi(u)\rVert_{L^q(\mathbb{R}^3)}\le C \lVert u\rVert_{L^{2p}(\mathbb{R}^3)}^2.$$ It is now clear that our hands are tied: the only way to get something meaningful is to have $q=2p$, which means that $q=\frac{3}{2}$. Thus the right functional setting for this problem is $L^\frac{3}{2}(\mathbb{R}^3)$-space.

Indeed, if we let $B_R\subset L^{\frac{3}{2}}(\mathbb{R}^3)$ denote the closed ball of radius $R$, we see from (4) that $\Phi(B_R)\subset B_R$ if $R< 1/C$. Then, again by (4), we see that $$\begin{split} \lVert \Phi(u)-\Phi(v)\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}&\le C \lVert u^2-v^2\rVert_{L^{\frac{3}{4}}(\mathbb{R}^3)} \\ &=C\lVert (u+v)(u-v)\rVert_{L^{\frac{3}{4}}(\mathbb{R}^3)} \\ &\le C \lVert u+v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}\lVert u-v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)} \\ &\le 2RC\lVert u-v\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)}. \end{split}$$ This means that the map $\Phi\colon B_R\to B_R$ is contractive if $R<\frac{1}{2 C}$.

As a final remark, let us observe that we have actually proven two facts:

1. the existence of a unique solution to (1) in small neighborhoods of the origin in $L^\frac{3}{2}(\mathbb{R}^3)$-space;
2. the fact that the sequence $$\begin{cases} u_{n+1}=(-\Delta)^{-1}\left( u_n^2 \right) \\ \lVert u_0\rVert_{L^{\frac{3}{2}}(\mathbb{R}^3)} \le R \end{cases}$$ converges in the $L^{\frac{3}{2}}(\mathbb{R}^3)$ topology to the solution to (1), no matter which initial condition $u_0$ we choose (provided that $R$ is sufficiently small).

Fact 2. justifies also the necessity to deal with convergence issues in $L^p$-spaces with $p\ne 2$.

Because $L^{p}$ spaces expose the subtle nature of arguments. You have reflexive, non-reflexive, separable, non-separable, algebra, Hilbert, Banach, etc.. And, interpolation works between such spaces because of the exponent. They're good spaces for testing conjectures. They're the original spaces that firmly established the need to separate a space from its dual, and they remain an important part of the foundation of Functional Analysis.

• why does interpolation works between such spaces because of the exponent? Jun 1 '19 at 16:21