Physical interpretation of Lebesgue norm $L^p$

I want to know precisely what $L^p$ represent really and physically. I know that $L^p$ norm of a function is an attempt to measure its width and height but what distinguish Lebesgue norms from $C^m$ norm for example or any other functional spaces norm?

• I would say "physically" can be a bit of a bad word at times in mathematics. I think you want a geometric interpretation of the $L^p$ norm. Commented Nov 14, 2017 at 2:28
• A clear and explicit geometric interpretation will be surely helpfull but what i want really is clear answer based on the fact that every mathematical object has its own translation in our world Commented Nov 14, 2017 at 2:34

The $L^p$ norms are a way to measure the "energy" of a function. The most imporant ones to start with are the case where $p=1, p=2, p=\infty$.

In general, norms are ways of somehow measuring "length" of some element of a vector space.

for $p=1$, the definition is the area under $|f|$ must be finite. This can lead to some interesting cases.

Consider the Dirichlet function $D(x)$ which is $1$ on the rationals and $0$ at the irrationals. $||f||_{L^1([0,1])}=0$, so despite the fact this function is nowhere continuous, we still have a sense of how much "area" is under the graph. In this case $0$ because "most" of the points are $0$.

Intuitively, the $L^1$ norm measures now "spiky" a function is. The $L^{\infty}$ norm measures how "wide" a function is, and $L^2$ is in some sense an averaging of height and width, the $L^p$ spaces with $2<p<\infty$ are increasingly "biased" to how "wide" $f$ is.

How are the $L^p$ norms different from $C^m$ norms:

Lots of ways. A desirable property of function spaces is that the space be "complete" with respect to the norm. Complete normed spaces are so useful they have their own name: Banach Spaces, and are the main objects studied in functional analysis.

Functions which are $C^{\infty}$ (i.e. smooth) have derivatives of all orders, and in particular, are $m$ times differentiable. It turns out $C^{\infty}$ is not only a subset of $L^p$ but also dense in it (except for $p=\infty$). The $C^m$ spaces are finicky when it comes to completeness under various norms, so as a "main" function space, they have some problems. As such, it is better to consider them as dense subspaces of $L^p$.

Why is it useful to consider them as dense subspaces instead? Here are two key reasons:

$1$. Closed subspaces of complete metric spaces (normed spaces are always metric spaces by taking the metric to be $d(x,y)=||x-y||$) are also complete. Since $C^m$ is closed, it follows that it is complete under the $L^p$ norm.

Really what we are doing is all the nice properties of continuous functions as a collection of objects are better studied when they are viewed as a subspace of $L^p$.

$2$. In the modern theory of PDE, the function spaces being used are the Sobolev spaces $W^{k,p}$ where $k$ denotes the $k-th$ weak derivative and $p$ is the corresponding $L^p$ norm.

Ideally, when studying PDEs, we want strong solutions (solutions which are $C^{\infty}$), however this is wishful thinking, and in many cases, finding strong solutions is extremely difficult. As such, the concept of "weak derivative" was introduced. As expected, if a function has a strong derivative, it has a weak derivative. This allows to create a weak formulation for $PDE$ where the PDE has a solution if we consider weak derivatives rather than strong.

If we tried to use $C^m$ to study $PDEs$ we would not be able to find weak solutions for $PDEs$, which is a cornerstone of the modern theory. In addition, the Sobolev spaces $W^{k,p}$ take into account both the norm (often interpreted as energy, like I mentioned in the first sentence) and the weak derivative.

the $C^m$ spaces under various norms are very restritive compared to the Sobolev spaces in PDE, and even just regular $L^p$ spaces.

Conclusion

The $C^m$ spaces are significantly less useful to people who study PDEs, but even in analysis, many notions such as "almost everywhere" continuous or differentiable don't make sense in the $C^m$ spaces under any norm, even under norms which make $C^m$ complete. The Sobolev spaces allow you to measure the energy of a function and smoothness/differentiability properties in a more general setting by allowing notions of "weak" differentiability, which doesn't make sense without the notion of "almost everywhere."

Much more can be said, but I hope this gives some intuition!

• Beautiful answer, +1. Do you want to edit the Dirichlet function. The way it is written now, the integral should be 0, not 1. Also, if you could add an illustration of what you mean by “spiky” and “wide”, that would also make the answer even better. Commented Nov 14, 2017 at 3:05
• edited! Thanks for the suggestion! Commented Nov 14, 2017 at 3:11
• Isn’t your spiky/wide also flipped? Higher p values favor spikier functions (I.e, spiky functions have higher p norm when p >1 even though it might have the same 1-norm as another function). Take f=char(0,10) and g=10 char(0,1). g is the more spiky function. Both have same 1-norm but g is higher 2-norm, 3-norm etc than f. Commented Nov 14, 2017 at 3:39
• I don't get the spiky/wide thing either. If I take $f(x)$ to be the characteristic function of the interval $[0,a]$, where $a > 0$, then the $L^{\infty}$ norm stays the same ($1$) no matter how wide I make $f$ by increasing $a$. Rather, I would say that the $L^{\infty}$ norm is a measure of height (in an almost-everywhere sense, of course).
– user169852
Commented Dec 1, 2017 at 21:46
• -1 $L^{\infty}$ is the measurable version of the $C^{0}$ norm. I don't see where "wideness" or "spikiness" enters into it. $L^{1}$ is basically a mass norm: how much mass is under the function $|f|$? You might ask where the sign enters into it. OK, let $f(x) \, dx$ be the (signed) density of charge near $x$, positive when there is a net positive charge, negative otherwise. Then $|f(x)| \, dx$ is the total (unsigned) density of charged particles. $L^{1}$ tells you how much "stuff" is there. This is also clear if you note its relationship to the total variation norm for measures.
– user81375
Commented Jan 13, 2018 at 19:00

My feeling is the most "physical" $L^{p}$ spaces are $L^{1}$, $L^{2}$, and $L^{\infty}$. I don't doubt there are cases when $L^{p}$ (with $p \notin \{1,2,\infty\}$) comes up in physics on its own (e.g. there is probably an application where there is an "energy" quantity of the form $\int_{X} |f(x)|^{p} \, \mu(dx)$ for some $p \notin \{1,2,\infty\}$), but I think often when it's used in math it's either for the sake of generality or because you can get better results (e.g. by assuming $p > 1$).

The $L^{1}$ norm measures "mass." It comes up naturally in cases where the function $f$ represents a particle density or probability density, in which case $f \geq 0$ and the $L^{1}$-norm reduces to the total mass. If $f,g$ are two particle or probability densities, then it's often helpful to study $\|f - g\|_{L^{1}}$, which can be understood as telling you how much the arrangement of "mass" differs. For exaample, if the particles of $f$ and $g$ are mostly concentrated in disjoint sets, then $\|f - g\|_{L^{1}}$ will be large. On the other hand, $\|f - g\|_{L^{1}}$ is small if the particles described by $f$ and $g$ are mostly in the same place. (More generally, the total variation norm of measures tells us the same thing. Note that the $L^{1}$ norm agrees with the total variation norm of "absolutely continuous" measures, if you're familiar with the concept, so this is no coincidence.)

The $L^{\infty}$ norm literally measures how big a function is (as seen by the reference measure --- remember $L^{p}$ spaces are relative to some fixed measure.) I don't remember the last time I came across a good "physical" example for the $L^{\infty}$ norm. My intuition is it measures how absolutely close functions are. Think about two curves that never get more than $\delta$ apart: these are close in $L^{\infty}$. On the other hand, if you have two curves whose trajectories diverge wildly, you expect the $L^{\infty}$ norm of their difference to be large.

The $L^{2}$ norm is the "energy" norm. Sometimes in physics a quantity like $\int_{X} |f(x)|^{2} \, \mu(dx)$ appears as an energy term: think of electrostatic energy in electrostatics or kinetic energy in quantum mechanics. From a mathematical point of view, it can become natural to then consider inner products $\int_{X} f(x) g(x) \, \mu(dx)$ since the energy is a quadratic form: if you want to minimize the $L^{2}$ norm subject in some subspace of $L^{2}$, then this sometimes boils down to $f$ having a certain orthogonality property. In the applications (e.g. signal processing), $\int_{X} f(x) g(x) \, \mu(dx)$ is sometimes interpreted as the correlation between $f$ and $g$. (It's very positive when $f$ and $g$ tend to be very similar, it's very negative when $f$ and $-g$ are very similar, and it's very small when $f$ and $g$ are totally different.) This can be a very tempting way of studying signals when you consider the Parseval relation in Hilbert spaces and the Fourier transform's nice relationship with $L^{2}$. Similarly, correlation comes up in quantum mechanics, where $\int_{X} f(x) g(x) \, \mu(dx)$ again tells you how similar the states $f$ and $g$ are. When $g$ is an eigenstate of an observable, it tells how much $f$ looks like $g$ rather than some other eigenstate and the square of the correlation even tells you the probability that $f$ will turn out to be the eigenstate $g$.

I don't think of the $C^{m}$ norms as physical norms. What do they do? They're $L^{\infty}$ norms for smooth functions, and they do more than the $L^{\infty}$ norm in so far as they ask that the function and its derivatives are close to some other function. In this sense, they're more geometrical and it's no wonder they don't get more mileage in PDE. Two smooth functions can be close in $L^{\infty}$ even if one of them has tons of small oscillations: think of the zero function and compare it to the function $t \mapsto \delta \sin(\delta^{-1} t)$ for small $\delta > 0$.) This is no longer true if they're close in $C^{1}$. I don't have a great intuition for curvature so I won't try to explain the difference between $C^{1}$ and $C^{2}$, or any of the higher $m$ for that matter.