TITLE: Physical interpretation of Lebesgue norm $L^p$
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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.