15
$\begingroup$

Can someone provide me with some material to make clear of the importance of $L^2$ space in engineering/physics?

Having a background in all the introductory mathematics courses offered in engineering, I find myself completely unable to appreciate the usefulness in knowing the definition of a $L^2$ space. Why bother define $L^2$? All physical construct I've seen so far, whether it is a signal or the wave function, naturally satisfies of square integrability. I have literally never encountered anything that is useful in engineering that do not satisfy this condition.

Therefore you can see why I am so curious as to why so many people are incline to say "suppose $f$ is $L^2$" before they prove something. That is like saying "suppose $x$ is in ${\mathbb R}$", which begs the question, what functional space would $f$ be a part of if not $L^2$?

Can someone show me some actual example on how the definition of a L2 space is invoked to prove something practical (i.e. calc, linear alg, ODE). If possible, can someone also show what would have been different if $f$ belonged to $L^1$ or other $L^p$s or some other spaces?

Many thanks.

$\endgroup$
7
  • 3
    $\begingroup$ Sine waves are not $L^2$. Nonzero constant functions are not $L^2$. Most solutions to Maxwell's equations are not $L^2$. $\endgroup$
    – user14972
    Jul 18 '14 at 4:39
  • 1
    $\begingroup$ Well "$x$" could be any number of things...$x\in\mathbb R$, .$x\in\mathbb C$, .$x\in\mathbb N$,... And so on. Most functions are not in $\ell^2$. These belong to a certain type of Hilbert space which is a very small part of a more broad class of functions in a Banach space. And not all application are square integrable. Take for example, scattering states in quantum mechanics. $\endgroup$ Jul 18 '14 at 4:52
  • 2
    $\begingroup$ I think a (the) reason is to do Fourier series, and to do fourier series, it makes sense to be in a Hilbert space, and $L^1$ is not a hilbert space but $L^2$ is special and has an inner product that makes it an hilbert spacec. $\endgroup$
    – abnry
    Jul 18 '14 at 4:55
  • $\begingroup$ @TylerHG: Don't you mean $L^2$? $\endgroup$
    – user99680
    Jul 18 '14 at 5:54
  • $\begingroup$ @Illegal Immigrant: Don't mean to nitpick , but one usually writes $L^(..)$, where ... is the region on which f satisfies the square-integrability condition. $\endgroup$
    – user99680
    Jul 18 '14 at 6:00
11
$\begingroup$

These are some reasons that I can see at the moment (In the case I recall something else it shall be added to this list):

1) In fact any signal in reality is a function in $L^2(I)$ where $I$ is a time interval, since its energy or power is finite, i.e. $$ \int_I |x(t)|^2 {\rm d}t < \infty $$

2) The Fourier series intrinsically means that any periodic signal can be approximated arbitrarily precise by some $\sin$ functions and $\cos$ functions, where they are an orthonormal basis for a $L^2$ space,...

3) Roughly speaking, $L^2$ space is the only functional space among $L^p$ spaces which is a Hilbert space, i.e. it has an inner product (and also complete)! One can imagine this spaces as a generalization of ${\mathbb R}^n$ to infinite dimensional cases. So many trends like finding minimum/maximum of function from ${\mathbb R}^n$ to ${\mathbb R}$ can be generalized to these spaces in a similar way...

4) In the case of dealing with infinite dimensional spaces like controlling a PDE, fractional systems,... the model and systems are defined in $L^2$...

5) ...

$\endgroup$
6
$\begingroup$

The Fourier transform $$ \hat{f}(s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}\,dt $$ is built from functions $e^{-ist}$ which are not in $L^{2}$ on the real line $\mathbb{R}$. This is an import example physically and mathematically. It tells you have "wave packets" that are in $L^{2}$ even though the components of the pack are not! The differentiation operator $D=\frac{1}{i}\frac{d}{dt}$ has no $L^{2}$ eigenfunctions on $\mathbb{R}$, but it has a wealth of non-$L^{2}$ eigenfunctions $e^{ist}$ from which general functions can be built. These non-$L^{2}$ eigenfunctions aren't physical, but they can used to approximate anything that is.

Wherever there is continuous spectrum in Quantum Mechanics found in unbound states, you have this situation. And you cannot escape continuous spectrum in Quantum Mechanics; continuous spectrum occurs in any system where any value in an interval can be an expected measurement. It's this interplay between non-$L^{2}$ eigenfunctions and $L^{2}$ packets that is critical, and it is related to uncertainty. Very important in nearly every free space problem, including atomic phenomena.

$\endgroup$
1
  • $\begingroup$ Very important points regarding the space: to be $L^2$ or not to be! :) $\endgroup$ Jul 18 '14 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.