28

I remember there was a tongue-in-cheek rule in mathematical analysis saying that to obtain the Fourier transform of a function $f(t)$, it is enough to get its Laplace transform $F(s)$, and replace $s$ by $j\omega$. Because their formula is pretty much the same except for the variable of integration.

And I know that this is not necessarily true (that's why I used the term tongue-in-cheek) e.g. take $f(t)=1$ to see the obvious difference. I read somewhere that although their formula is somehow similar, their result is not necessarily similar because Laplace transform is a function and Fourier transform is a distribution. For example, the Dirac delta, $\delta(\omega)$ is a distribution and not a function.

So this led me to wonder:

  1. What is the difference between a function and a distribution? (preferably in layman's terms)
  2. Why the Laplace transform is a function and Fourier transform is a distribution? I mean, they are both infinite integrals. So what am I missing?
polfosol
  • 9,923
  • 2
    And every $L^1_{loc}$ function is a distribution, so the distributions are a larger set than the functions for which $\int_a^b f(t) e^{-st}dt$ is well-defined. – reuns Oct 27 '16 at 19:47
  • Just to clarify things for possibly-naive readers: presumably you literally mean to ask in Q2 why the Laplace transform of a_thing is a function, while the Fourier transform of that_thing is (only/merely) a distribution. – paul garrett Oct 27 '16 at 22:51
  • 1
    I would recommend you read Rudin's Functional Analysis ch. 6, the best introductory treatment of distributions known to me, and your questions will be answered. Analysis in general, and distribution theory in particular (a tool for analysis), are logical and involve no mysterious ideas, but you just need to understand step by step what is going on. A few hints: (a) each locally $L^1$ function gives rise to a distribution, (b) not all distributions can be so expressed, (c) $L^1$ functions (and some others) have Fourier transforms which are also functions. – ForgotALot Oct 28 '16 at 01:28

5 Answers5

18

A distribution is also a function (mapping), but its "input" are also functions and not "numbers". This is what a distribution is in layman's terms. A precise definition would be rather complicated (basically the topology on the test functions is rather difficult to define).

To clarify: when people say that "$\delta$ is not a function", then they mean there is no function $\delta:\mathbf{R}\rightarrow\mathbf{R}$ such that $$\int_{-\infty}^\infty \delta(x)f(x)dx=f(0)$$ for all $f\in\mathcal{F}$ where $\mathcal{F}$ is a certain vector space of functions. By definition $\delta$ is the function $\delta:\mathcal{F}\rightarrow\mathbf{R}$ by $\delta(f)=f(0)$. So this "not is a function" relies on a more narrow interpretation of "function", i.e. that functions are those whose domain is (say) a set of real numbers.

Why the Laplace transform is a function and Fourier transform is a distribution? I mean, they are both infinite integrals. So what am I missing?

I think you mean that the Laplace transform of $1$ and the Fourier transform of $1$, right? Well the integral defining the Fourier transform of $1$ does not converge!

syzygy
  • 763
  • So as you say, a distribution is also a function, but why they say the Dirac delta is a distribution and not a function? – polfosol Oct 27 '16 at 18:42
  • @polfosol I have (hopefully) clarified this in my edit. I think it is mainy a historical reason, a hundred years (say) ago the concept of a function was more narrow than today – syzygy Oct 27 '16 at 18:47
  • 3
    A distribution is not always a function in the conventional sense. In particular $\delta$ and $\delta'$ are not. $\qquad$ – Michael Hardy Oct 27 '16 at 18:59
  • 2
    Viewing distributions as linear functionals on the space of test functions is only one way of "coding" the concept of distribution. Another is as a kind of limit of ordinary functions, and another is as convolution quotients. One should not say that they are linear functionals on the space of test functions, but rather that that's one way of coding them, and thereby showing logical consistency of operations done with them. $\qquad$ – Michael Hardy Oct 27 '16 at 19:01
  • 1
    @MichaelHardy I am using function in the modern terminology, i.e. mapping (domain and codomain can be arbitrary sets). we are in 2016 remind you – syzygy Oct 27 '16 at 19:03
  • 1
    I think it's more considerate towards laymen to not call distributions functions but rather functionals. Of course function could refer to any map but it's not what one imagines – plot, derivative, etc. – The Vee Oct 27 '16 at 19:04
  • 3
    @syzygy : As I said, coding them as linear functionals on a space of test functions is only one point of view; one shouldn't say that they are the encoding. And yes, we're in the stone age, the year 2016. $\qquad$ – Michael Hardy Oct 27 '16 at 19:06
  • @MichaelHardy as a matter of fact most somewhat advanced (functional analysis say) books do what you say one shouldn't do – syzygy Oct 27 '16 at 19:24
11

Disclaimer: In the following, “function” refers to the classical map from $(\mathbb{R})$ to $\mathbb{C}$. Many book authors do the same, and so presumably did yours.

A distribution in a more general concept than a function. Some distributions correspond to functions (although they are still different objects, if you look deep enough) so many authors just use the same notation for those, like $\sin x$. But there are many more distributions which behave like no function could.

(Most strikingly, they may not have values in points on the real axis. You must take $\delta(\omega)$ as nothing but a symbol with known properties: $\delta(0)$ can't be evaluated not only because it would not be finite or something but because there's no such thing as evaluating a $\delta$ in what is a singular point to start with.)

Your integral is a good example. You can indeed write an integral representation of the Fourier transform, and if you can successfully calculate the integral (plus some boring further assumptions), then the function you obtain can be thought of as the distribution that is the “real” result, in the sense of the first paragraph. But the Fourier transform is also well-defined in many cases where the integral would diverge.

It's more natural to define Fourier transform in terms of distributions, because it allows many tricks people would be doing anyway, and because restricting oneself to normal functions would mean losing or obscuring many interesting cases from real world use*) **). For Laplace transform you get a strong enough theory in functions alone, so there's no need to make things harder by imposing an unnecessarily general formalism that takes its own module to explain.

*) Also because it beautifully reflects many internal properties and symmetries the transform has.

**) You can also define Fourier transform on the subspace of functions known as $L^2$ (important for quantum mechanics, for example) and stay within that realm. It's a slightly different assumption resulting in a slightly different theory. For example, the constant $1$ is not a $L^2$ function and would not have any Fourier transform in that version.

The Vee
  • 3,151
  • 2
    No, just no. https://en.wikipedia.org/wiki/Function_(mathematics) look at that definition. A distribution is merely a special case. – syzygy Oct 27 '16 at 19:05
  • 1
    That's deliberate. See my comment to your answer. – The Vee Oct 27 '16 at 19:05
  • 1
    @syzygy Clarified in the answer. – The Vee Oct 27 '16 at 19:08
  • 1
    All answers are great and informative. But I choose your answer because it is more intuitive and closer to layman's terms – polfosol Oct 28 '16 at 11:59
  • 1
    I disagree with "Many book authors do the same, and so presumably did yours." though. Any first year math or physics student is familiar with the general definition of a function – syzygy Oct 28 '16 at 17:00
10

Short answer by analogy: distributions are to functions as real numbers are to rational ones. The real number set can be defined as all rational numbers with the limits of all convergent infinite sequences of rational numbers added in, so distributions would be the set of all functions plus all infinite sequences of functions that "converge" in some sense. The Dirac delta function, which is the classic example of a distribution, has many equivalent definitions just like there are many different ways to calculate $\pi$ or $\operatorname{e}$. One frequently used definition is: $$\delta(x) = \lim_{\sigma\rightarrow 0} \frac{1}{\sigma\sqrt{2\pi}} \operatorname{e}^{-\frac{1}{2} \left(\frac{x}{\sigma}\right)^2}.$$

Formally, just like how every real calculation done is performed on rational numbers, the limit is only taken after it's argument has been used in an integral. As a practical matter, though, just like taking integrals and derivatives, we can short circuit that formal process in most cases.

Sean Lake
  • 1,757
  • 10
  • 23
  • By definition of tempered distributions, with $\varphi_\epsilon(x) = \frac{e^{-x^2 / \epsilon^2}}{|\epsilon| \sqrt{\pi}}$ we have $ \displaystyle T(x) = \lim_{\epsilon \to 0} T \ast \varphi_\epsilon(x)$ where the limit is in the sense of distirbutions, and $T \ast \varphi_\epsilon(x)$ is a Schwartz function. For the non-tempered distributions, it is the same but with a $C^\infty_c$ cut of $e^{-x^2/\epsilon^2}$. – reuns Oct 27 '16 at 20:11
  • I confess, I have no idea what you wrote, @user1952009. Please provide links when using jargon (eg. "tempered distributions" "Schwartz function"). – Sean Lake Oct 27 '16 at 20:13
  • Do you know what means $\varphi \in C^\infty_c$, a test function ? A Schwartz function is a larger set of $C^\infty$ functions, allowing non-compact support and requiring $\varphi(x) = o(x^{-k})$ for every $k$. And a tempered distribution is a distribution whose growth rate at $x \to \infty$ is low enough for extending $\langle T, \varphi \rangle$ to $\varphi$ any Schwartz function. – reuns Oct 27 '16 at 20:15
  • 1
    And sorry but when answering to a question on the Fourier/Laplace transform of distributions, you should know the Schwartz functions and the tempered distributions. – reuns Oct 27 '16 at 20:21
  • 1
    @user1952009 I disagree with you, irrespective of whether Sean Lake knows or not what a tempered distribution is, his analogy of illustrating the connection between a normal function and a distribution (Lighthill's generalised function) while comparing it to the rational vs. irrational numbers is actually quite clever. – hyportnex Oct 27 '16 at 22:06
  • @hyportnex You disagree with what ? I didn't criticize his answer, I wanted to show that without a rigorous definition of the (tempered) distributions, you can't make sense to $T(x) = \lim_{\epsilon \to 0} T \ast \varphi_\epsilon(x)$ implicitely mentioned at the end of his answer – reuns Oct 27 '16 at 22:13
  • @user1952009 you criticized him for not knowing what tempered distributions were. I said that was OK with me, for his answer was still quite clever. Rigor is nice but no circuit was ever analyzed with test functions be it Schwartz or Weiss and EEs are quite adapt using $\delta$s when the need arise, so check your Heaviside for further details. – hyportnex Oct 27 '16 at 22:20
  • @hyportnex No, he criticized me for not giving all the links, and what you say at the end is 100% false. Example : in electrical signal processing, you have the idealized capacitor $\frac{1}{j\omega C}$, but in real life you only have an approximation to it. Now tell me how do you define a "good" approximation to $\frac{1}{j\omega C}$ without a rigorous definition of the (tempered) distributions ? – reuns Oct 27 '16 at 22:34
  • @user1952009 This is moving away from mathematics, but in short, and staying within the realm of linear circuits, Maxwell's equation guarantee that anything (finite size and passive) can be represented as either parallel connected series resonators and a radiating structure (antenna) or a series connected parallel resonators and antenna (yes, analogous to an expansion). Normally the dominant term (impedance representation) has either a pole at zero (capacitor) or at infinity (inductor) and the true resonator parts with the radiator can be ignored. If not, you have a lot more work. – hyportnex Oct 27 '16 at 22:46
  • @hyportnex The correct concept is that your approximate capacitor $\tilde{C}$ acts on signals $s(t)$ passing through it, and with $C$ the ideal capacitor, you want that $|C s(t)-\tilde{C} s(t)| < \alpha |s(t)|$ for certain class of signals, and some norm $|.|$ on those. – reuns Oct 27 '16 at 22:54
  • 1
    Apart from other discussion of (arguably legitimate) details... I'd claim that this answer's notion that "distributions" are "simply" limits of ordinary/classical functions is exactly the most useful viewpoint to take... for operational purposes, anyway, and I don't care so much about ... um... inoperational (dysoperational?) viewpoints. :) – paul garrett Oct 27 '16 at 22:55
  • What I wrote was not a criticism of you, @user1952009, but an admission of my own limitations and a request for more information. Yes, by calling your terminology "jargon" I'm assuming that a lot of users of Fourier and Laplace transforms will be like me: having an operational knowledge of how to use them without knowing all the rigorous details and names. – Sean Lake Oct 27 '16 at 23:25
  • 1
    @SeanLake Yes but those are not details : you should look at the rigorous definition of distributions, and remember that distributions are defined rigorously by how they act on some set of functions : the continuous or $C^1$ or $C^\infty$ functions, depending on the order of the distribution. – reuns Oct 27 '16 at 23:38
  • A better way to write this would be $\delta = \lim_{\sigma\to0} \left(\backslash x \mapsto \frac1{\sigma\sqrt{\pi}} e^{-(\tfrac{x}{\sigma})^2}\right)$. For while it's valid to introduce the Dirac distribution as a limit, I'd avoid suggesting this limit can be evaluated pointwise. – leftaroundabout Oct 28 '16 at 12:56
6

We look only at the bilateral Laplace transform, the unilateral Laplace transform being the Laplace transform of $f(t) 1_{t > 0}$.

The Laplace transform of a function (or a distribution) $f(t)$ is $$F(s) = \int_{-\infty}^\infty f(t) e^{-st}dt \overset{def}= \lim_{A \to -\infty} \lim_{B \to +\infty}\int_A^B f(t) e^{-st}dt$$

If this limit of integrals converges for $s = s_0$ and $s=s_1$, then it converges for every $Re(s) \in (Re(s_0),Re(s_1))$, and $F(s)$ is analytic there.

This is why we require $F(s)$ to be well-defined (and hence analytic) on some strip $Re(s) \in (a,b)$. In that case, the Fourier transform of $f(t) e^{-\sigma t}$ is well-defined and is $=F(\sigma+i\omega)$ for any $\sigma \in (a,b)$.

Conversely, if the Fourier transform of $f(t)$ is well-defined only as a (tempered) distribution, then the Laplace transform of $f(t)$ isn't defined on $Re(s) =0$. Indeed, because of the definition of tempered distributions and because the Fourier transform isn't well-defined for non-tempered distributions, only two cases are possible if the Fourier transform of $f(t)$ is well-defined (as a distribution) : $F(s)$ is well-defined on $(0,a)$, or $F(s)$ is well-defined nowhere.

A last case exists : when $F(s)$ admits an analytic continuation beyond $Re(s) \in (a,b)$, in that case you have to remember that it isn't the Laplace transform of $f(t)$ anymore. For example $\frac{1}{s}$ is the Laplace transform of $1_{t > 0}$ on $Re(s) > 0$, and the Laplace transform of $-1_{t < 0}$ on $Re(s) < 0$

reuns
  • 79,880
  • Note in $\lim_{A \to -\infty}\lim_{B \to \infty}$, it can be $\lim_{n,m \to \infty} \int_{A_n}^{B_m} f(t) e^{-st}dt$ where $A_n,B_m$ are two sequences such that $A_n \to -\infty,B_m \to \infty$. For example the Laplace transform of $f(t) =\sum_n \delta'(t-n) e^{-n^2}$ is well-defined, even if $\int_{-n}^n f(t) e^{-st}dt$ isn't for any $n \in \mathbb{N}$ – reuns Oct 27 '16 at 19:50
  • Although not so layman's terms but nice. Why don't you edit your answer instead of posting comments? – polfosol Oct 27 '16 at 20:15
  • @polfosol because it is a technical detail – reuns Oct 27 '16 at 20:17
  • Actually, if one needs to take Fourier transforms of arbitrary distributions for some reason, it is possible, but what are produced are things in the dual space to the Paley-Wiener space, the latter being exactly the collection of Fourier transforms of test functions. So the symmetry is lost, certainly. But, for example, this allows us to (correctly) identify the Fourier transform of $e^x$ as a Dirac delta at a suitable point off the real line, etc., which does actually reasonably match intuition. – paul garrett Oct 27 '16 at 21:06
  • @paulgarrett It don't know this. Maybe I should, for example I have some problems with the "inverse Fourier transform" of $\frac{1}{\Gamma(\sigma+i\omega)}$ – reuns Oct 27 '16 at 21:17
4

Suppose $f$ is a well behaved continuous function and $f(0)=0$. Then $\displaystyle \int_{-\infty}^\infty f(x)\delta(x)\,dx = 0$, no matter which function $f$ is and if $\delta$ were a function in the conventional sense of the word, that can happen only if $\delta(x)=0$ whenever $x\ne0$. And what is $\delta(0)$? If $\delta$ were a function in the conventional sense, then $\delta(0)$ would be some number, if so $\delta(x)$ would be $0$ if $x\ne0$, and some particular number if $x=0$. But if you integrate $\displaystyle \int_{-\infty}^\infty f(x)\delta(x)\,dx$ and if $\delta$ is as just described, you would get $0$ rather than $f(0)$. So it gets said that $\delta(0)$ is a sort of "infinity". But this is an "infinity" that can be multiplied by, for example, $3.2$, so that $\displaystyle\int_{-\infty}^\infty f(x) \Big( 3.2 \delta(x) \Big) \, dx = 3.2f(0)$.

And what of the fact that $\displaystyle\int_{-\infty}^\infty f(x) \delta'(x)\,dx = -f'(0)$? What number or what kind of "infinity" could be assigned as the value of $\delta'(0)$ that would make that true?

A function $f$ in the conventional sense takes an input $x$ which is some number and returns an output $f(x)$ which is a number. The "distributions" or "generalized functions" $\delta$ and $\delta'$ don't do that.

Michael Hardy
  • 1
  • I think I am getting somewhere. So all functions are distributions, but a distribution is not necessarily a function, right? Now what about the second part of my question? – polfosol Oct 27 '16 at 18:57
  • Correct. $\qquad$ – Michael Hardy Oct 27 '16 at 19:01
  • When you mean "function" as a function between Eucidlean spaces yeah, otherwise it is wrong – syzygy Oct 27 '16 at 19:05
  • @syzygy : If you want to insist that a generalized function "is" a linear functional on a space of test functions, I think you'll end up with an up-hill battle. $\qquad$ – Michael Hardy Oct 27 '16 at 19:07
  • @MichaelHardy do you want me to cite the definition of a distribution from standard books? like reed/simon, rudin, schwarz, etc.? – syzygy Oct 27 '16 at 19:25
  • 2
    Minor nitpick (with sometimes major impact): Only locally integrable functions are distribution. E.g. be careful when trying to interpret $|x|^{-1}$ as a distribution. – Dirk Oct 27 '16 at 19:26
  • @Dirk that's clear and I think each of the guys that wrote an answer avoided this technicality, basically because the OP is looking for a layman's answer – syzygy Oct 27 '16 at 19:28
  • @syzygy : I am well aware of that standard "definition". Linear functionals on a space of test functions are one way of coding the concept of distribution. Another is a kind of limit of ordinary functions. Yet another is as convolution quotients. Do you claim to be able to prove that the first way is the only right way. – Michael Hardy Oct 28 '16 at 04:40