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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?
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    $\begingroup$ 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. $\endgroup$
    – reuns
    Oct 27 '16 at 19:47
  • $\begingroup$ 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. $\endgroup$ Oct 27 '16 at 22:51
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    $\begingroup$ 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. $\endgroup$
    – ForgotALot
    Oct 28 '16 at 1:28
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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!

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    $\begingroup$ A distribution is not always a function in the conventional sense. In particular $\delta$ and $\delta'$ are not. $\qquad$ $\endgroup$ Oct 27 '16 at 18:59
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    $\begingroup$ 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. $\qquad$ $\endgroup$ Oct 27 '16 at 19:01
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    $\begingroup$ @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 $\endgroup$
    – syzygy
    Oct 27 '16 at 19:03
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    $\begingroup$ 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. $\endgroup$
    – The Vee
    Oct 27 '16 at 19:04
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    $\begingroup$ @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. $\qquad$ $\endgroup$ Oct 27 '16 at 19:06
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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.

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    $\begingroup$ No, just no. en.wikipedia.org/wiki/Function_(mathematics) look at that definition. A distribution is merely a special case. $\endgroup$
    – syzygy
    Oct 27 '16 at 19:05
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    $\begingroup$ That's deliberate. See my comment to your answer. $\endgroup$
    – The Vee
    Oct 27 '16 at 19:05
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    $\begingroup$ @syzygy Clarified in the answer. $\endgroup$
    – The Vee
    Oct 27 '16 at 19:08
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    $\begingroup$ All answers are great and informative. But I choose your answer because it is more intuitive and closer to layman's terms $\endgroup$
    – polfosol
    Oct 28 '16 at 11:59
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    $\begingroup$ 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 $\endgroup$
    – syzygy
    Oct 28 '16 at 17:00
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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 $Re(s) < 0$

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  • $\begingroup$ 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 $n \in \mathbb{N}$ $\endgroup$
    – reuns
    Oct 27 '16 at 19:50
  • $\begingroup$ Although not so layman's terms but nice. Why don't you edit your answer instead of posting comments? $\endgroup$
    – polfosol
    Oct 27 '16 at 20:15
  • $\begingroup$ @polfosol because it is a technical detail $\endgroup$
    – reuns
    Oct 27 '16 at 20:17
  • $\begingroup$ 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. $\endgroup$ Oct 27 '16 at 21:06
  • $\begingroup$ @paulgarrett It don't know this. Maybe I should, for example I have some problems with the "inverse Fourier transform" of $\frac{1}{\Gamma(\sigma+i\omega)}$ $\endgroup$
    – reuns
    Oct 27 '16 at 21:17
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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.

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  • $\begingroup$ 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 $e^{-x^2/\epsilon^2}$. $\endgroup$
    – reuns
    Oct 27 '16 at 20:11
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    $\begingroup$ 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. $\endgroup$
    – reuns
    Oct 27 '16 at 20:21
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    $\begingroup$ @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. $\endgroup$
    – hyportnex
    Oct 27 '16 at 22:06
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    $\begingroup$ 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. :) $\endgroup$ Oct 27 '16 at 22:55
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    $\begingroup$ @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. $\endgroup$
    – reuns
    Oct 27 '16 at 23:38
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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.

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  • $\begingroup$ 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? $\endgroup$
    – polfosol
    Oct 27 '16 at 18:57
  • $\begingroup$ Correct. $\qquad$ $\endgroup$ Oct 27 '16 at 19:01
  • $\begingroup$ When you mean "function" as a function between Eucidlean spaces yeah, otherwise it is wrong $\endgroup$
    – syzygy
    Oct 27 '16 at 19:05
  • $\begingroup$ @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. $\qquad$ $\endgroup$ Oct 27 '16 at 19:07
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    $\begingroup$ 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. $\endgroup$
    – Dirk
    Oct 27 '16 at 19:26

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