Introduction

To begin with we consider a calculus problem that you may have seenin your exam:

Let $f$ be a continuousfunction on $[0,\infty)$ that $\lim_{x \to \infty} f(x)=l$. Prove that\[\int_0^\infty \frac{f(ax)-f(bx)}{x}\mathrm{d}x = (f(0)-l)\ln\frac{b}{a}\]

And we solve this problem as follows. Put $g(x)=f(x)-l$, then $\lim_{x \to \infty}g(x)=0$. Consider thetwo variable function $F(x,y)=-g'(xy)$ and the range $D=\{(x,y):x \ge 0, a \le y \le b\}$, wehave this result: \[\begin{aligned}\iint_D F(x,y)\mathrm{d}x\mathrm{d}y &=\int_0^\infty\mathrm{d}x\int_a^b -g'(xy)\mathrm{d}y \\ &= \int_0^\infty\frac{g(ax)-g(bx)}{x}\mathrm{d}x \\ &= \int_a^b \mathrm{d}y\int_0^\infty -g'(xy)\mathrm{d}x \\ &= \int_a^b\frac{g(0)}{y}\mathrm{d}y \\ &=g(0)\ln\frac{b}{a} \\\end{aligned}\] Substituting $g(x)$ with$f(x)-l$ gives exactly what we want,isn't it? Well, the more analysis you learn, the more absurdthis proof has been you will realise. If you write this in anexam you will get $0$ mark no matterwhat. There are two major mistakes:

Can we change the order of integration? We have no idea. But it iscertain that we cannot change the order with ease, and we have some counterexamples.
Is this function even differentiable? We also have no idea.It is almost certain that $f$ is not (the probability that $f$ is differentiable is $0$), see this post tolearn why if you have some background in functional analysis.

For a good proof, please turn to math.stackexchange.This is not easy at all.

The problem is, it is really unfair that in somecircumstances we have to axe out all properties of differentiation. Ifyou are studying differential equations, and a non-differentiablefunction pops up, you have no way to go. Sometimes, chances are that youeven have no idea whether a function is differentiable.

So this post is written. We introduce the concept of (Schwartz)distribution (a.k.a. generalisedfunctions), where differentiation is significantly extended, toobtain derivative in a generalised sense. Roughlyspeaking, after distribution being introduced, differentiation can bedone with absolute ease.

A function ofbad blood among physicists

In fact, physicists have been using distribution long beforemathematicians established formal theories. For example the $\delta$ function introduced byDirac that you may have met in Fourier transform: \[\delta(x) = \begin{cases}\infty &\quad x=0, \\ 0&\quad\text{others} .\end{cases}\] And it is required that \[\int_{-\infty}^{\infty}\delta(x)\mathrm{d}x=1.\] But this does not make any sense in calculus. Von Neumann, inhis book on quantum physics, warned against the theory using thisfunction, and dismissed this function because this was a "fiction". Notso pleasant. He tried with a lot of effort to demonstrate that, quantumphysics could live without such a "fiction". As you can imagine, thisfunction may have created some bad blood between von Neumann andDirac.

LaurentSchwartz however, managed to be a peacemaker. He developed thetheory of distribution (which is exactly what we are talking about inthis post), and the "fiction" became an easy "fact". Years later, hebecame the 1950 Fields Medalist (one of the most prestigiousmedal/awards in mathematics) at the age of 35 with reason

Developed the theory of distributions, a new notion of generalizedfunction motivated by the Dirac delta-function of theoretical physics.(Source)

As you can see later, thanks to Schwartz, the twisted $\delta$ function is well-defined and isreally plain and elegant. So von Neumann didn't need to be angrylater.

On backgrounds

By concept I mean, I will try to include basic ideas(without many proofs though they can be delivered), so that the seriousstudy of it can be simpler (it can be really tough!). It is not possiblethat you can solve problems on distributions after reading thispost.

There will be two parts. Part one focus on motivation and what isgoing on. I will try to make it readable to many people having finishedcalculus or more ideally undergraduate analysis and linear algebra,though rigour is not always guaranteed. It would be better if you knowsome differential equation theory, but that's not a must. If you alreadyhave the background to read part 2, then part 1 is much easier for youand therefore is served as a good source of intuition andmotivation.

If you still need to understand differentiation in single-variablecalculus, then you have no need to struggle on generaliseddifferentiation at an early point. It does not help. The requirements oflinear algebra are vector spaces, subspaces and linear maps. You shouldknow that integration and differentiation are linear maps. This is agraduate course topic, it is not realistic to assume reader to have noidea about calculus and linear algebra.

The second part will be much more advanced, and you are expected tohave some background in topological vector spaces (functional analysis).Both parts cannot be considered as a lecture note but they may help youfind where you are when you study this concept seriously.

Part I - Integration byParts

Throughout, we consider functions on $\mathbb{R}$ with real value. Thesetheories can be generalised to $\mathbb{R}^n$ with complex value wherepartial derivative can take part in, but we are not doing that here. Atthe end of the day, these work would not be a big deal.

Motivation: thevector space of distributions

In calculus, a lot of functions we study are smooth (for example,$y=\sin{x}$), and we write $C^\infty$ as they are infinitelydifferentiable. This is a vector space and this vector spacedifferentiation can be done with absolute ease. For given $f \in C^\infty$, we have $f',f'',\cdots,f^{(k)}$ welldefined for all $k = 1,2,\cdots$. Butin vector spaces like $C^2$, $C^1$, or even $C$, differentiation can only be done withcaution: we may only have $f''$ and no $f^{(3)}$, or even $f'$ does not exist. We don't feellike this kind of caution. Hence we introduce the concept ofdistribution which is also known as generalisedfunctions. We want a space where we can still dodifferentiation with absolute ease. We may need to modify ourdefinition of differentiation such that it works on every continuousfunctions (but it shall not lost its meaning within $C^\infty$). Bearing these in mind, we haveseveral settings or expectations for distributions:

Every continuous function should be (considered as) a distribution.(So we can take derivatives for all continuous functionswithout to many worry. Unlike the calculus problem at thebeginning.)
The "modified differentiation" should make sure that the "modifiedderivative" of a distribution is still a distribution. In other words,distributions are "infinitely differentiable" (which makes differentialequation theory much easier). In the language of algebra, the "modifiedderivative" should be an endomorphism.
The usual formal rules of calculus should hold. For example in thenew sense we should still have $(fg)'=f'g+g'f$. (Our modifieddifferentiation should not go to far.)
Convergence properties should also be available. (Validating thisrequires more theories so this can only be mentioned in part 2.)

Let's write our desired distribution as $\mathscr{D}'$, and all continuousfunctions $C$. All $C,C^\infty,\mathscr{D}'$ areconsidered as real vector spaces and we should have \[C^\infty \subset C \subset \mathscr{D}'\] in the sense of subspaces.

Whatis distribution and extended differentiation exactly

Here is a breakdown of these concepts. You will see terminologies anddefinitions later.
A smooth, continuous or more generally, locally integrable function,give rise to a bounded linear functional. The converse is not guaranteedto be true, but we pretend it to be true, so allbounded linear functionals give rise to distributions,a.k.a. generalised functions (this name is nice because wepretend the converse to be true). Whenever you areasked what is generalised function, you can say, it is a linear map, andsometimes it can be determined by a normal function.
For these distributions or generalised functions, we modify thederivative with respect to integration by parts. The modified derivativecannot be put down explicitly but we don't care, because integration byparts doesn't give us many problems. Whenever you are asked howthe derivative of a non-differentiable function is given, you can say,it is given by pretending nothing wrong in integration byparts.

We now try to understand what we really what about distribution. Westart our study through integration, because differentiationdoes not work. Given $f \in C\subset \mathscr{D}'$, we first need to make sure $\int f\phi$ is well-defined, forsome $\phi\in C^\infty$,because we want to do integration by parts, which involves somedifferentiation, and we may make use of it.

If $f$ is not even a continuousfunction, we still need to consider some $\phi$ in the same manner, or our extensionwould be abrupt.

Let's talk about these $\phi$ alittle bit, with respect to integration by parts. Consider the bumpfunction \[\phi(x) = \begin{cases} \exp(\frac{1}{(x-a)(x-b)}) & \quad a < x< b, \\ 0 &\quad \text{ otherwise. }\end{cases}\] On $(a,b)$, we have $ $. Onthe boundary $a$ and $b$ we have $\phi(x)=0$ but that shouldn't be aproblem, because they are the alpha and omega. Points outside $[a,b]$ have no contribution to the valueof this function. For some obvious reason we call $[a,b]$ the closure of $(a,b)$. In general, given a real-valuedfunction $f$, we call the closure ofthe set of points where $f(x) \ne 0$the support of $f$.As you can tell, the support of $\phi$ is $[a,b]$.

If $\phi$ has unbounded support(the support of a function $f$ is theclosure of the set of points $x$where $f(x) \ne 0$), then we may needto discuss limit at infinity. But we don't want improper integrals atall. Hence the support of $\phi$ arealways assumed to be closed and bounded subset of $\mathbb{R}$ It is closed because it isdefined to be a closure. These closed and bounded sets are calledcompact sets. If you are not familiar with topology, it is OKat this moment to consider compact sets as bounded closed interval $[a,b]$.

The test function space $\mathscr{D}$ is defined to be all $C^\infty$ functions with compact support.This is indeed a vector space and the verification is a good excise onboth linear algebra and calculus. What about $\mathscr{D}'$? Here we demonstrate howthings are extended.

For each $f \in C$ (which contains$C^\infty$), we have a functional (afunctional is a linear map between a vector space and its base field,here is $\mathbb{R}$. Nothingspecial, just a different name that has been used by mathematicians fordecades!) \[\begin{aligned} \Lambda_f: \mathscr{D} &\to \mathbb{R}, \\ \phi &\mapsto \int f\phi.\end{aligned}\] This functional is bounded for all $\phi \in \mathscr{D}$ because if $\phi$ has support $K$, then \[|\Lambda_f(\phi)|=\left|\int_K f\phi\right| \le \left(\int_K |f|\right)\sup_{x \in K}|\phi|.\] A continuous function on a compact set is always bounded (proof),hence the integral on the right hand side is always bounded. If ittouches infinity a lot of problems are also touched.

In general, a bounded linear functional $\Lambda:\mathscr{D} \to \mathbb{R}$ iscalled a distribution, which forms $\mathscr{D}'$ exactly. Since everycontinuous function $f$ gives rise toa unique bounded functional $\Lambda_f$, we consider $C$ as a subspace of $\mathscr{D}'$. Such a function giverise to a functional, which is called distribution. The converse is notgenerally true, but we pretend it to be true (we pretend thefunctional gives rise to a function anyway), which makes our studyeasier, hence the name generalised function iswell-deserved.

Distribution derivative

Differential operator $D$ in $C^\infty$ should be extended naturallyinto $\mathscr{D}'$ naturally.There are many ways to extend a linear function. For example theidentity map $i:\mathbb{R} \to\mathbb{R}$ has at least two ways to be extended into $\mathbb{R}^2$:

$I:\mathbb{R}^2 \to \mathbb{R}^2$by $(x,y) \mapsto (x,y)$.
$\pi:\mathbb{R}^2 \to \mathbb{R}$by $(x,y) \mapsto x$.

The restriction of these two maps on $\mathbb{R}$ is the same as $i$.

But if we extend $D$ in severalways, things would be messy. Originally derivative is defined in thesense of limit, but for a non-differentiable function, we cannot dothat. We need an extension that makes most sense: it is by validatingintegration by parts. It seems like we are developingsome advanced concepts, but still we need to make use of elementaryones.

For $f(x)=\sin{x}$ and $\phi \in \mathscr{D}$, we have \[\Lambda_{f'}(\phi)=\int f'\phi = \int \phi\cos{x} =\underbrace{\phi\sin{x}|_{-\infty}^{\infty}}_{\text{zero}} -\int\phi'\sin x=-\Lambda_f(\phi')\] The derivative of $f$ isassigned to the derivative of $\phi$.Again we are using integration by parts. If $f$ is not assumed to be differentiable, wepretend it is, skip the body and jump to the resultimmediately. For example, $f(x)=|x|$is not differentiable, but we do that anyway: \[\int |x|'\phi = -\int |x|\phi'.\] In general for $f \inC^\infty$, we have (this can be verified by some computation)\[\Lambda_{D^k f}(\phi)=\int D^k f \phi = (-1)^k \int fD^k\phi = (-1)^k\Lambda _f(D^k\phi).\] Differentiation for distributions (on top of $C^\infty$ functions) should be in the sameshape, hence we define the $k$-th distributionderivative of a distribution $\Lambda$ by \[D^k\Lambda: \phi \mapsto (-1)^{k}\Lambda(D^k\phi).\] Since all $\phi$ areassumed to be of $C^\infty$, thereare no problem with this formula and this differentiation is defined forall $\Lambda$. We don't care aboutfirst order limit on a continuous but not differentiable function. Whatmatters here is the differentiation on test functions.

Why integration by parts

Try to recall what you have learnt about integration by parts. Wehave \[\int uv' = \int (uv)' - \int u'v\] because \[(uv)' = u'v+uv'.\] Therefore, if our generalisation of differentiation (though wedo not know how to do yet) pays respect to integration by parts, then wecan still work on product rule of differentiation, hence the usualformal rules of calculus would not go too far. If our extensionconflicts with integration by parts, then the ordinary meaning ofdifferentiation is damaged.

Let's sum up what has happened. We have obtained an inclusion \[C^\infty \subset C \subset \mathscr{D}'.\] Every distribution is infinitely differentiable becausefunctions in $\mathscr{D}$ are. If$f \in C^\infty$, then the $k$-th derivative can be understood in boththe sense of ordinary differentiation and the sense of distributionbecause it is given by \[\phi \mapsto (-1)^k\int f \phi^{(k)} = \int f^{(k)}\phi\quad \forall\phi \in \mathscr{D}.\] This is independent to the choice of $\phi$. If $h$ is a function such that $\int h\phi = \int f^{(k)}\phi$, then $h=f^{(k)}$.

If $f$ is merely continuous, stillwe can write the $k$-th derivative as\[\phi \mapsto (-1)^{k} \int f \phi^{(k)} \quad \forall \phi \in\mathscr{D}.\]

At this point, whether $f$ isdifferentiable or not is not of our concern. Since $\phi$ is smooth, the formula above iswell-defined. In general we don't even care whether $f$ is continuous or even integrable, aslong as it gives rise to a bounded linear functional,which can be guaranteed by being locally integrable. A functionis locally integrable if $\int_K|f|<\infty$ for all compact $K\subset \mathbb{R}$. In particular, $K$ can be taken to be any bounded closedinterval. As long as $f$ islocally integrable (for example, differentiable, continuous, or simplybounded), we can assign derivative in the new sense(integration by parts).

Product rule ofdifferentiation

We want something like $(fg)'=f'g+fg'$. To avoidconfusion we use $D$ to denote thederivative on distribution and $f'$ to denote the derivative in theordinary sense. This is pretty hard but for a multiplication of a $C^\infty$ function and a distribution itis not that hard. Suppose $\Lambda \in\mathscr{D}'$ and $f \inC^\infty$. We define their 'product' by \[(f\Lambda)(\phi) = \Lambda(f\phi).\] We have another distribution and derivative follows in anatural way: \[\begin{aligned} D(f\Lambda)(\phi) &=-(f\Lambda)(\phi') \\ &= -\Lambda(f\phi') \\\end{aligned}\] Meanwhile \[\begin{aligned}(f'\Lambda+fD\Lambda)(\phi) &=\Lambda(f'\phi)+D\Lambda(f\phi) \\ &=\Lambda(f'\phi)-\Lambda(f'\phi+f\phi') \\ &=-\Lambda(f\phi').\end{aligned}\] Things still work in this aspect.

We haven't verify convergence yet, but that requires much moreknowledge on functional analysis, so we don't do that here but in part2. Fortunately, things would go in an intuitive way.

Dirac delta distribution

Consider the linear functional on $\mathscr{D}$ by \[\delta(\phi)=\phi(0).\] This is bounded and is in fact our rigour definition of Dirac$\delta$ function (Von Neumann canrelax then!). It does have the required property. Say, if werealise this function as integration (informally) as \[\delta(\phi)=\int \delta\phi=\phi(0) \quad \forall \phi \in \mathscr{D},\] then $\delta$ can indeed beconsidered as a function whose support is the origin, and theintegral over $\mathbb{R}$ is $1$.

The derivative of $\delta$ is well-presented as well. Note$\delta'(\phi)=\delta(\phi')$,hence we have \[\delta'(\phi)=\phi'(0).\]

So much for part 1. If you don't have many background in functionalanalysis, then part 2 is not recommended, as you have no idea what isgoing on at all. It is not feasible to make part 2 to be readable tomore people.

Part II - Topologyand Calculus - a Overview

Here we provide some basic facts of test functions and distributions,assuming the reader some background in functional analysis. No proof isdelivered because if I do this post can be as long as I want. I hope byorganising facts here I can help you realise what is going on before youdrown yourself in details of a proof. It is recommended to see the tableof content on the right hand side first if you are on PC.

Topology

In brief, test functions are smooth functions with compact support.By the support of a function we mean theclosure of the set $\{x:f(x) \ne0\}$. Let $K$ be a compactset in $\mathbb{R}$, then $\mathscr{D}_K$ denotes a subspace of $C^\infty$ whose support lies in $K$. Since a closed subset of a compact setitself is compact, we see all functions in $\mathscr{D}_K$ have compact support.

Test function space is defined by \[\mathscr{D} := \bigcup_{K \text{ compact}}\mathscr{D}_K.\] And the distribution space $\mathscr{D}'$ is defined to be thedual space of $\mathscr{D}$, i.e. thespace of continuous linear functionals of $\mathscr{D}$. But if we don't know thetopology of $\mathscr{D}$, we cannotproceed. Here is how we attempt to establish the norm.

Topology attempt 1 - norm

Consider the norm for $\phi \in\mathscr{D}$ for all $N=0,1,2,\cdots$ by \[\| \phi \|_N = \sup_{x \in \mathbb{R}; n \le N}|D^nf|.\] This induces a local base \[V_N = \left\{ \phi \in \mathscr{D}_K:\|\phi\|_N \le \frac{1}{N} \right\}\quad (N=1,2,3,\cdots).\]

And we get a locally convex metrisable topology on $\mathscr{D}$.

If this topology makes $\mathscr{D}$ a Banach space, then it wouldbe fantastic - a lot of Banach space technique can be used. However,this topology is too small to be complete. One simply need toconsider this sequence: \[\psi_m(x)=\phi(x-1)+\frac{1}{2}\phi(x-2)+\cdots+\frac{1}{m}\phi(x-m)\] where $\phi \in\mathscr{D}_{[0,1]}$ and $\phi>0$ on $(0,1)$. This sequence is Cauchy but thelimit has no bounded support hence does not lie in $\mathscr{D}$.

Topology attempt 2 -enhancement

This time we do an enhancement on the previous topology,which makes $\mathscr{D}$ a locallyconvex topological space, which is complete and has the Heine-Borelproperty (closed and bounded set is compact and vice versa). We stillneed the topology defined in our first attempt. It is broken into threesteps:

For each compact set $K$, let$\tau_K$ denote the subspace topologyof $\mathscr{D}$ defined in attempt1.
Let $\beta$ be the collection ofall convex balanced set $W \subset\mathscr{D}$ such that $\mathscr{D}_K\cap W \in \tau_K$ for all compact $K$. (A set $W$ is balanced if $\alpha{W} \subset W$ for all $|\alpha| \le 1$.)
The new topology $\tau$ isdefined to be the collection of all unions of sets of the form $\phi + W$ with $\phi \in \mathscr{D}$ and $W \in \beta$.

This is the topology we want, and one can indeed verify that $\tau$ is a topology, with local base $\beta$. This topology has the followingproperties:

$\tau$ makes $\mathscr{D}$ a locally convex topologicalvector space.
$\mathscr{D}$ has the Heine-Borelproperty.
In $\mathscr{D}$, every Cauchysequence converges.

Locally, the topology of $\mathscr{D}_K$ is the same as $\tau_K$. Hence we can still useproperties of these norms if we want. In fact, this $\tau_K$ makes $\mathscr{D}_K$ a Fréchet space, i.e.locally compact and complete metric space.

Continuity and category

We cannot discuss continuity without topology. But still continuityhas to be treated carefully. For example the space $L^p([0,1])$ with $0<p<1$ is weird: the dual space istrivial, due to its topology: the only two open convex sets are emptyset and itself. Fortunately we have the following, which is quiteintuitive.

Suppose $\Lambda$ is a linearmapping of $\mathscr{D}$ into alocally compact convex space $Y$(which can be $\mathbb{R}$, $\mathbb{C}$ or $\mathscr{D}$ itself). Then the followingare equivalent:
$\Lambda$ is continuous. (We careabout the behaviour of $\mathscr{D}'$)
$\Lambda$ is bounded. (You musthave learnt the equivalence of 1 and 2 already)
$\phi_i \to 0$ in $\mathscr{D}$ implies $\Lambda\phi_i \to 0$ in $Y$.
The restriction of $\Lambda$ toevery $\mathscr{D}_K$ iscontinuous.

In particular, it follows that the differential operator $D^n$ is continuous for all $n$. We also have some knowledge of thebehaviour of $\mathscr{D}'$now:

If $\Lambda$ is a linearfunctional on $\mathscr{D}$, then thefollowing are equivalent:
$\Lambda \in\mathscr{D}'$.
To every compact set $K$ therecorresponds a nonnegative integer $N$and a constant $C<\infty$ suchthat the inequality
\[|\Lambda\phi| \le C \|\phi\|_N\]
holds for every $\mathscr{D}_K$.

Consider the Dirac distribution on $x$ given by \[\delta_x(\phi)=\phi(x)\quad \phi \in \mathscr{D}.\] This is indeed a distribution. The case when $x=0$ gives us the Dirac function inphysics. Note \[\mathscr{D}_K = \bigcap_{x \in K^c}\ker\delta_x,\] $\mathscr{D}_K$ is aclosed subspace of $\mathscr{D}$. Since $\mathscr{D}_K$ is also nowhere dense, andthere is a countable collection of $K_i\subset \mathbb{R}$ (for example $K_i=[-i,i]$) such that $\mathscr{D} = \bigcup \mathscr{D}_i$ (ofthe first category), and $\mathscr{D}$ itself is complete, by Baire's Category Theorem, $\mathscr{D}$ is not metrisable. This is aflaw of the topology of $\mathscr{D}$, though is not thattroublesome.

Calculus of distributions

We have shown that every $C^\infty$ functions can be considered as adistribution. In general, for a function $f$ one only need to require that $f$ is locally integrable,i.e. for every compact set $K$ wehave \[\int_K |f|<\infty.\] If we define $\Lambda_f:\phi\mapsto \int f\phi$, we see \[|\Lambda_f(\phi)|\le \left( \int_K |f| \right)\sup|\phi|, \quad \phi \in\mathscr{D}_K.\]

In particular, at the very least, all $L^1$ functions can be considered asdistributions.

On the other hand, if $\mu$ is apositive measure on $\mathbb{R}$ with$\mu(K)<\infty$ for all compact$K$, then \[\Lambda_\mu:\phi \to \int \phi d\mu\] also defines a distribution.

Absolute continuity

We know the fundamental theorem of calculus in $L^1$ only hold when the function $f$ is absolutely continuous. TheCantor function $f$ is differentiablealmost everywhere on $[0,1]$ but\[\int_0^1 f'(x)\mathrm{d}x = 0, \quad f(1)-f(0)=1.\] This restriction still makes sense here. Pick $f$ to be a left-continuous function withbounded variation. Then it can be shown that \[D\Lambda_f = \Lambda_\mu\] where $\mu([a,b))=f(b)-f(a)$. Hence $D\Lambda_f=\Lambda_{Df}$ if and only if$f$ is absolutelycontinuous.

Convergence (uniform?)

We consider the weak*-topology of $\mathscr{D}'$ by \[\Lambda_i \to \Lambda: \lim_{i \to \infty}\Lambda_i\phi = \Lambda\phi\quad \forall \phi \in \mathscr{D}.\] Then fortunately this limit operator commutes withdifferential operator in a natural way, which may remind you of uniformconvergence. In fact, \[\Lambda_i \to \Lambda \implies \Lambda \in \mathscr{D} \text{ and}D^k\Lambda_i \to D^k\Lambda \quad \forall k=1,2,\cdots.\] To prove this one needs Banach-Steinhaus theorem. Hereconcludes our four requirements of distributions.

Convolution

Convolution plays an important role in Fourier analysis, and here ishow to invite distribution to the party.

Normally for two $L^1$ functions$f,g$ we define \[(f \ast g)(x)=\int_\mathbb{R}f(y)g(x-y)\mathrm{d}y.\] We can create more symbols to make life easier:

$\tau_xu(y)=u(y-x)$.
$\check{u}(y)=u(-y)$.

It follows that $\tau_x\check{u}(y)=\check{u}(y-x)=u(x-y)$.Hence \[(f \ast g)(x) = \int_\mathbb{R} f(y)(\tau_x\check{g})(y)\mathrm{d}y.\] It shows that $g \to (f \astg)(x)$ is actually a linear functional of $\Lambda_f$, $\tau_x$ and $g\mapsto \check{g}$. But $\Lambda_f$ itself can be a distribution,hence we define convolution for a distribution and a smooth function by\[L \ast \phi(x) = L(\tau_x\check{\phi}), \quad L \in \mathscr{D}',\phi \in \mathscr{D}.\] Convolution can be characterised in a natural way. In fact,for any $T:\mathscr{D} \to C^\infty$,if \[\tau_x T = T\tau_x,\] then there is a unique $L \in\mathscr{D}'$ such that \[T\phi = L\ast \phi.\] As you can imagine, this setting creates a lot of potentialsfor Fourier transform.

References and FurtherReading

Walter Rudin, Functional Analysis, Second Edition. (Part IIof the book)
Peter Lax, Functional Analysis. (Appendix B)
StanfordEncyclopedia of Philosophy Archive (Fall 2018 Edition), QuantumTheory: von Neumann vs. Dirac.