# How do distributions generalize functions?

So I've been reading Grafakos's book on Fourier analysis and the chapter dedicated to distributions. And although it provides a lot of details, I came out of it without understanding the essence of the topic. I wanted to understand how the notion of a distribution generalizes/extends the idea of functions. I know that some functions define distributions but that only gives me a correspondence without necessarily knowing how the distribution behaves similar to the function. For instance, the Dirac delta (distribution) can be imagined as a pulse at the origin, or as the limit of the functions $$\frac{n}{2}\chi_{[-\frac{1}{n},\frac{1}{n}]}$$ (explained below). The action of the delta on $$f$$ should produce $$f(0)$$, so we think of a possible "function" $$\phi$$, integrating $$f$$ against which would give $$f(0)$$. Looking at the Riemann sums for the integral, we can ask that $$h\phi(0)f(0)=f(0)$$ for every small $$h$$. This suggests that the delta is an infinitely large pulse at the origin. Again, delta is the derivative of the Heaviside function. Now the Heaviside function can be approximated by the sequence $$f_n(x)=\begin{cases} 0, \:\:x<-\frac{1}{n}\\ \frac{n}{2}x+\frac{1}{2}\: \:\:|x|\leq\frac{1}{n}\\ 1,\:\: x>\frac{1}{n} \end{cases}$$ Whose derivatives $$f_n'(x)=\begin{cases} 0, \:\:x<-\frac{1}{n}\\ \frac{n}{2}\: \:\:|x|<\frac{1}{n}\\ 0,\:\: x>\frac{1}{n} \end{cases}$$ are (almost everywhere) equal to $$\frac{n}{2}\chi_{[-\frac{1}{n},\frac{1}{n}]}$$.

Both the above arguments (however informal they may be) convince me that the Dirac delta distribution has some likeness to a traditional function.

Is there a book/lecture that would allow me to view all distributions as generalizations of functions? Thank you all in advance.

• I understand your question and respect it. However, may I ask why do you want to see distributions as functions? A distribution is a linear map defined on a space of functions. Is it this point that you have issues to understand. Because that is by essence distributions. Jul 18 at 7:52
• @mathcounterexamples.net No, I do not wish to see distributions "as" functions. I wish to see a resemblance to functions (as I've discussed for the Dirac delta). They are called "generalized functions" after all. Jul 18 at 7:56
• This is mainly because for any function $f$ the linear map $\phi_f : g \mapsto \int fg$ is a distribution. However any distribution cannot be written in that way. When you’re confortable with this fact, everything is fine. No? Jul 18 at 7:59

Every locally integrable fucntion on $$\Bbb{R}^n$$ induces a distribution on $$\Bbb{R}^n$$ via integration: i.e we have an injective linear mapping $$\iota:L^1_{\text{loc}}(\Bbb{R}^n)\to\mathcal{D}'(\Bbb{R}^n)$$, given by taking $$[\iota(f)](\phi):=\int_{\Bbb{R}^n}f(x)\phi(x)\,dx$$. Note that this mapping $$\iota$$ is merely an injection, not a surjection (case in point the dirac delta $$\delta$$ is not in the image of $$\iota$$). What this says is that every locally integrable function gives rise to a unique distribution, but not every distribution arises form a locally integrable function. So, the fact that we only have an injective, but not surjective mapping is why the concept of a distribution is a strict generalization of the concept of a locally integrable function.

I'm not sure what you mean by $$\delta$$ has some likeness to a traditional function. It can be expressed as a limit (with respect to the weak* topology on $$\mathcal{D}'(\Bbb{R}^n)$$, as opposed to pointwise convergence) of locally integrable functions, but why should this mean that $$\delta$$ ought to be/behave like a usual function?

Just because a bunch of quantities behave a certain way, why should we expect that a "limiting object" ought to behave in the same way? For example, not every limit of rational numbers is again a rational number.

For motivating the definition of a distribution as a continuous linear functional on the space $$\mathcal{D}(\Bbb{R}^n)$$ of test functions, I found Amann and Escher's Analysis Volume III very helpful. Take a look at pages 177-179, and just read the text, and skip all the theorems if you wish on a first reading. The motivation for distributions as "measurement devices" was a very helpful analogy for me.

• That is why I said "behaves like a function" instead of "is a function". Looking at the Dirac delta and the term "generalized functions" I had reasons to believe that such resemblance can be found for all distributions. Jul 18 at 8:07
• Is it difficult to prove that the regular distributions are dense in the space of distributions? Jul 18 at 8:30
• @HritRoy it's a two-step process, but getting the definitions and some of the details right can be tricky. Given any distribution on an open set $U$, say $F\in\mathcal{D}'(U)$, we approximate $F$ by a sequence of distributions $G_j$ of compact support, which increases to $U$. Next, we take an approximate identity of smooth functions $\{\psi_t\}_{t>0}$ (i.e something which approximates the dirac delta), and then take the convolution $G_j*\psi_t$. Then for $j$ very large and $t>0$ very small, $G_j*\psi_t$ will be a smooth compactly supported function which approximates $F$ well. Jul 18 at 8:40
• See Folland's real analysis, or probably Rudin's functional analysis for more details. Jul 18 at 8:40
• @HritRoy That's a nice way to put it but being slightly more precise with the wording: for distributions one should abandon pointwise notions, so I'd simply say "A distribution is something which has continuously varying 'weighted-averages' (roughly speaking of course)". I avoid the term "smooth" because it's not appropriate here; continuity is. I added the adjective "weighted" because in the case of a usual function $f$, we regard $\int f\phi$ as an average of $f$ weighted by the function $\phi$. Jul 18 at 9:13

You don't necessarily need a book to understand how distributions generalize functions: Basically any test function is itself a distribution.

For instance, you may consider the test function space of (complex valued) compactly supported smooth functions $$\mathcal{D} = C_0^\infty(\mathbb{R})$$, which is a subspace of $$L^2(\mathbb{R})$$. So scalar products between any $$\phi, \psi \in \mathcal{D}$$ are well-defined via $$(\phi, \psi) = \int_\mathbb{R} \phi(x)^* \psi(x) \; dx$$

(with $$^*$$ being ocmplex conjugation). Now, you may equally interpret $$\phi$$ as a linear map, $$\mathcal{D} \to \mathbb{C}$$, which sends $$\psi \in \mathcal{D}$$ to $$(\phi, \psi) \in \mathbb{C}$$. Such a linear map is a distribution (i.e.,$$\phi \in \mathcal{D}'$$), if it has some nice boundedness properties (see Wikipedia on distributions). And one can indeed verify them for any $$\phi \in \mathcal{D}$$.

So any test function $$\phi \in \mathcal{D}$$ is in particular a distribution $$\phi \in \mathcal{D}'$$, meaning the distributions $$\mathcal{D}'$$ generalize the test functions $$\mathcal{D}$$.

As you noticed correctly, not every distribution is a test function, for instance the Dirac delta distribution with peak at $$x_0 \in \mathbb{R}$$, called $$\delta_{x_0}(x)$$ would have to be zero everywhere except at $$x_0$$, if it was a function. Now, the set $$\{x_0\}$$ has Lebesgue measure 0 (or in other words, it is null). And the theory of $$L^p$$-spaces tells us that a function which is everywhere zero, except for a null set, is equivalent to a function being zero everywhere. So we cannot make sense of $$\delta_{x_0}(x)$$ as element of any $$L^p$$-space (such as $$L^2$$). And since $$\mathcal{D} \subset L^2(\mathbb{R})$$, it s neither a test function. I.e. $$\delta_{x_0} \in \mathcal{D}'$$, but $$\delta_{x_0} \notin \mathcal{D}$$.

A good reference for further reading on distributions is The Analysis of Linear Partial Differential Operators by Lars Hörmander, Volume I (2nd edition), chapters 1 and 2.