2
$\begingroup$

In a book I'm reading (functional analysis of Stein and Shakarshi), they say that a distribution $F$ is not be given by assigning value of $F$ at most point, but will instead be determinated by its average taken with smooth function. So if we think a function $f$ as a distribution $F$, we determine $F$ by the quantity $$F(\varphi )=\int_{\mathbb R}f(x)\varphi (x)dx,$$ where $\varphi $'s range over an appropriate test function.


1) Something I don't get, when they say : "if we think a function $f$ as a distribution $F$", does this mean that $f$ and $F$ are the same ? How can we think $f$ as $F$ ? (or $F$ as $f$ ). But $f$ and $F$ looks very different...

2) A little bit farther for example, it says that if $f$ is smooth, then $$\int f'\varphi =-\int f\varphi '.$$ Therefore, if $f\in L^1(\mathbb R)$ locally, we can define the derivative of $f$ as the distribution as $F'(\varphi )=-\int f\varphi '$. So $F'$ is considered as the dérivative of $f$ or not ? I don't understand. It's the same ?

3) Also, it says that $F$ is not a function. For me it's a function from $$\mathcal C_0^\infty(\Omega )\to \mathbb R.$$ So in what isn't it a function ? For example, let $$F(\varphi )=\int_{\mathbb R}f\varphi ,$$ where $f\in L^1$ and $\varphi $ smooth. In what isn't it a function ? $F(\varphi )$ is well defined (I guess). May be someone could explain ?

$\endgroup$
  • $\begingroup$ Welcome to math.SE. Could you let us know which book you are reading? That would help people give more informed answers, and make the post more useful in the future. $\endgroup$ – Carl Mummert Nov 10 '18 at 16:18
  • $\begingroup$ Let $(f_n)$ be a sequence of functions ($C^\infty$, continuous, locally integrable as you wish). When do we have that $\lim_{n \to\infty} \int_{\mathbb{R}} f_n(x)\varphi(x)dx$ converges for any $\varphi \in C^\infty_c$ ? Answer : when $(f_n)$ converges in the sense of distributions. $\endgroup$ – reuns Nov 10 '18 at 20:26
  • $\begingroup$ @CarlMummert: Functional analysis of Stein and Shakarshi $\endgroup$ – sam Nov 11 '18 at 11:07
  • $\begingroup$ @reuns: And so ? in what this convergence is benefit ? $\endgroup$ – sam Nov 11 '18 at 11:08
  • $\begingroup$ Your problem is more linguistic than mathematical. $F$ is not a function does not mean $F$ can never be expressed as an ordinary function. Sometimes it can be, as is shown in your example; sometimes it's not so. $\endgroup$ – Cave Johnson Nov 11 '18 at 11:19
3
$\begingroup$

The book is pretty misleading, a distribution $F$ is indeed a linear function $F: C^{\infty}_0 (\Omega) \to \mathbb R$.

Given any $f \in C_0^{\infty}(\Omega)$ we can define a corresponding distribution $F_f$ by

$$ F_f(\varphi) = \int f \varphi \; \text{d} \varphi $$

This is common enough that people often abuse notation and write $f$ for both the function $f \in C_0^{\infty}(\Omega)$ and the distribution $F_f: C_0^{\infty}(\Omega) \to \mathbb R$. $f$ and $F_f$ are indeed very different objects, but people will use $f$ to refer to both. Normally this won't cause much confusion as it will be quite clear which they are referring to by what they're acting on.

This all gets at the motivation behind defining what a distribution is. The classic example of a distribution is the Dirac Delta $\delta_0$. You've probably already met this at some point in your studies but in a very informal way, where you've said it's a function such that $$ \int \varphi(x) \delta_0(x)\; \text{d}x = \varphi(0) $$ Of course this is actually nonsense, no such function exists. But we really want to pretend like such a function exists.

Instead we can formalise this idea as a distribution. Considering $\delta_0$ as a distribution $C^{\infty}_0(\mathbb R) \to \mathbb R$ defined by $\delta_0(\varphi) = \varphi(0)$ makes perfect sense. However we still want to keep the intuition that "$\delta_0$ is a function such that $\int \varphi(x) \delta_0(x)\; \text{d}x = \varphi(0)$", despite that fact that such a function does not exist.

For example, how would we define the derivative of $\delta_0$? Let's pretend that $\delta_0$ is in fact a $C^{\infty}_0(\mathbb R)$ function. Then using integration by parts we have $$ \int \varphi(x) \delta'_0(x) \; \text{d}x = - \int \varphi'(x) \delta(x)\; \text{d}x = - \varphi'(0) $$ (since the supports are compact, all boundary terms disappear). Now $\delta_0'(f) = -\varphi'(0) = -\delta_0(\varphi')$ is a perfectly legitimate distribution, and this leads us to an idea to define the derivative of a distribution $F$ as $F': C^{\infty}_{0}(\Omega) \to \mathbb R$ defined by $$ F'(\varphi) = - F(\varphi') $$ This is an idea that permeates distribution theory. Whenever we want to manipulate a distribution, we pretend as if the distribution is actually a $C^{\infty}_0(\Omega)$ function and try and get to something which we can define via the distributions.

One thing I haven't mentioned is that not every linear function $F: C_0^{\infty}(\Omega) \to \mathbb R$ is a distribution, they have to satisfy a continuity condition. With this condition, as reuns mentions below, every distribution $F$ can actually be written as a limit of $C^{\infty}_0(\Omega)$ functions $f_n$ "understood" as distributions $F_{f_n}$. Hence pretending a distribution is a $C^{\infty}_0(\Omega)$ function is not as silly as it might sound.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "The book is pretty misleading." Why? Didn't the book say the same line "a distribution is a continuous linear functional $C_0^\infty(\Omega)\to\mathbb R$" as you do? $\endgroup$ – Cave Johnson Nov 11 '18 at 11:23
  • $\begingroup$ The person asking the question hadn't specified the book when I posted this answer, according the question the book had said that $F$ is not a function $\endgroup$ – bitesizebo Nov 12 '18 at 1:21
2
$\begingroup$

What is a distribution?

It is linear map from $C_0^\infty(\Omega)$ to $\mathbb R$ which is continuous with respect to a very particular topology (here there are some informations about it).

Why it's not really a function?

Because for who say that a distribution is not a function, the word "function" means "real-valued function of a real variable".

1) Something I don't get, when they say: "if we think a function $f$ as a distribution $F$", does this mean that $f$ and $F$ are the same?

No. This means that the (real-valued) function (of a real variable) $f$ can be identified with a very particular distribution, namely, the distribution $T_f:C_0^\infty(\Omega)\to\mathbb R$ defined by $$T_f(\varphi)=\int_{\Omega}f(x)\varphi(x) \;dx\tag{1}$$ which usually is represented by $F$. Here, "can be identified" means that the map $f\mapsto T_f$ is linear, continuous and injective.

How can we think $f$ as $F$?

We use the formula $(1)$.

(or $F$ as $f$).

In general this is not possible: there are distributions which are not defined by a (real-valued) function (of a real variable). This is the meaning of the sentence "some distributions are not functions". Explicitly: "the distribution $T$ is not a function" means that there is no locally integrable function $f$ such that $$T(\varphi)=\int_\Omega f(x)\varphi(x)\;dx,\quad \forall \ \varphi\in C_0^\infty(\Omega).$$

But $f$ and $F$ looks very different...

They are different, but there are analogous cases in mathematics: a rational number (ratio $p/q$) is different from a real number (Dedekind cut). However, we say that every rational number is a real number, but a real number is not necessarily rational. The same ideia is used in the sentence "every locally integrable function is a distribution but a distribution is not necessarily a function". This analogy is presented by Schwartz (who created the theory of distributions) in this book, p. 212.

2) A little bit farther for example, it says that if $f$ is smooth, then $\int f'\varphi =-\int f\varphi '.$ Therefore, if $f\in L^1(\mathbb R)$ locally, we can define the derivative of $f$ as the distribution as $F'(\varphi )=-\int f\varphi '$. So $F'$ is considered as the dérivative of $f$ or not? I don't understand. It's the same?

Under the described identification, not every function has a derivative which is a function, but all functions have derivatives which are distributions, and every distribution has a derivative which is a distribution. These are words of Schwartz in the same book and page mentioned above. Therefore, every locally integrable function is infinitely differentiable in the distributional sense.

3) Also, it says that $F$ is not a function. For me it's a function from $\mathcal C_0^\infty(\Omega )\to \mathbb R.$ So in what isn't it a function. For example, let $F(\varphi )=\int_{\mathbb R}f\varphi,$ where $f\in L^1$ and $\varphi $ smooth. In what isn't it a function? $F(\varphi )$ is well defined (I guess). May be someone could explain?

This was already explained, and you are right: it is indeed a function, but it is not a function of a real variable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think "every locally integrable function is infinitely differentiable in the distributional sense" is better, $C^\infty$ starts to reference continuity which is sort of irrelevant here. $\endgroup$ – Ian Nov 16 '18 at 2:27
  • $\begingroup$ @Ian Edited, thanks. $\endgroup$ – Pedro Nov 16 '18 at 2:31

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.