The unitary pulse function (or sample function) is defined as follow:

Let's $\newcommand\R{\mathbb R}d_1:\R\to\R$ be a positive intrgrable function such that $$\int_{-\infty}^{\infty}d_1(x)dx=1.$$ (The usual example is $d_1(x)=1$ for $x\in(0,1)$ and $d_1(x)=0$ otherwise.)

Now: let $d_n:\R\to\R$ be defined as $d_n(x)=nd_1(nx)$. The unitary pulse function is then defined as $$d=\lim_{n\to\infty}d_n.$$

Some properties:

  1. $d(x)=0$ for $x\in\R\setminus\{0\}$.

  2. $\int_{-\infty}^{\infty}f(x)d(x-s)dx=f(s)$ (hence the name “sample function”).

However $d$ is not a function. At least $d$ is not a function in the space of real numbers as $d(0)\notin\R$. We might say that $d(0)=+\infty$ in the closure $\R_{-\infty}^{+\infty}=\R\cup\{-\infty,+\infty\}$; however $d$ is scalable. v.g. $$\int_{-\infty}^{\infty}f(x)\cdot(3d(x))dx=\int_{-\infty}^{\infty}(f(x)\cdot3)d(x)dx=3f(0)$$ which would suggest $3d(0)\ne d(0)$, unlike $\infty\in\R_{-\infty}^{+\infty}$ where $3\infty=\infty$.

So. How can I characterize $\R_d$ the co-domain of $d$?

This $\R_d$ would have the following characteristics:

  1. $d$ is a function $d:\R\to\R_d$.
  2. $\R\subset\R_d$ (which means that $f:\R\to\R$ then $f:\R\to\R_d$).
  3. $\forall a\in\R:a\cdot d(0)\in\R_d$.
  4. $a\cdot d(0)=d(0)\implies a=1$

Is this space $\R_d$ described in literature?

Instead of extending de co-domain $\R$ into a higher space to make $d$ a function, is it better to extend the spaces of functions $\R^{\R}$ to make $d$ a member of the new space?

  • $\begingroup$ It seems to be more common to treat $d$ as a functional on the space of test functions. $\endgroup$
    – user99914
    Dec 10, 2013 at 5:02
  • 1
    $\begingroup$ There's basically no way to make this idea work, and one can give a proof that statement (depending on how you want to formalize it). The accepted way to handle this is the theory of distributions, or functionals on a certain space of functions. I can provide more information if you would like. $\endgroup$
    – Potato
    Dec 10, 2013 at 5:04

1 Answer 1


It doesn't quite work out, algebraically speaking, to extend $\mathbb{R}$ into a larger space to serve as a codomain for $d(x)$. We could try just adjoining some infinities onto $\mathbb{R}$, but, while this can work out topologically, it's hard to find a sensible way to do algebra.

However, we usually don't need to do algebra with a "function" like $d$: often, we only care about how it behaves with integration, bringing us to the idea of a distribution, or generalized function.

To get a space of distributions, we start with some nice space $F$ of functions $\mathbb{R}\mapsto\mathbb{R}$, often the space of smooth, compactly supported functions, and let a distribution be a linear map $F\mapsto F$. We interpret, intuitively, the evaluation of a distribution $D$ at a function $f$ as the "integral" $\int f(x) D(x) dx$. For example, your $d$ would be the functional sending a function $f$ to $f(0)$, and we can embed the space of integrable functions $\mathbb{R}\mapsto\mathbb{R}$ into the space of distributions with usual integration.

Note that, by this definition of a generalized function, we don't care about the values taken on by the function itself, only its behavior under integration. Although you can sometimes define the evaluation of a distribution at a point, essentially locally inverting the embedding of normal functions into distributions, this doesn't always work. For example, although, for your $d$, saying $d(x)=0$ for $x\ne0$ makes sense, there is no way of saying what $d(0)$ is.


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