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I'm having a little bit of trouble understanding Dirac Delta, or rather, finding a proper definition. I understand the way it is "found" by using Fourier transforms on a function, and that it isn't really a function but a distribution. I also get the idea that I shouldn't look at it as a common Riemann integral. But then, why are there limits of functions that represent the same mathematical object? Specifically I can't see how the following would be true: $$\int_{-\infty}^{\infty}\delta(x)dx = \int_{-\infty}^{\infty}\lim_{\sigma\to 0}\frac{1}{\sqrt{2\pi\sigma²}}e^{\frac{-x²}{2\sigma²}}dx = 1$$ If I think about the evolution of amplitude and width of the Gaussian as $\sigma$ approaches zero it kind of makes sense, but then, that's the "intuitive" and not rigorous definition of the Delta (being 0 everywhere except at $x=0$ where it is "$\infty$"). To summarize, is there a proper definition of the Delta that would allow me to show the relation above? Or any explanation as to "why the limit has the same properties"? Thanks!

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    $\begingroup$ The equation in your question is an informal way of giving an intuition for the delta function. You can make the delta "function" rigorous by using the theory of distributions or by using Lighthill's concept of generalized functions, but either way you're talking about a book length exposition. $\endgroup$ – Brian Borchers Aug 14 '18 at 1:44
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Generally, if $G_i$ is a sequence of distributions, their limit (if it exists) is given pointwise; i.e. by the formula

$$ \left( \lim_i G_i \right) [f] = \lim_i \left( G_i[f] \right) $$

for every test function $f$.

The problem begins when we introduce the faux-integral notation for evaluating a distribution. Recall that

$$ \int_{-\infty}^{\infty} F(x) f(x) \, \mathrm{d}x := F[f]$$

is how we define the meaning of the integral-like notation on the left. Using this, the limit above is written

$$ \int_{-\infty}^{\infty} \left( \lim_i G_i \right)(x) f(x) \, \mathrm{d} x = \lim_i \int_{-\infty}^{\infty} G_i(x) f(x) \, \mathrm{d} x$$

Notice how the limit has magically transported from inside the integral to outside the integral.

The notation problem becomes serious when we use another form of shorthand. For any (sufficiently nice) function $h$, let me introduce the notation $\widehat{h}$ to mean the distribution defined by the formula

$$ \widehat{h}[f] := \int_{-\infty}^{\infty} h(x) f(x) \, \mathrm{d} x $$

Note that the integral on the right is an ordinary integral of functions. Suppose that each of the $G_i$ in the example above is of the form $G_i = \hat{g_i}$. Then the integral becomes

$$ \int_{-\infty}^{\infty} \left( \lim_i \hat{g}_i \right)(x) f(x) \, \mathrm{d} x = \lim_i \int_{-\infty}^{\infty} g_i(x) f(x) \, \mathrm{d} x$$

The final abuse of notation is that people often don't decorate $g$ at all when they do this, and so we have the horribly notated that fact

$$ \int_{-\infty}^{\infty} \lim_i g_i (x) f(x) \, \mathrm{d} x = \lim_i \int_{-\infty}^{\infty} g_i(x) f(x) \, \mathrm{d} x$$

where almost nothing on the left hand side means what it looks like it means.

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  • $\begingroup$ Thanks, I think this makes it easier to understand. At least the abuse of notation seems clear and now I see where the problem comes from. But I have a doubt left: how can we be sure that a limit actually represents the delta? $\endgroup$ – Solfreludio Aug 14 '18 at 13:50
  • $\begingroup$ @Solfreludio: By actually working out the limit of the integral. The basic idea is that for tiny $\sigma$ and sufficiently nice $f$ (e.g. you might be using Schwartz functions, as your "test functions"), the integrand is negligible outside of a tiny interval, and $f$ is nearly constant on that interval. You play the real analysis game of bounding the errors, and deduce the limit of the integral converges to $f(0)$. $\endgroup$ – Hurkyl Aug 14 '18 at 15:25
  • $\begingroup$ Note, incidentally, that if you are using a space of distributions whose test functions are Schwartz functions (or if they are smooth functions with compact support), then $\int_{-\infty}^{\infty} \delta(x)\, \mathrm{d}x$ is actually undefined; the constant function $1$ is not a test function, and so it doesn't make sense to convolve it with the dirac delta. $\endgroup$ – Hurkyl Aug 14 '18 at 15:27
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On a certain class $F$ of real functions that is also a vector space over $\Bbb R$ we have may have a class $G$ of linear operators mapping $F$ to $\Bbb R,$ where each $g\in G$ is identified with a real function $g^*$ such that $g(f)=\int_{\Bbb R}f(x)g^*(x)dx.$ And we have the linear operator $D(f)=f(0),$ which (usually) is not in $G.$ But $D$ is (usually) a point-wise limit of members of $G,$ in that there is a sequence $(g_n)_n$ in $G$ such that for every $f\in F$ we have $f(0)=\lim_{n\to \infty}g_n(f)=\lim_{n\to \infty}\int_{\Bbb R}f(x)g_n^*(x)dx.$

There are advantages to writing $D$ as if it belonged to $G$ and as if there were some function $\delta=D^*$ such that $f(0)=D(f)=\int_{\Bbb R}f(x)\delta (x)dx.$ Some of the formulas involving integration remain valid, and it easier to discuss $D$ and members of $G$ together.

Reference topic: Heavide Calculus /Heaviside Operational Calculus.

The $F$ and $G$ above vary , depending on the topic, context, and application.

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  • $\begingroup$ Thank you, I will check the references. The fact that it's not an integral and yet some properties are the same its still a bit confusing to me. Specially when showing some proof of these properties by using integration. $\endgroup$ – Solfreludio Aug 14 '18 at 13:53
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The way I was taught, the delta function was defined by the expression:

$$ \int_{-\infty}^\infty \delta(x)f(x)dx = f(0) $$

(For continuous $f$.)

This then explains why $\int_{-\infty}^\infty \delta(x)dx = 1$, just let $f(x)=1$. To make your other integral work, it would only really make sense to have the limit on the outside of the integral, like so:

$$ \lim_{\sigma\to0} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}} e^{-x^2/2\sigma^2} dx $$

$$ =\lim_{\sigma\to0} 1 = 1 $$

EDIT: In general, it is probably best to put the limit on the very outside of whatever expression using the delta function you are computing. As you decrease $\sigma$, $\frac{1}{\sqrt{2\pi\sigma^2}} e^{-x^2/2\sigma^2}$ approximates the delta function better and better, so in theory, the expression should approach its true value.

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  • $\begingroup$ FYI, what's happening is that if we have functionals (e.g. $f \mapsto f(0)$), then we define the limit of functionals pointwise; $(\lim_i F_i)(f) = \lim_i( F_i(f)) $. The problem comes when we write evaluation using this faux-integral notation: the definition of limit becomes $$\lim_i \int F_i(x) f(x) dx = \int (\lim_i F_i)(x) f(x) dx$$ Extra confusion arises when $F_i$ is a functional of the form $f \mapsto \int g_i f dx$, and we further abuse notation to write $g_i(x)$ in place of $F_i(x)$. $\endgroup$ – Hurkyl Aug 14 '18 at 2:50
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The proper definition (and by proper I mean what we did in my class) is of a rectangle that gets thinner and thinner, but longer and longer. Say we define $d_c(x)$ as the following

$$ d_c(x)=\begin{cases} 0 & x < -\frac{1}{2c} \\ c & -\frac{1}{2c} \leq x \leq \frac{1}{2c} \\ 0 & x > \frac{1}{2c} \end{cases} $$

As you can see, this makes a rectangle. Solving for the area we get $(\frac{1}{2c}+\frac{1}{2c})c = 1$ no matter what $c$ value we say. We then define the Dirac Delta function as

$$ \delta(x) = \lim_{c\rightarrow\infty}d_c(x) $$

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  • $\begingroup$ The problem with this typical sort of exposition is that it looks like "define a sequence of functions and take the limit", which is very much the wrong thing to do here. This is only a correct calculation if you're in on the secret of what it all "really means" -- and the OP wants to know the secret. $\endgroup$ – Hurkyl Aug 14 '18 at 2:46

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