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I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function $$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$ then it is readily shown that $$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define $$\delta(x) = \lim_{a \to 0} \delta_a (x).$$ Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations $\delta(x) = 0$, if $x = 0$ $$\int_{-\infty}^\infty \delta (x) d x = 1.$$

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    $\begingroup$ Defining the delta function at $x=0$ is meaningless as a math concept. To be rigorous use Schwartz distribution idea. The delta function is a shorthand way of expressing the concept and is meaningful only under an integral sign. $\endgroup$ – herb steinberg Mar 15 at 21:49
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    $\begingroup$ Re: "he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$": That's actually even more wrong than $\delta (0) = \infty$; on the contrary, $\delta (0) \to 0$ as $x \to 0$. $\endgroup$ – ruakh Mar 16 at 6:01
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    $\begingroup$ FWIW, my Physics Profs presented it with the correct definition, i.e. a function whose integral = 1 but which is zero everywhere except at x= 0 ; then went on to say "physicists found this very useful while mathematicians shrieked in horror until they decided it could be the limit of a space of functions..." . Granted this is a highly biased presentation of the function's history. $\endgroup$ – Carl Witthoft Mar 16 at 12:30
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    $\begingroup$ As a further note: if one defined $\delta(0)=\infty$, how would one then distinguish between $\delta$ and $2\delta$? $\endgroup$ – Greg Martin Mar 16 at 16:25
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    $\begingroup$ Sadly, the text you are quoting has the definition of $\delta$-function that is plain wrong. "Letting $a\to 0$", we don't see that $\int_{-\infty}^\infty \delta(x)\,dx=1.$ What we do see is $\lim_{a\to 0}\int_{-\infty}^\infty \delta_a(x)\,dx=1,$ which is very different from the claimed $\int_{-\infty}^\infty \lim_{a\to 0} \delta_a(x)\,dx=1.$ Luts Lehmann's answer below has a convincing counterexample to such naive exchange of limit and integral. $\endgroup$ – Kostya_I Mar 16 at 22:18

11 Answers 11

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No. Defining $\delta(0)=\infty$ is not at all right, because it has to equal $\infty$ in just the "right way". You can define this as a distribution, which ignoring all technicalities, just means that it is an "evaluation linear map". I.e, given an appropriate vector space $V$ of functions $\Bbb{R}^n\to\Bbb{C}$, we consider $\delta_a:V\to\Bbb{C}$ defined as $\delta_a(f):= f(a)$. In other words, $\delta_a$ is an object which eats a function as input and spits out the value of the function a the point $a$.

Afterall, this is literally what we want it to do: when we write $\int_{\Bbb{R}^n}\delta_a(x)f(x)\,dx=\int_{\Bbb{R}^n}\delta(x-a)f(x)\,dx=f(a)$, we literally mean that $\delta_a$ is an object such that whenever we plug in a function $f$, we get out its value at $a$. Of course, writing it in this way inside an integral is a priori just nonsense, because there is no function $\delta_a:\Bbb{R}^n\to\Bbb{C}$ for which the above equality can hold true.

So, in summary, the dirac delta is a function, but it's just that the domain of the dirac delta is a space $V$ of functions, and the target space for $\delta_a$ is $\Bbb{C}$. In short, it is the "evaluation at $a$ map".


As a side remark: the dirac delta is not in any way magical/esoterical once you think of it as an evaluation map. The concept of a function as being a mapping from one set to another set is (from our luxurious perspective of having hindsight) a completely standard concept. So, in this regard the dirac delta is simply a function/mapping. The only difference is that the domain is a space of functions.

Furthermore, the concept of evaluation maps is a very basic concept in linear algebra (e.g., if you study the isomorphism between a finite-dimensional vector space and its double dual you'll see exactly what I'm talking about).

Now, the "difficulty" which comes with this is the question of "how to do calculus with these new types of objects". What I mean is in the ordinary setting of discussing functions $f:\Bbb{R}\to\Bbb{R}$, we have a notion of convergence/limits (i.e a topology), we have a notion of derivative (the study of differential calculus), and we have the notion of anti-derivatives/finding primitives etc. Extending these ideas to the more general setting is where the heart of the matter lies, and to fully appreciate that one should study more closely Laurent Schwartz's theory of distributions.

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    $\begingroup$ I think your first sentence hits the nail on the thumb. "... it has to equal $\infty$ in just the "right way"." I think that's what I've been getting at when I have argued that "we don't actually know what $\infty$ means in this context". $\endgroup$ – Prime Mover Mar 15 at 21:57
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    $\begingroup$ @PrimeMover Exactly, because in the usual arithmetic in the extended real numbers, we have $2\times \infty=\infty$. So, if one isn't clear and accurate with the definitions, you can arrive at nonsensical equalities like $f(0)=2f(0)=4f(0)=10000000000000f(0)$ and so on. I hope my last paragraph also points you towards some of the issues one has to be careful of. For example, you talk about $\lim\limits_{\epsilon \to 0}F_{\epsilon}=\delta$, viewed as a pointwise equality, but strictly speaking one should define a topology on $V$, and only after that can we talk about limits in that space. $\endgroup$ – peek-a-boo Mar 15 at 22:10
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    $\begingroup$ At this stage we haven't gone into the analysis of what happens in a general vector space. We've just got as far as defining $\delta$ on $\mathbb R$. Hence I can only understand part of what you wrote. I have some studying to do. $\endgroup$ – Prime Mover Mar 15 at 22:15
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    $\begingroup$ @PrimeMover in that case, you can just think of my last paragraph as "we need to be careful with definitions so that we can systematically extend calculus from real numbers to more general situations, and this is where a lot of the meat is". $\endgroup$ – peek-a-boo Mar 15 at 22:17
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    $\begingroup$ @CarstenS I agree with Christian here, there's actually a sense in which $\delta(x) = 0$ for $x\ne 0$ since the singular support of $\delta$ is just the singleton set $\{0\}$. Outside this set, it can be identified with a smooth function, which is the constant zero function in this case. $\endgroup$ – BigbearZzz Mar 16 at 16:11
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Adding to the existing answers and comments, I think a good way to argue against the slogan "$\delta(0)=\infty$" is to point out how limiting it is with respect to developing intuition for other things of the same "type" as $\delta$ itself. The $\delta$-"function" is, as others have said, not a function; rather, it's just something which can be integrated, which is not quite the same thing!

Making this rigorous leads to a lot of very interesting mathematics, to which the whole "$\delta(0)=\infty$" slogan is (in my opinion at least) a conceptual obstacle. Even if it doesn't get in the way of how one works with $\delta$ specifically, ignoring subtleties early on will just make things harder when they become central later.

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    $\begingroup$ When I was a young pup and I first saw the notation $\lim f(x) = \infty$ I was very upset because I had learned that "things can't equal infinity". It took me a while to accept that it was ok in this case, because my professor had in fact $defined$ what it means for a limit to equal infinity. So we had made sense of what it meant to write that. I actually learned the theory of distributions from the same professor some 5 years later, and we made sense of the definition that OP provided as an appropriate distributional limit. I find the two situations very analogous. $\endgroup$ – Prince M Mar 16 at 5:53
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    $\begingroup$ Not sure why I shared that, but I agree with you. Tackling those subtleties early on, and understanding some simplified or unorthodox notation was certainly better than accepting them and being unclear later on. $\endgroup$ – Prince M Mar 16 at 5:55
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    $\begingroup$ I think the reason it's so confusing it's because it's not a function, or a distribution -- it's a concept. Something that, by itself, doesn't make sense and, as such, is very hard (or impossible) to define. But since a definition is required, to strip the smoke and mirrors, mathematical notations are used, and it's these that are either a distribution, or an integral, or what have you, and it's only together with these that using Dirac in any calculations is possible. But, by itself, Dirac is an impossibility, just like Heaviside. As you say: "something which can be integrated". +1 $\endgroup$ – a concerned citizen Mar 17 at 10:45
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The $\delta$ "function" is not really a function at all. We define $\delta$ as a sort of special notation. When we write

$\int \delta(x) f(x) dx$

This is just "syntactic sugar" (shorthand) for $f(0)$. It has many of the same properties that an ordinary function does - for example,

$\int \delta(x) (f(x) + g(x)) dx = \int \delta(x) f(x) dx + \int \delta(y) g(y) dy$

But do not mistake $\delta$ for a "real function". It is not one.

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    $\begingroup$ I believe an accurate name is distribution. I wonder why in English Dirac's Delta is called a function – it's not the case in many other languages. $\endgroup$ – user1079505 Mar 17 at 17:01
  • $\begingroup$ It's a function almost everywhere. Mostly a function? A function save for one point? Statistically certain to be a function? $\endgroup$ – user121330 Mar 17 at 22:22
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I'm not sure why it hasn't yet been pointed out that your definition of $δ$ is wrong. You certainly cannot defined it by pointwise limits as you attempted to do. Otherwise you will indeed get $δ(0) = ∞$ but it will also be completely useless and not the desired dirac-delta (e.g. you would get $2·δ(0) = ∞$ as well).

Here is another way to think of $δ$. Instead of taking limits 'prematurely', you simply refrain from taking the limit for $δ$ at all. That is, you treat $δ$ as a suitable sequence of functions, such as $(F_{1/n})_{n∈ℕ^+}$, meaning $δ$ is literally nothing more than the sequence $(F_1,F_{1/2},F_{1/3},\cdots)$. In doing so, you can then rigorously manipulate expressions involving $δ$. We first define arithmetic and integrals to apply point-wise on a sequence. For example, for sequences $p,q$ we define $p·q$ to be the point-wise product of $p$ and $q$, namely the sequence $(p_0·q_0,p_1·q_1,p_2·q_2,\cdots)$. And given sequence $r(x)$ for every $x∈ℝ$, we define $\int_ℝ r(x)\ dx$ to be $\big( \int_ℝ r(x)_0\ dx, \int_ℝ r(x)_1\ dx, \int_ℝ r(x)_2\ dx, \cdots \big)$.

Now look. For any continuous $g : ℝ→ℝ$, we have that $\int_ℝ g(x)·δ(x)\ dx$ is also a sequence, namely $\big(\int_ℝ g(x)·F_1(x)\ dx, \int_ℝ g(x)·F_{1/2}(x)\ dx, \int_ℝ g(x)·F_{1/3}(x)\ dx, \cdots \big)$ and we can easily show that this sequence tends to $g(0)$. Your $F$ works for this, but if you want further nice properties you may need smoother ones than the ones you chose (e.g. infinitely differentiable).

~ ~ ~ ~ ~ ~ ~

[Edit: Here is my response to the later edit that quotes from a text.]

The text's definition of $δ$ is indeed completely wrong. As many people here (not just myself) have already pointed out, if you define $δ(x) = \lim_{a→0} δ_a(x)$ then definitely $δ(x) = 0$ everywhere except when $x=0$, and $δ(0) = ∞$ (using the affinely-extended reals), and definitely $\color{blue}{\int_{-∞}^∞ δ(x)\ dx = 0}$ (using the Lebesgue integral), NOT $1$, contrary to that text's erroneous claim. Note that if you do not use the affinely-extended reals, then the definition of $δ(0)$ is simply meaningless, and the whole wrong business stops there. And if you do not use the Lebesgue integral or something capable of handling isolated infinities, then you cannot even integrate $δ$. But there is no meaningful way that you can ever get $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$ from your text's definition, and here is why:

If $\color{red}{\int_{-∞}^∞ δ(x)\ dx = 1}$, then $\int_{-∞}^∞ δ(2x)\ dx = \int_{-∞}^∞ δ(x)\ dx = 1$ because $δ(2x) = δ(x)$ for every $x∈ℝ$ (by your text's definition of $δ$). But also $\int_{-∞}^∞ δ(2x)\ dx = \frac12 · \int_{-∞}^∞ δ(x)\ dx = \frac12$ by a simple stretch, so contradiction. This shows that if you use your text's definition of $δ$, then you do not have the desired properties for $δ$ at all.

~ ~ ~ ~ ~ ~ ~

If, in contrast, you use the sequence-based definition I gave you, you get no such nonsense. In this section I will again use my definition. See, $\int_{-∞}^∞ δ(2x)\ dx$ is truly equal with $\frac12 · \int_{-∞}^∞ δ(x)\ dx$, because each of them is a sequence and they have identical terms! The latter is $(\frac12,\frac12,\frac12,\cdots)$, which makes complete sense.

Furthermore, if you choose your $F$ to be a sequence built from a function that is infinitely differentiable and nonzero only on a finite interval, such as a bump function (as briefly mentioned here), then you get further nice properties. For instance, $δ'$ (i.e. the term-wise derivative of $δ$) would be well-defined, and for any $g : ℝ→ℝ$ with continuous derivative and any $y∈ℝ$ we have:

$\int_ℝ δ'(x)·g(y-x)\ dx$
$ = \big[ δ(x)·g(y-x) \big]_{x=-∞}^∞ + \int_ℝ δ(x)·g'(y-x)\ dx$
  which converges to $g'(y)$.

To explain the last line, consider a suitable example for $δ$: Let $δ_0$ be an infinitely-differentiable function on $ℝ$ that is nonzero on $[-1,1]$ such that $\int_ℝ δ_0(x)\ dx = 1$. For each $k∈ℕ^+$ let $δ_k : ℝ→ℝ$ defined by $δ_k(x) = δ_0(x·2^k)·2^k$ for every $x∈ℝ$, so we have $δ_k$ is nonzero on $[-1/2^k,1/2^k]$ and $\int_ℝ δ_k(x)\ dx = 1$. Intuitively, $δ_k$ is nonzero on a shrinking interval as $k→∞$, but its integral over $ℝ$ remains exactly one.

Then $\big[ δ(x)·g(y-x) \big]_{x=-∞}^∞$ is a constant $0$ sequence because $\big[ δ_k(x)·g(y-x) \big]_{x=-∞}^∞ = 0$ for each $k∈ℕ$. And $\int_ℝ δ(x)·g'(y-x)\ dx$ is a sequence that converges to $g'(y)$ because:

$\int_ℝ δ_k(x)·g'(y-x)\ dx$
$= \int_{-1/2^k}^{1/2^k} δ_k(x)·g'(y-x)\ dx$
$= \int_{-1}^1 δ_k(x/2^k)/2^k·g'(y-x/2^k)\ dx$
$= \int_{-1}^1 δ_0(x)·g'(y-x/2^k)\ dx$
$∈ \int_{-1}^1 δ_0(x)·[m_k,M_k]\ dx = [m_k,M_k]$
  where $m_k,M_k$ are the min,max of $g'(y-x/2^k)$ over all $x∈[1,1]$.

The result follows since continuity of $g'$ implies that $m_k,M_k → g'(y)$ as $k→∞$.

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    $\begingroup$ @PrimeMover: No, according to your question you too were wrong; you wrote "in fact $δ$ is precisely defined by means of the limit definition as given above", which is false. $\endgroup$ – user21820 Mar 16 at 13:37
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    $\begingroup$ I give up. I'll never get my head round mathematics. Every time you think you've got a handle on what you're trying to understand, someone clever tells you you're wrong. The whole exercise is utterly pointless. $\endgroup$ – Prime Mover Mar 16 at 13:58
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    $\begingroup$ @PrimeMover: It's not impossible. Just read and think very carefully through what I wrote in my post, because it is a complete explanation of how to rigorously deal with the dirac-delta, without requiring any complicated mathematics that may not be accessible to your current level. The point is to keep the sequence of approximating functions instead of replacing it by its (pointwise) limit, so that $\int_ℝ ( g(x)·δ(x) )\ dx$ is (naturally defined as) the sequence $\big( \int_ℝ ( g(x)·F_{1/n}(x) )\ dx \big)_{n∈ℕ}$, which is not at all $\int_ℝ \lim_{n→∞} ( g(x)·F_{1/n}(x) )\ dx$. $\endgroup$ – user21820 Mar 16 at 14:09
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    $\begingroup$ @Kostya_I: Exactly, thanks for saying it! $\endgroup$ – user21820 Mar 16 at 15:51
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    $\begingroup$ But it is generally not true that $$\lim_{\epsilon\to 0} \int F_\epsilon(x)G(x)\,dx = \int\lim_{\epsilon\to 0} F_\epsilon(x)G(x)\,dx,$$ which is why the definition is not rigorous. These books never really use $\delta$ other than inside the integral, so they don't really need to define what it is, they just formulate a rule to calculate any expression involving $\delta$ inside the integral sign, for example, in Fourier/Laplace transform. The actual definition of $\delta$ just abstracts out this rule. $\endgroup$ – Kostya_I Mar 16 at 16:03
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I actually disagree with other answers; yes, it is natural to argue that the $\delta$-function does have an infinite value at the origin. If you think of a function as a density on the real line (for example, mass density), then the "physical" definition of density would be $$ \lim_{\epsilon\to 0} \frac{m([-\epsilon,\epsilon])}{2\epsilon}, $$ the mass of the interval divided by the length of the interval. The $\delta$-function would be a density of a point mass at the origin; that is, the mass will always be $1$ and the density will be infinite.

The critiques raised in the other answers boil down to the remark that this does not tell the whole story, or that you cannot define the $\delta$-function this way. This is absolutely correct, but it is a common feature of infinity; by "replacing" something with infinity you forget some information that may or may not be relevant.

For example, if you have a sequence $a_n\to +\infty$, you may sometimes wish to "forget" the sequence, retaining only the information that $\lim a_n=+\infty$. This is very useful for solving some problems (for example, calculating $\lim (a_nb_n)$ when $b_n$ is bounded from below by a positive constant) and not useful for others (for example, calculating $\lim (a_nb_n)$ when $b_n\to 0$). The scope of usefulness basically boils down to "extending" $\mathbb{R}$ with $\pm\infty$ and partially extending the operations and relations (for example, $x+\infty=\infty$ for any $x\neq -\infty$) while leaving them undefined in some other cases (as in $\infty\cdot 0$).

So, the main problem with defining the $\delta$-function (or, rather, $\delta$-density) is the following: the main use of the density is that you can recover the mass from it by integration. If you try to do it with the $\delta$-density, you naturally run into the classical $+\infty\cdot 0$ problem (infinite density times zero length), which shows that you must have retained more information than just the infinite value an the origin. But to me, it is not per se an argument against the identity $\delta(0)=+\infty$, similarly to the fact that the $+\infty\cdot 0$ indeterminancy does not invalidate the use of infinity altogether.

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  • $\begingroup$ Dirac's $\delta$ doesn't even act on the real numbers. The domain of $\delta$ is some space of functions (say, smooth functions with compact support), hence it doesn't even make sense, really, to talk about the values taken by $\delta$ on the real line. $\endgroup$ – Xander Henderson Mar 16 at 18:12
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    $\begingroup$ @XanderHenderson of course you are right, but it's quite common in mathematics to ask what is a natural way to define something that is not defined :). $\endgroup$ – Kostya_I Mar 16 at 19:12
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    $\begingroup$ But in this case, I don't think that it is reasonable to ask how $\delta(0)$ is defined, since $\delta$ doesn't act on numbers at all. Historically, the study of distributions came about in an effort to make the style of approximation in this question rigorous. Folk like Sobolev and Schwartz recognized that thinking about $\delta$ as a function on the reals was the wrong idea and developed a theory to put the Dirac $\delta$ on solid theoretical ground. I feel like trying to give meaning to $\delta(0)$ is a step backwards. :\ $\endgroup$ – Xander Henderson Mar 16 at 19:18
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    $\begingroup$ @XanderHenderson, on many occasions, the fact that an abstract functional can actually be represented by an honest function plays an important role in the theory (Du Bois Reymond, $L^p$ and Sobolev duality, elliptic regularity, ...) so I don't agree that this line of thought is a step backwards. Also, $\delta$ is nicer than just any generalized function; it can be viewed as a measure for example. But I see your point. $\endgroup$ – Kostya_I Mar 16 at 20:31
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I spent ages fighting with this stuff. Let me spare you some grief. It gets taught horribly almost everywhere. I have had assignments and textbooks that contained utter gibberish pretending to be coherent questions.

First of all, there is no such thing as a value of a Dirac delta. I cannot stress that enough. There is no such thing as a value of a Dirac delta. (I never call it a function, because it isn't one). A function is defined by its values. A Dirac delta is defined by its integrals.

If you have any doubt about Dirac delta not being a function, then consider: what would be the values of $5\delta(x)$? How could you tell the difference between that and $\delta(x)$? Answer: by its alleged values, you can't. Values do not give anything like enough information. Values are enough to define a function. Therefore, the Dirac delta is NOT a function.

Now, there IS a convoluted way in which you can sort of say that "$\delta(x) = 0$ for $x \neq 0$", but it is highly misleading. What they are saying is that there is a generalized distribution, the $0$ distribution, whose integrals all equal $0$. If you take the integral of $\delta(x)$ over any interval that does not include $x = 0$, then you get the number $0$. Two generalized distributions are equal if they have the same values of their integrals. So you can say their claim, but it is like saying "dogs meow" when by "dogs" you mean "fish" and by "meow" you mean "swim." It is incredibly misleading.

What I find ironic is that you can define a perfectly ordinary function:$$I = \int_{a}^{b}\delta(x-c) dx$$ This is a function of three variables: $a,b,$ and $c$. It is not a function of $x$. Such a function $I$ can be defined for any function $f(x)$, but there are functions $I$ which do not come from integrating any function. In this example, $I(a,b,c) = 1$ if $a<c<b$ and $0$ otherwise. I don't know why they don't just teach it that way.

But remember: calling a Dirac delta a function is almost exactly like calling $\sqrt{2}$ a fraction. If you ask various questions about $\sqrt{2}$ while insisting it is a fraction, you might be able to get away with it, but if you ask what the numerator of $\sqrt{2}$ is, it becomes obvious that something is very wrong. $\sqrt{2}$ is the limit of many sequences of fractions; the Dirac delta is the limit of many sequences of functions.

Consult Lighthill's little book on distributions if you want more details. Good luck!

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  • $\begingroup$ First thing I'm going to have to do is study exactly what is meant by a "distribution". I am coming to this from the study of Laplace transforms and Fourier transforms on real functions, where no word on "distributions" is made. The integral definition is sometimes used, and sometimes the integral definition is "derived" from the definition as a limit of the rectangle function. So (apart from having a vague notion of a "distribution" as something that arises in probability theory, and then the question arises as to "Which distribution are you talking about?) ... $\endgroup$ – Prime Mover Mar 17 at 6:40
  • $\begingroup$ ... I don't unfortunately completely grok what is actually meant when the word "distribution" is bandied around. So as I say, the best I can glean from any of this discussion so far is: "Dirac Delta Functions is: not this, and it's not this, and it's not this either. Clear? Oh yes, and it's also not this, and you can't define it like this either, you have to define it using distributions." So at this stage I am going to have to abandon ship, and go and study more deeply in prob theory and measure theory before I have a hope of understanding the basics. $\endgroup$ – Prime Mover Mar 17 at 6:44
  • $\begingroup$ But I will hunt down that Lighthill (if it's the one I think you mean it's £24 on Amazon, deep joy) and add it to my stack of books I need to study. Till then I'm going to have to shelve all this, and leave alone those webpages defining $\delta$ incorrectly until I know exactly what I'm talking about. $\endgroup$ – Prime Mover Mar 17 at 6:50
  • $\begingroup$ Your statements about $\delta$ and $5\delta$ are also valid for the zero function, are you saying the constant zero function is not a function either? $\endgroup$ – syntonym Mar 17 at 10:36
  • $\begingroup$ @syntonym, no, because 5 times 0 is 0. Those ARE the same function. But $\delta$ is supposed to have an area of 1 and "$5\delta$" is supposed to have an area of 5, so they are not supposed to be the same thing, even though they would have the same alleged values. $\endgroup$ – RobertTheTutor Mar 17 at 15:42
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Peek-a-boo mentioned in his answer "$\delta(0)$ has to equal $\infty$ in just the right way", and it can be illuminating to see what can go wrong. Lets look at a different function which also "converges pointwise" to $\delta$:

$$ G_\epsilon(x) = \begin{cases} \epsilon & x = 0 \\ 0 & x \neq 0\end{cases}$$

Now $\lim_{\epsilon \to \infty} G_\epsilon(0) = \infty$ so we would expect that this is the Dirac delta function. But if we integrate over $G_\epsilon$, it behaves very differently:

$$\int f(x) G_\epsilon(x) dx = 0$$

Which can be seen by using the Riemann definition and never using $0$ as an evaluation point:

$$\int f(x) G_\epsilon(x) dx = \sum f(x_i) (x_i-x_{i+1}) = \sum f(x_i) (0-0) = 0$$

(alternatively it is a well known theorem from Lebesgue integration that changing a single point cannot change the value of the integral)

So defining the Delta distribution simply via stating that it is infinite at 0 does not work, as having two different functions ($F_\epsilon$ and $G_\epsilon$) "converging pointwise" to this definition leads to different values when integrating over them. It matters "how you converge to $\infty$".

Why do we then even care analysing $F_\epsilon$, if defining distributions that way is tricky (compared to the "direct way" of defining a distribution as a function from functions to a number)? Functions are dense in their distributions, which means that every distribution can be approximated by such a series of functions. Thus it is very natural to look at approximations of distributions.

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To make an even more ridiculous example, consider the function sequence $$ \phi_n(x)=\begin{cases} 0& \text{for }~n|x|<1\\ n/2& \text{for }~1\le n|x|<2\\ 0& \text{for }~2\le n|x|. \end{cases} $$ This has the required properties for an "approximation of unity" (delta being the unit element for convolution) of being

  • non-negative,
  • having integral one, and
  • pointwise converging to zero for $x\ne 0$,

so that for any continuous $f$ $$ \int_{-\infty}^\infty f(x)\phi_n(x)dx=\int_1^2\frac{f(s/n)+f(-s/n)}2ds~~\xrightarrow{n\to\infty}~~f(0). $$ But in the pointwise limit $\phi_n(x)\to 0$ everywhere, also at $x=0$, there is no infinite value.

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  • $\begingroup$ I'm not clear on what your point, here, is. What is this a more ridiculous example of? Also note that your integral converging to $f(0)$ relies on it being a continuous $f$, but the Dirac delta function does not rely on this. $\endgroup$ – Glen O Mar 19 at 3:09
  • $\begingroup$ It is an approximation of the behavior of the Dirac delta with regular functions, however the pointwise limit is zero everywhere, there is no point at which there could be a value $\infty$. The point of the question was to characterize the delta "function" with this infinite value. $\endgroup$ – Lutz Lehmann Mar 19 at 6:48
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The answer to the OP question is no, but the reason depends on what definitions you are using for the delta function and for integration. I don't think any of the existing answers properly address this, so let me try to sketch the spectrum of possibilities:

  • In the Lebesgue theory of integration (the one most mathematicians use), redefining a function at a point does not affect its integral. Hence, the value of $\delta$ at $0$ doesn't matter, and so the definition suggested in the OP cannot generate an object that has the right integration properties.
  • In the Riemann theory of integration, the situation is worse: The upper and lower Darboux sums will not converge at all for the suggested $\delta$ of the OP, so that the proposed delta function definition is not integrable at all.
  • In non-standard analysis (NSA), there is a way to define a delta function which is zero away from the origin and infinite at the origin, and which has the right integration properties (total mass 1). There are in fact infinitely many such delta functions, with various infinite values $\delta(0)$ and corresponding infinitesimal widths. NSA allows this because it incorporates infinite and infinitesimal numbers in a coherent way. However, because there are many infinite numbers, the symbol $\infty$ is not used. You have to pick a specific infinite number, not just any arbitrary one. So in this case again the suggested definition of the OP is meaningless.

There are (to my knowledge) three ways to define a delta function properly:

  1. The standard method is via distributions or measures.
  2. A definition frequently used in physics and engineering is to interpret the delta function as a shorthand for a limiting process involving approximations to identity. This is sometimes called a "nascent delta function".
  3. In NSA, you can define a delta function in the above-mentioned way as a function with infinitesimal width and infinite height, chosen in such a way that the total mass is 1.
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    $\begingroup$ The NSA objects are still non-standard functions with no standard object close-by. As functionals they also give non-standard values, none of them gives the same value as the standard Dirac delta functional. $\endgroup$ – Lutz Lehmann Mar 19 at 6:51
  • $\begingroup$ @LutzLehmann If $\delta(x)$ is a delta function in the NSA sense, then for well-behaved $f$, the standard part $^{\circ}(\int \delta(x) f(x)dx) = f(0)$, as you would expect. So yes, the straight integral gives non-standard values (as it must), but by taking the standard part you recover a standard Dirac delta. $\endgroup$ – Yly Mar 19 at 7:13
  • $\begingroup$ Yes, that is then the same as the physics approach, only that the limit process is replaced by the standard part construction. The difference is between $a_n\to a$ and $°(a_N)=a$ for (all) $N\approx\infty$. $\endgroup$ – Lutz Lehmann Mar 19 at 7:24
  • $\begingroup$ @LutzLehmann Firstly, the NSA approach is of course equivalent to a standard approach at some level. This horse has been beaten to death so I don't really want to get into it, but to recapitulate: The advantage of NSA is (1) it is, to some palates, more attractive; (2) it sometimes is more efficient than standard methods; and (3) most importantly, it provides a different intuition than standard approaches. For instance, (though I'm not an expert) I gather that the NSA perspective on distributions leads naturally to Colombeau algebra for multiplying them, which is hard in the standard approach. $\endgroup$ – Yly Mar 19 at 7:45
  • $\begingroup$ You can make sense of Colombeau? When I looked into it, it was some strange voodoo to me that essentially negated the solution concept because almost everything could be a solution. $\endgroup$ – Lutz Lehmann Mar 19 at 7:51
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Yes, many world famous mathematicians (such as Kovalski, Borodchuk, and Vladimirich) have defined the Dirac delta function as infinity this way. In many optimization, field theory, and wavelet problems this is useful. For example, when computing the inverse of a Borodchuk field, if the Diract delta is infinity, then the field is finite at singularities. Similarly, Kovalski and Vladimirich have shown that having infinite wavelet amplitudes can lead to simpler Gamma-extensions. There are many other use cases, too.

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I think the confusion here may be due to the difference between definition and description.

A good definition of the Dirac delta function is "the function/distribution $\delta(x)$ that satisfies $\delta(x)=0$ for $x\neq0$ and $\int_a^b \delta(x)\ dx = 1$ if $a<0<b$." - one may also define it as a limit of functions.

When describing this function, it is reasonable to state that the value at $x=0$ can be thought of as infinite. And thus, as a description, it is reasonable to say $$ \delta(x)=\begin{cases} \infty & :x=0\\0 & :x\neq 0 \end{cases} $$

This allows one to understand that the value is zero away from $x=0$, but at $x=0$, in order for the function to have its other properties, the value at $x=0$ must be infinite. However, this is not sufficient to operate as a definition, as it does not establish the properties that make it the delta function. Indeed, as "infinity" is not, in itself, a well-defined value, the description isn't a definition because it isn't well-defined.

It is conceivable to turn the description into a proper definition, of sorts, if one can somehow introduce the concept of the "special infinity" that has the necessary properties to make the description work as a definition. For example, one might plausibly define $\infty_{dx}$ as an infinite number that can be thought of as the inverse of the width $dx$ that appears in the integration symbol. It would require more careful definition, but with the right structure of the definition (and likely, using some nonstandard analysis), it would become a reasonable concept. And then the concept of $2\infty_{dx}$ is perfectly sane as the value of $2\delta(0)$

But that would be like defining $a\times b=e^{\ln(a)+\ln(b)}$ as the base definition of multiplication (rather than merely an example of a property that can be found). Sure, it's workable, but it's also an overcomplicated solution to a simple problem.

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