Is it meaningful to define the Dirac delta function as infinity at zero?

Question

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 $x = 0$ $$\int_{-\infty}^\infty \delta (x) d x = 1.$$

Answer 1

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".

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).

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[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.

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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 $k→∞$.

Answer 6

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.

Answer 7

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.

Answer 8

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!

Answer 9

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.

Answer 10

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.

Answer 11

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.