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The $\delta$ function is defined as a distribution, and is thus an element of $\mathcal{D}'(U)$ for some open set $U \in \mathbb{R}^n$. In other words, it is a function in a dual space, and hence we may write $\delta: C_c^\infty(U) \rightarrow \mathbb{R}$ where $C_c^\infty(U) = \mathcal{D}(U)$ is the set of smooth functions over $U$ with compact support.

What has been bothering me is that I often see expressions such as $$\int \delta(x)e^x dx.$$ But the function $e^x$ does not have compact support over any open set. Likewise, but less common, I also have seen the $\delta$ function applied to functions that are not infinitely differentiable.

How are such uses of the delta function (and distributions in general) justified?

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    $\begingroup$ You can also see the Delta distribution as a functional on the space of all continuous functions on an open set. $\endgroup$
    – PhoemueX
    Commented Dec 11, 2022 at 7:20
  • $\begingroup$ See also Distributions with compact support $\endgroup$ Commented Dec 11, 2022 at 7:22
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    $\begingroup$ Any distribution is a linear functional on some space of test functions. The whole space of continuous functions can serve as test functions, then there are no derivatives. The space of fast-falling smooth functions gives tempered distributions that serve to explore the Fourier transform. $H^{-1}$ are the bounded functionals on $H^{1}$ etc. $\endgroup$ Commented Dec 11, 2022 at 8:11
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    $\begingroup$ @LutzLehmann I always thought that by definition the domain of a distribution is $C^\infty(U)$ and the domain of tempered distributions is the dual of a Schwartz space. However, based on your comment it seems distributions can have more general domains? $\endgroup$
    – CBBAM
    Commented Dec 11, 2022 at 8:14
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    $\begingroup$ The language changes at some point, one speaks of duality pairs, and leaves the name of "distribution" to the classical ones, as enumerated in the answer. $\endgroup$ Commented Dec 11, 2022 at 8:19

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Given any sets $A,B$, and any point $a\in A$, I can always decide that I want to define the function $\delta_a:F=\text{Functions}(A,B)\to B$, $\delta_a(f):= f(a)$. I can call this an evaluation at $a$ mapping, or I can call this the Dirac delta centered at $a$, or I can give it any other name I like. Point is, I can define it because it makes sense. The question is if this is useful?

Well, Dirac deltas are mostly useful in vector spaces, so it would be a good idea to consider the target $B$ to be a vector space. Also, we want to deal with this in analysis, so the most common choice would be $A$ an open set in $\Bbb{R}^n$, and $B=\Bbb{C}$ (but again, if you're going to be fancier, you may want to allow the domain $A$ to be something else, and you may want to allow $B$ to be any Banach space). Also, for the purposes of analysis, I usually don't just want to consider all functions $A=U\to\Bbb{C}$, I probably only want to consider some of them, and I probably want to equip a topology on these function spaces. Some common choices are

  • $A=U$ open in $\Bbb{R}^n$, $B=\Bbb{C}$, $F=C_b(U)$, the bounded complex-valued continuous functions on $A$ with say the supremum norm, and $\delta_a:C_b(U)\to\Bbb{C}$ defined as above. Then, you can show $\delta_a$ is continuous and linear (bounded continuous functions is an important space, e.g in probability, and also in many areas of analysis; look up the Riesz representation theorem for measures and integrals).
  • You can consider $A=U$ open in $\Bbb{R}^n$, $B=\Bbb{C}$, $F=\mathcal{E}(U)=C^{\infty}(U)$, the space of smooth functions equipped with say the topology of uniform convergence on compact subsets of all derivatives (so giving you a locally convex topological vector space rather than a Banach space). You can again consider $\delta_a:\mathcal{E}(U)\to\Bbb{C}$, defined as above. You can show that it is again continuous and linear with respect to this topology. In fact, continuous linear maps on this larger space of functions can be identified with distributions of compact support (see for example Folland's real analysis text for more information).
  • You can also consider it as a mapping $\delta_a:\mathcal{S}(\Bbb{R}^n)\to\Bbb{C}$, as a linear mapping on Schwartz functions. You can show it is continuous and linear here as well... the name given here is that it is a tempered distribution, i.e an element of $\mathcal{S}'(\Bbb{R}^n)$ (so the theory of Fourier transform applies here).
  • You can also consider it as a mapping $\delta_a:\mathcal{D}(U)\to\Bbb{C}$, where it is again continuous and linear... i.e a distribution in the usual sense of Schwartz.

The last three are reflecting the idea that, on $\Bbb{R}^n$, $\mathcal{D}\subset \mathcal{S}\subset \mathcal{E}$ (in the sense of there is a natural inclusion, which is continuous in the appropriate sense). So their topological duals satisfy $\mathcal{E}'\subset\mathcal{S}'\subset\mathcal{D}'$.


There are of course a whole bunch of other spaces on which you can consider things (e.g manifolds, vector bundles blablabla). The question though is for what purpose are you using it. Beyond a certain point it becomes pretty irrelevant whether you want to call it a distribution, or a functional or a linear form, or whatever, because people will freely interchange terms. One has to use context to decipher the intended meaning.

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  • $\begingroup$ Thank you for your comment. I see your point that we may define the $\delta$ function however we like. However, in some cases wouldn't this still fail the definition of a distribution being a map $C_c^\infty(U) \rightarrow \mathbb{R}$? Thus if we define the $\delta$-function to be an evaluation map it will certainly be a linear functional but not a distribution. $\endgroup$
    – CBBAM
    Commented Dec 11, 2022 at 8:19
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    $\begingroup$ @CBBAM see my last paragraph which I just edited. It doesn't matter what you decide to call it. After some stage, people will take liberties with using terms beyond their original use (part of this is of course historical). $\endgroup$
    – peek-a-boo
    Commented Dec 11, 2022 at 8:20
  • $\begingroup$ Thank you, that last paragraph really cleared up my confusion. I did not know there are multiple definitions of a distribution. $\endgroup$
    – CBBAM
    Commented Dec 11, 2022 at 8:22
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    $\begingroup$ @CBBAM it's not so much that there are multiple definitions of a distribution (that's of course part of it, depending on subject, e.g probability and differential geometry have some other definitions), rather beyond a certain point mathematicians become more human and hence are willing/eager to tolerate overload of terminology, and they'll use same words to mean different things, because they're human so they'll use whatever comes to mind first, and they're also knowledgable, so they know based on context what the intended meaning is (and if not, they'll ask:) $\endgroup$
    – peek-a-boo
    Commented Dec 11, 2022 at 8:25
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    $\begingroup$ While I'm on this subject, see Terry Tao's On “compilation errors” in mathematical reading, and how to resolve them. There he talks about theorems, but the same can be said about definitions. $\endgroup$
    – peek-a-boo
    Commented Dec 11, 2022 at 8:29

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