I am currently studying distribution theory, with an eye towards PDEs, and have a question on regarding locally integrable functions as distributions.

Suppose $u\in L^1_{\mathrm{loc}}(\Omega).$ Then we can map $u$ to the space of distributions $D'(\Omega)$ by $$T(u)=I_u\in D'(\Omega),$$ where the distribution $I_u$ is defined by $$\langle I_u,\phi\rangle:=\int_{\mathbb{R}^n}\!f(x)\phi(x)\ \mathrm{d}x.$$ Given this definition, it is technically not correct to say $u\in D'(\Omega).$ Nonetheless, many textbooks consider $u$ to be a distribution. Salsa (PDEs in Action, pp. 375) even goes as far as to say $L^1_{\mathrm{loc}}(\Omega)\hookrightarrow D'(\Omega)$, which seems to assume $L^1_{\mathrm{loc}}(\Omega)\subset D'(\Omega)$. Furthermore, this would imply $$|| u||_{D'(\Omega)}\le C ||u ||_{L^1_{\mathrm{loc}}(\Omega)},$$ but what meaning can be assigned the the left hand side of the above equation if $u\not\in D'(\Omega)$?

As far as I can tell, when one writes $L^1_{\mathrm{loc}}(\Omega)\hookrightarrow D'(\Omega)$, to mean that the mapping $T:L^{1}_{\mathrm{loc}}(\Omega)\to D'(\Omega)$ is injective, and $\mathrm{im}(T)\hookrightarrow D'(\Omega)$?

The only thing I can think of is that this is just an abuse of notation, tolerated so one can write $$\Delta u=f,$$ and understand $u$ as a distribution. This would be more comfortable and convenient than writing $$\Delta I_u=f.$$

Am I thinking along the right lines? Am I crazy? I have done some searching, but most sources just seem to brush this off by saying they "regard" (?) $u$ as a distribution. I would very much appreciate any help offered.

  • $\begingroup$ Note that your norm inequality doesn't make sense since the topology on the distributions is a weak-*-topology and not normable. However you can indeed easily show that the embedding is continuous. $\endgroup$ Feb 23, 2017 at 19:36

2 Answers 2


I don't think this question is about distributions, but about notation. Do you think $f\in C^\infty([0,1])$ is also an element of $f\in L^2([0,1])$? This is technically incorrect, as elements of $L^2([0,1])$ are equivalence classes of functions. However, there is a natural map $i:C^\infty([0,1])\rightarrow L^2([0,1])$ by sending a function to its equivalence class. It is cumbersome to write always $i(f)$ here, so we get used to writing $f$ for the thing in $L^2([0,1])$. A more elementary example is $\mathbb{Q}\subset \mathbb{R}$. If one goes into the definitions of the elements of $\mathbb{R}$ (as sequences in $\mathbb{Q}$, or Dedekind cuts, or whatever), it does not really make sense to say $\mathbb{Q}\subset\mathbb{R}$. However there is an obvious injective function, and not writing this function makes our live easier. For example we can talk about $\frac{1}{2}+\pi$ instead of $i(\frac{1}{2})+\pi$. You have to agree the first formula is better.

Exactly the same thing is happening here. There is an injection that you call $T:L^1_\mathrm{loc}(\Omega)\rightarrow D'(\Omega)$. We also write $u$ for its image under $T$.

  • $\begingroup$ Thank you for your answer, although I don't think your analogy carries through for distributions. Since $1/2$ is a number, it makes sense to write $1/2\in \mathbb{Z}$ or $1/2\in \mathbb{R}$. Now suppose $u\in L^1_{\mathrm{loc}}(\Omega)$. Then $u$ is a real valued function, which takes values from $\Omega$ and maps them to real numbers. The space $D'(\Omega)$ consists of distributions (linear functionals). Since $u$ is not a linear functional, strictly speaking it does not make sense to make statements like $u\in D'(\Omega)$, unless one is abusing notation (which Herme's justifies below). $\endgroup$
    – runner.87
    Feb 22, 2017 at 21:16
  • $\begingroup$ I do think my analogy goes through. What is a real number for you? For me this is an equivalence class of cauchy sequences of rational numbers. You can map any rational number to such an equivalence class: For $q\in \mathbb{Q}$ take the equivalence class of the constant sequence $a(n)=q$. An element of $\mathbb{Q}$ is different than an element of $\mathbb{R}$ (even though it is rational). You are more accustomed to this, which is why you do not worry about this. $\endgroup$
    – Thomas Rot
    Feb 23, 2017 at 8:49
  • $\begingroup$ @ThomasRot I wouldn't say a real number is an equivalence class, since, as you have already mentioned, there are several other ways to introduce real numbers. For me, the real numbers are more or less a "concept". To go away from realizations makes such discussions about identifications obsolete. $\endgroup$ Feb 23, 2017 at 19:34
  • $\begingroup$ And all come equipped with some canonical embedding of the rationals... $\endgroup$
    – Thomas Rot
    Feb 23, 2017 at 19:44

Indeed, that mapping $T : L^1_{loc}(\Omega) \to D'(\Omega)$ defined as $T(u) = I_u$ is injective, embedding the set of locally integrable functions on $\Omega$ into the set of distribution of $\Omega$.

As you pointed out, when we formally have a map $T : A \hookrightarrow B$, (meaning $T$ is injective), it is current to identify $ a \in A$ with $T(a) \in B$, i.e. writing $a =: T(a)$. There might be several reasons for that, though mainly for the two following (+ a reason at infinity):

  1. It's simpler. Sure, taking your example, we could write $I_u$ instead of $u$, or worse, something like $\varphi \mapsto \int_{\mathbb{R}^n} u(x)\varphi(x)\, \text{d}x$. But we tend to go for the lightest pack of notations, swerving away from heavy uses of indexes and the likes. The drawback might sound like a loss of clarity, but it eventually pays off considering the amount of words and esoteric symbols saved.

  2. It means something. In the very case of distribution, such a notation not only simplifies the overall writing quality, but also carries a lot of meaning: that the distribution are constructed as a somehow generalization of functions, more precisely that of locally integrable functions. Therefore, by letting $u =: I_u$, we put emphasis toward this very fact. As you'll see, the basic theory of distribution is built as to make a coherent continuation of the usually theory on $L^1_{loc}(\Omega)$ (an example being the derivative of a distribution defined such that $\partial^{\alpha}I_u = I_{u\prime}$)

$(\infty)$- The aesthetic of mathematics. Not to enter a philosophical debate on aesthetic, I believe most mathematicians will agree that this identification conveys the intertwining of mathematical efficiency with mathematical beauty.


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