1
$\begingroup$

I'm approaching the study of distributions, and together with many notes, I'm following L.Hormander book "linear partial differential operators".

In page $5$ he writes

In view of the identification of an absolutely continuous measure with its density function, which is customary in integration theory, this means in particular that a function $f \in L^1_{loc}(\Omega)$ is identified with the distribution

$$\phi \to \int \phi f \ \text{d}x$$

This distribution will also be denoted by $f$.

Well my question is: is there a proof for that? Why a function $f$ is / can be identified with that distribution?

$\endgroup$

1 Answer 1

2
$\begingroup$

The fact that this defines a distribution should be rather straightforward from the definition since $$\left|\int \phi(x)\,f(x)\,dx\right|\leq\sup_{x\in K}|\phi(x)|\int_K |f(x)|\,dx$$ where $K$ is the support of $\phi$. So that if $\phi_n\rightarrow\phi$ in $\mathcal{D}$ then $$\int\phi_n(x)\,f(x) dx\longrightarrow\int \phi(x) f(x)\,dx$$

You can identify the distribution with the function because of the following fact: if $f\in L^1_{loc}(\Omega)$ and $$\int \phi(x) f(x)\,dx=0$$ for every $\phi\in\mathcal{D}$ then $f=0$ a.e. This shows that the mapping $L^1_{loc}(\Omega)\rightarrow\mathcal{D'}(\Omega)$ which assigns $f$ to the distribution you've defined is one-to-one, and we can consider $L^1_{loc}(\Omega)$ as embedded in the space of distributions.

$\endgroup$
1
  • $\begingroup$ Wow that was as fast as clear. Thank you so much! $\endgroup$ Sep 30, 2020 at 20:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .