Mathematics Asked by Hans-André-Marie-Stamm on December 23, 2020

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?

The fact that this defines a distribution should be rather straightforward from the definition since $$left|int phi(x),f(x),dxright|leqsup_{xin K}|phi(x)|int_K |f(x)|,dx$$ where $K$ is the support of $phi$. So that if $phi_nrightarrowphi$ in $mathcal{D}$ then $$intphi_n(x),f(x) dxlongrightarrowint phi(x) f(x),dx$$

You can identify the distribution with the function because of the following fact: if $fin L^1_{loc}(Omega)$ and $$int phi(x) f(x),dx=0$$ for every $phiinmathcal{D}$ then $f=0$ a.e. This shows that the mapping $L^1_{loc}(Omega)rightarrowmathcal{D'}(Omega)$ 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.

Answered by Olivier Moschetta on December 23, 2020

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