2
$\begingroup$

I am currently reading Brezi's Book:"Functional Analysis, Sobolev Spaces and PDE's" and do not understand one step in the proof of the following lemma.

Let $f\in L^{1}_{loc}(I)$ be such that $$\int_I f\phi^{'}=0, \ \ \forall \phi \in C^{1}_c(I).$$ Then there exists a constant $C$ such that $f = C$ a.e. on $I$.

Proof. Fix a function $\psi \in C_c(I)$ such that $\int_I \psi = 1$. For any function $w\in C_c(I)$ there exists $\phi \in C^{1}_c(I)$ such that $$\phi^{'}=w-(\int_I w)\psi.$$ Indeed, the function $h=w-(\int_I w)\psi$ is continuous, has compact support in I and also $\int_I h=0$. Therefore $h$ has a (unique) primitive with compact support in I. We deduce from our assumption on $f$ that $$\int_I f[w-(\int_I w)\psi]=0,\quad \forall w \in C_c(I),$$ i.e., $$\int_I [ f-(\int_I f)\psi ]w=0,\quad \forall w \in C_c(I).$$ And therefore $f-(\int_I f\psi)=0$ a.e. on $I$ and we choose $C=\int_I f\psi$.

I don't understand the last step, replacing the $f$ and $w$ in the integral. I would really appreciate someone explaining that step even if it is really simple.

$\endgroup$

1 Answer 1

1
$\begingroup$

We have $$\int_I f(s) [ w(s)-(\int_I w(t) dt )\psi (s) ]ds =\int_I f(s) w(s) ds -\int_{I\times I } w(t) \psi (s) f(s) dt ds=\int_I f(t) w(t) dt -\int_{I\times I } w(t) \psi (s) f(s) dt ds=\int_I [ f(t)-(\int_I f(s) \psi (s) ds)]w(t)dt$$

$\endgroup$

You must log in to answer this question.

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