1

Let $\Omega$ be an open bounded domain and let $\varphi\in C^{\infty}_c(\Omega)$. Let $f, g:\Omega\to\mathbb{R}$ be two smooth functions. Consider the equation $$\int_{\Omega} (f(x)\varphi(x) +g(x)\varphi^{\prime}(x)) dx=0.$$

My question is: there is relation between $f(x)$ and $g(x)$? It is true that $f(x)=g(x)=0$ a.e.?

About me the answer to the last question is no. Firstly, since $\varphi\in C^{\infty}_c(\Omega)$, thus so it is $\varphi^{\prime}$. Moreover, I think that we can only say that $f(x)=-g(x)$ a.e. and they have not to be necessarily $0$.

Could someone please tell me if am I wrong and/or help me to answer also the first question?

Thank you in advance!

C. Bishop
  • 3,418
  • Is $\varphi\in C_c^\infty(\Omega)$ arbitrary? – xpaul Mar 06 '22 at 16:50
  • 4
    If I remember correctly, $\int_{\Omega} (f(x)\varphi(x) +g(x)\varphi'(x)) dx = \int_{\Omega} (f(x) -g'(x)\varphi(x)) dx$, if that is zero for all test functions $\phi$ then $f-g'=0$. – Martin R Mar 06 '22 at 16:54
  • Same comment as @xpaul: You should place a $\forall \varphi \in C_c^{\infty}$ in front of your formula. – Jean Marie Mar 06 '22 at 16:56
  • Use linearity of integral operator, to separate. Then use integration by part. (And you will got what Martin R said.) – kolobokish Mar 06 '22 at 16:59
  • Please, as I said, modify your text by saying that your formula is valid for all $\varphi$. If it is valid for a particular $\varphi$ you can conclude nothing at all. – Jean Marie Mar 06 '22 at 19:05
  • I have downvoted your question because you don't attempt to improve your text. I will erase the downvoting once you do it. – Jean Marie Mar 07 '22 at 19:48

2 Answers2

3

It follows from Green's first identity (the higher dimensional equivalent of integration by parts), that $$ \tag{$*$} \int_{\Omega} (f(x)\varphi(x) +g(x)\varphi^\prime(x)) dx = \int_{\Omega} (f(x) -g^\prime(x))\varphi(x) dx = 0\, . $$

This is for example satisfied if the support of $f-g^\prime$ and the support of $\varphi$ are disjoint. So one can not conclude much about $f$ and $g$ if $(*)$ holds for some $\varphi\in C^{\infty}_c(\Omega)$.

But if $(*)$ holds for all test functions $\varphi\in C^{\infty}_c(\Omega)$ then necessarily $f-g^\prime = 0$.

Explanation of $(*)$: Choose $U \subset \Omega$ such that the support of $\varphi$ is contained in $U$. Then $$ \int_\Omega (g(x)\varphi'(x) - g'(x) \varphi(x)) \, dx = \int_U(g(x)\varphi'(x) - g'(x) \varphi(x)) \, dx\\ = \int_{\partial U} g(x) \varphi(x) \cdot \mathbf{n} \, dS = 0 $$ since $\varphi$ vanishes on the boundary and outside of $U$.

Martin R
  • 113,040
  • Martin R, thank you for your answer. So, in general, it is false that $f=g$ a.e., isn't it? Moreover, could you please better explain why $f$ does not have multiplication neither $\varphi$ nor $\varphi^{\prime}$? It is not clear for me by using the link you quoted. Thank you. – C. Bishop Mar 06 '22 at 17:36
  • @C.Bishop: Sorry, it is unclear to me what you are asking. If the integral is zero for all test functions then $f=g'$, and vice versa. What does “f does not have multiplication ...” mean? – Martin R Mar 06 '22 at 17:39
  • Martin R, it is not clear the first "=" in $(*)$. – C. Bishop Mar 06 '22 at 17:41
  • @C.Bishop: Since $\varphi$ has compact support, you can choose $U \subset \Omega$ such that the support of $\varphi$ is contained in $U$. Then apply Green's identity and note that that the integrals $\int_{\partial U}$ vanish. See for example the second formula here: https://en.wikipedia.org/wiki/Integration_by_parts#Green%27s_first_identity – Martin R Mar 06 '22 at 17:43
  • Martin R, sorry again, bu why it vanishes in the boundary? – C. Bishop Mar 06 '22 at 17:56
  • @C.Bishop: Because $\varphi$ is zero on the boundary. – Martin R Mar 06 '22 at 18:20
  • @J.G: Thanks for your edit. There was indeed a missing closing parenthesis. For the notion of derivatives, g' is equivalent to g^\prime, compare https://tex.stackexchange.com/q/87134/28519. – Martin R Mar 06 '22 at 18:24
  • @MartinR I'll remember to be lazier with my $'$s in the future. – J.G. Mar 06 '22 at 18:30
  • @C.Bishop: I have added more details. – Martin R Mar 06 '22 at 19:02
2

Another way to prove this would be distribution-theory.
$\int_{\Omega}g(x)\varphi'(x) \,dx = -\int_{\Omega}g'(x)\varphi(x) \,dx\: \forall \varphi(x) \in \mathscr{C}^{\infty}_c$
where by g' I mean the "distributional-derivative".

Then we get that $\int_{\Omega}f(x)\varphi(x) \,dx = \int_{\Omega}g'(x)\varphi(x) \,dx$
This means precisely that f and g' are equal in "the distributional sense". If we assume that $f,g' \in L^{1}_{loc}$ then we can apply the fundamental lemma of calculus of variations.

See Proof of fundamental lemma of calculus of variation.

And get $f(x) = g'(x)$ almost everywhere. So you see that smoothness of $f,g$ is a very strong assumption that is not needed at all.

(P.S: I deleted the earlier post because I missed the prime over $\varphi$ there)

Paul Joh
  • 697