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I have a question on a variation of the fundamental lemma .

If $\int_\Omega f(x) g(x)=0$ and $f, g $ are $C^0\Omega$ functions and $\int_\Omega g(x)=0 $

then is it possible that there exist some constant $c$ such that $f(x)=c$ for all $x\in \Omega$

I tried to use mollification on one of the function and throw derivative on the mollified function but that doesn't give me anything . I am wondering if the question makes sense ?

Theorem
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  • I don't understand: yes it's possible that $f$ is constant (nothing prevents it). – Davide Giraudo Oct 31 '12 at 10:44
  • @DavideGiraudo : I am trying to prove it but i am not able to do it . Can you give me some guidelines to start off with proof . . – Theorem Oct 31 '12 at 20:56
  • One should specify the hypothesis of the lemma. Suppose that $\Omega$ is a bounded interval $(a,b)$ and let $f \in C^0({\overline{\Omega}})$ verify $\int_\Omega f(x) g(x)=0$ for all $g \in C^0({\overline{\Omega}})$ such that $\int_\Omega g(x)=0$. Then there exists a constant $c \in \mathbb{R}$, such that $f(x)=c$ for all $x \in \overline{\Omega}$. – derthomas Jan 31 '17 at 02:04
  • It seems that the correct interpretation of the problem is the one given by H.H.Rugh below. If this is indeed the case, you should edit your question and reformulate, otherwise somebody might waste time answering the wrong question. – Giuseppe Negro Feb 04 '17 at 11:23
  • Why isn't the answer accepted? – ViktorStein Mar 17 '21 at 12:29

1 Answers1

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The formulation of the question is not very clear. Here is my interpretation: If the integral $\int_\Omega fg$ vanishes whenever $g\in C_c(\Omega)$ and $\int_\Omega g=0$ then indeed the function $f$ must be constant.

We assume here that $\Omega$ is an open subset of ${\Bbb R}^n$. In the following let $\psi\in C_c({\Bbb R}^n)$ be a continuous function with compact support, say in the unit ball $B(0,1)$ in ${\Bbb R}^n$, such that $\int_{B(0,1)} \psi(x) \; d^n x=1$. It is easy to construct such a function and one may even do it so that $\psi$ is $C^\infty$. If $a\in {\Bbb R}^n$ and $\epsilon>0$ then the function $$ \psi_{a,\epsilon}(x) = \frac{1}{\epsilon^n} \psi \left( \frac{x-a}{\epsilon}\right) $$ will be continuous and have support in $B(a,\epsilon)$. The normalisation ensures that $\int\psi_{a,\epsilon}=1$. When $f\in C(\Omega)$ and $a\in \Omega$, then for $\epsilon>0$ small enough, $B(a,\epsilon)\subset \Omega$ and as $f$ is continuous at $a$ we see that $$ f(a) = \lim_{\epsilon\rightarrow 0} \int f(x) \psi_{a,\epsilon}(x) \; d^n x.$$

Now, let $a,b\in \Omega$ and consider the function $g_\epsilon= \psi_{a,\epsilon}-\psi_{b,\epsilon}$. For $\epsilon>0$ small enough we have $\int_\Omega g_\epsilon = 0$, so by the hypothesis on $f$ we deduce:

$$ 0 = \lim_{\epsilon\rightarrow 0}\int_\Omega f \cdot g_\epsilon = \lim_{\epsilon\rightarrow 0} \int_\Omega f(x) (\psi_{a,\epsilon}(x)-\psi_{b,\epsilon}(x))\; d^n x = f(a)-f(b)$$ So as $f(a)=f(b)$ for any two points $a,b\in \Omega$ we conclude that $f$ is constant.

A similar though more technical argument works if $f$ is only assumed locally integrable rather than continuous, with the conclusion that there is $c\in {\Bbb R}$ so that a.e. $f=c$.

H. H. Rugh
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