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Let $f:(a,b) \rightarrow \mathbb{R}$ be locally integrable and such that \int_a^ b f(x) \phi'(x)dx=0 \textrm{ for each } \phi \in C_0^\infty(a,b). How to show, without help of distribution theory, that $f=const$ a.e.?

I noticed that every constant functions $f=c$ satisfies this condition, because for $\phi \in C_0^\infty(a,b)$ we have $\phi(a)=\phi(b)=0$ and \int_a^b c\phi'(x)dx=c\phi(b)-c\phi(a)=0.

I know also that similar condition $\int_a^ b f(x) \phi(x)dx=0 \textrm{ for each } \phi \in C_0^\infty(a,b)$ implies that $f=0$ a.e.

Thanks.

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    @sos440 You can put your comment as an answer since it solves the problem.2012-02-07

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Let $\theta\in C_0^{\infty}(a,b)$ such that $\int_{(a,b)}\theta(t)dt=1$. Then for a fixed $\varphi\in C_0^{\infty}(a,b)$, we put $A\varphi(x)=\int_a^x\varphi(t)dt-\int_a^x\theta(s)ds\int_a^b\varphi(t)dt$. Then $A\varphi\in C_0^{\infty}(a,b)$ and (A\varphi)'=\varphi(x)-\theta(x)\int_a^b\varphi(t)dt so 0=\int_{(a,b)}f(x)(A\varphi)'(x)dx=\int_{(a,b)}f(x)\varphi(x)dx-\int_{(a,b)}\int_a^b\varphi(t)dt\theta(x)f(x)dx, hence $\forall \varphi\in C_0^{\infty}\quad \int_{(a,b)}f(x)\varphi(x)dx=C\int_a^b\varphi(x)dx,$ where $C=\int_{(a,b)}f(x)\theta(x)dx$. Hence $f$ is almost everywhere constant.

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    Thanks. It is a nice answer.2012-02-07