# A caracterisation of f=0 by integrals

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[SOLVED] A caracterisation of f=0 by integrals

1. Homework Statement
Does anyone know how to show, or know a book that proves the implication

$$\left(\int_0^1f\varphi = 0 \ \ \forall \varphi \in C_c^1([0,1])\right)\Rightarrow f=0$$

for f in L²([0,1]) and where $C_c^1([0,1])$ denotes the C^1([0,1]) functions whose support is contained in (0,1).

Thanks.

3. The Attempt at a Solution

I tried using the density of $C_c^1([0,1])$ in L^1 to obtained a sequence $\varphi_n$ that converges pointwise a.e. to the caracteristic function of [0,1] and then plugging-in the convergence theorems (Fatou, motone and dominated) but I eventually aknowledged that this would not work. Well, at least I got

$$\int_0^1f\leq 0$$

out of Fatou.

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morphism
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Just to be clear, isn't $C_c^1([0,1])$ the set of compactly supported continuously differentiable funtions on [0,1]?

Anyway, here are some thoughts. Consider an open subset U of [0,1], whose characteristic function is $\chi$. Get a sequence $\{\varphi_n\}$ in $C_c^1([0,1])$ that converges a.e. to $\chi$. We can assume that $M = \sup_n \| \varphi_n \|_\infty < \infty$ (just construct them properly). So $f\varphi_n \to f\chi$ a.e., and $\{f\varphi_n\}$ is majorized by M|f|, which is integrable. Hence, by the dominated convergence theorem, $\int_U f = 0$ for all open subsets U of [0,1]. Now use the regularity of the Lebesgue measure to conclude that f=0 a.e.

Homework Helper
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Just to be clear, isn't $C_c^1([0,1])$ the set of compactly supported continuously differentiable funtions on [0,1]?
Dunno if this is standard convention or not but my professor uses $C_c^1([0,1])$ to denote the set of compactly supported continuously differentiable funtions on (0,1). So for instance, the characteristic function of [0,1] is not in $C_c^1([0,1])$ according to my prof.

Anyway, here are some thoughts. Consider an open subset U of [0,1], whose characteristic function is $\chi$. Get a sequence $\{\varphi_n\}$ in $C_c^1([0,1])$ that converges a.e. to $\chi$. We can assume that $M = \sup_n \| \varphi_n \|_\infty < \infty$ (just construct them properly). So $f\varphi_n \to f\chi$ a.e., and $\{f\varphi_n\}$ is majorized by M|f|, which is integrable. Hence, by the dominated convergence theorem, $\int_U f = 0$ for all open subsets U of [0,1]. Now use the regularity of the Lebesgue measure to conclude that f=0 a.e.
Neat argument!

Say, have you seen my other similar question ?