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Let $\Omega \subset \mathbb{R}^n$ ($n \geq 1$) be a bounded open domain and $f \in L^\infty(\Omega)$ possibly changes the sign. Assume that the set $$ \Omega^+ := \{x \in \Omega: f(x) > 0 \} $$ has positive measure. Is it true that there exists non-negative $\varphi \in C_0^1(\Omega)$, such that $$ \int_\Omega f \, \varphi \, dx > 0 ~? $$ Obviously, if $\Omega^+$ has nonempty interior (after redefinition on a set of measure $0$), then the answer is Yes. But I'm afraid of the case $\Omega^+$ is a "bad" set, in the sense that its interior is empty (like fat Cantor set).

Thanks.

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  • $\begingroup$ $C^1_0$ means differentiable with compact support? $\endgroup$
    – fgp
    Mar 30 '14 at 12:47
  • $\begingroup$ @fgp Yes, just differentiable with compact support. $\endgroup$
    – Voliar
    Mar 30 '14 at 12:51
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For $g(x) = \chi_{\Omega^+}$, it is clear that $$ \int_{\Omega} f g = \int_{\Omega^+}f = R> 0 \text{,} $$ but of course in general $g \notin C^1_0$. However, since $\Omega$ is bounded, $g \in L^1(\Omega)$ because $\int_{\Omega} g \leq \int_{\Omega}1 < \infty$. Since $C^1_0(\Omega)$ is a dense subset of $L^1(\Omega)$, there is thus some $\varphi \in C^1_0(\Omega)$ with $$ \|\varphi - g\|_1 < \frac{R}{2\|f\|_\infty} $$ and thus with $$ \int f(\varphi - g) \leq \|f\|_\infty\cdot\|\varphi - g\|_1 \leq \frac{R}{2} \text{.} $$ But then $$ \int_{\Omega} f\varphi = \int_{\Omega} fg + \int_{\Omega} f(\varphi - g) \geq R - \frac{R}{2} = \frac{R}{2} > 0 \text{.} $$

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  • $\begingroup$ Excellent! Thank you very much! $\endgroup$
    – Voliar
    Mar 30 '14 at 14:25

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