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Denote with $C_0(\Omega)$ the subspace of countinuous functions with compact support, $\Omega\subseteq\mathbb{R}^n$ open. Define an application on $C_0(\Omega)$ as $||\cdot||\colon C_0(\Omega)\to\mathbb{R}_+$ $$||f||_1:=\int_\Omega |f|\;d\lambda_n$$ where $\lambda_n$ is the Lebesgue measure.

I must be prove that $||\cdot||_1$ is a norm on $C_0(\Omega)$. The triangle inequality and positivity is ok. Remain to prove that $$||f||_1=0\iff f=0\;\text{on}\;\Omega.$$

$(\Leftarrow)$ is obvious; I am having difficulty proving $(\Rightarrow)$.

Suppose that $$||f||_1=0\Rightarrow f=0\;\text{a.e in support}$$

Can I conclude?

Jack J.
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1 Answers1

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If $f(x_0)\neq 0$ for some $x_0\in\Omega$ then since $f$ is continuous there must be a neighborhood $U$ of $x_0$ where $f(x)\neq 0$ for all $x\in U$, and now for an $\varepsilon>0$ there is a ball $B_{\varepsilon}(x_0)\subset U$ .Clearly $$0<\int_{B_{\varepsilon}(x_0 )}|f|d\lambda_n\leq\int_{\Omega}|f|d\lambda_n$$

Peter Melech
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  • I don't understand what you mean – Jack J. Apr 23 '20 at 14:03
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    the space $C_0(\Omega)$ is a space of continuous functions, not a space of equivalence classes of functions as the Lebesgue spaces and thus a continuous function which is not zero must take non-zero values on set of positive measure – Peter Melech Apr 23 '20 at 14:05
  • Can I conlcude from this? – Jack J. Apr 23 '20 at 14:13
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    yes because if the function was non-zero for some $x\in\Omega$ then because of the continuity it should be non-zero on some set with positive lebesgue measure. But you already know that the function $f$ is 0 a.e. ! – Chaos Apr 23 '20 at 14:16
  • Why the integral is made on $B$ and not on U directly? – Jack J. Apr 23 '20 at 14:46
  • It could also be valuated on $U$ directly, just wanted to make clear there is a set of positive measure, using that the open balls form a basis for the topology on $\mathbb{R}^n$ – Peter Melech Apr 23 '20 at 17:00