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Let $G$ be locally compact group prove that

$$L_{0}^{1}(G)=\left\{f\in L^{1}(G): \int_G f(g) dm(g)=0 \right\}$$ is a closed ideal in $ L^{1}(G)$ with codimension one

I am grateful for any help

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Hint: If $f,g\in L^1(G)$ it's easy to verify from the definition that $$\int_Gf*g=\int_G f\int_Gg.$$

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  • $\begingroup$ @Dear David C. Ullrich, thanks, could you tell me how to do I prove closed? $\endgroup$ – user62498 Apr 29 at 5:21
  • $\begingroup$ @user62498 Come on now! Say $I$ is that set. It should be obvious that $I$ is the kernel of a bounded linear functional. Or if that's too abstract, assume $f_n\in I$, $f_n\to f$ and show $f\in I$. $\endgroup$ – David C. Ullrich Apr 29 at 16:32
  • $\begingroup$ thanks, your comment is definitely an improvement over mine I do it $\endgroup$ – user62498 Apr 30 at 13:47

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