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Let $G$ be a locally compact group and $\phi$ be in $ L^{\infty}(G)$ that annihilates $I$, where $I$ is a closed ideal of $ L^1(G)$, so by duality we have: $$\int_G f(y)\phi(y)dy=0$$ for all $f\in I$ …

Really, I prove that if $f\in I$, then $\check{f}\in I$ where $I$ is a closed ideal of $L^1(G)$ and $\check{f}(x)=f(x^{-1})$ for every unimodular group $G$. Therefore since $\phi$ annihilate $I$. so …

For abelian case, close ideal of $L^1(G)$ is equal to translation invariant subspace of $L^1(G)$.[ please see: W.Rudin, Harmonic analysis on semigroup] In general you can see [ Folland, Harmonic analy …