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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 $f\in I$.

$\mathbf{QUESTION}$: Let $G$ be a locally compact abelian group. Is it true that if $\phi$ annihilates $I$, then $\int_G f(-y)\phi(y)dy=0$ for all $f\in I$? Is the convers true?

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  • $\begingroup$ If $\phi$ annihilates $L^1(G)$, then $\phi$ is zero itself. $\endgroup$
    – user1688
    Commented Apr 16, 2017 at 12:32
  • $\begingroup$ Thanks you prof. Corbennik for your comment. You are right. Infact $\phi$ annihilates $I\subset L^1(G)$. $\endgroup$
    – M.fouladi
    Commented Apr 16, 2017 at 12:42
  • $\begingroup$ Oh shucks, I misread the question. Well for that one you may choose an open set $U$ and let $I$ be the ideal of all $f$ that vanish on $U$. Then $\phi$ must simply vanish on $U$, but not on $-U$. $\endgroup$
    – user1688
    Commented Apr 16, 2017 at 13:08
  • $\begingroup$ The comments on your previous question mathoverflow.net/questions/266727 seemed to indicate that Rudin is defining the duality pairing between $L^1(G)$ and $L^\infty(G)$ in a way different from the one you use at the start of your question. Hence I don't understand the motivation for your actual question $\endgroup$
    – Yemon Choi
    Commented Apr 16, 2017 at 18:40

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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 $$\int f(y^{-1})\phi(y)dy=0$$

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