Let $G$ be a locally compact group and $\phi$ be in $ L^{\infty}(G)$ that annihilates $I\subset L^1(G)$, 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?

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?

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

$\mathbf{QUESTION}$: Let $G$ be a locally compact abelian group. Is it true that if $\phi$ annihilates $L^1(G)$, then $\int_G f(-y)\phi(y)dy=0$ for all $f\in L^1(G)$? Is the convers true?

Let $G$ be a locally compact group and $\phi$ be in $ L^{\infty}(G)$ that annihilates $I\subset L^1(G)$, 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?