I have a question regarding the theory of Fourier transforms in locally compact abelian groups.
Let $f \in L^p(G)$ we the Fourier transform of $f$ is the function $\widehat f$ on $\widehat G$, the Pontryagin dual of $G$ defined by: $$ \widehat f(\xi) = \int_G \xi(x)f(x)dx. $$ Moreover for $f \in L^p(\widehat G)$ we can define the inverse Fourier transform $$ \check f(x)=\int_\widehat G \xi(x)f(\xi)d\xi. $$
There is a Hausdorff-Young inequality:
Let $1 \leq p \leq 2$ and $p^{-1}+q^{-1}=1$. If $f \in L^p(G)$ then $\check f \in L^q(\widehat G)$, moreover, $$ \| \widehat f\|_q \leq \|f\|_p. $$ My question if the reverse order of this inequality is also true, more precisely, let again $1 \leq p \leq 2$ and $p^{-1}+q^{-1}=1$, but this time we start with $f \in L^q(\widehat G)$ and we want to prove that $\check f \in L^p(G)$ and the inequality: $$ \| f\|_q \leq \| \check f\|_p \quad \text{ holds.} $$ This would be true using the Fourier Uniqueness Theorem if I could prove that for all $\xi \in \widehat G$: $$ \widehat{\check f}(\xi) = f(\xi), $$ however I'm not being able to prove this.