I know that Hausdorff–Young inequality says that If $ f \in L^p(\mathbb{R}^d)$, and $1\leq p\leq 2$ then
\begin{equation*} \| \hat{f}\|_{L^{q}(\mathbb{R}^d)}\leq \|f\|_{L^p(\mathbb{R}^d)}, \quad \frac{1}{p}+\frac{1}{q}=1. \end{equation*} where $\hat{f}$ is the Fourier transform of $f$.
I am reading a proof where the reverse is used and is attributed to the same theroem, that is given $1 \leq q \leq 2$ and $\hat{f} \in L^q(\mathbb{R}^d)$ , it is claimed that
\begin{equation*} \| f\|_{L^{r}(\mathbb{R}^d)}\leq \|\hat{f}\|_{L^q(\mathbb{R}^d)}, \quad \frac{1}{r}+\frac{1}{q}=1. \end{equation*}
Is don't see why the reverse should hold. Any idea?
The $(q,r)$ in the second statement is the $(p,q)$ of the first statement. So, if $\hat{f}\in L^q$ and $1\leq q\leq 2$, then by the first statement, $\|\hat{\hat{f}}\|_{L^r}\leq \|\hat{f}\|_{L^q}$. But now recall that the double Fourier transform is essentially $f(-x)$, so their $L^r$ norms are the same (this is obvious in the case when $f$ is Schwartz; in general you can argue by density for example).
So, to be clear, this is not a reversed Hausdorff-Young inequality. It is exactly the Hausdorff-Young inequality, just applied to $\hat{f}$ instead of $f$. The reverse of Hausdorff-Young would mean the first statement with $(p,q)$ interchanged, and this is a false inequality (except when $p=q=2$ of course); there are explicit counterexamples.