Let $p^{\prime}, q^{\prime}$, and $s^{\prime}$ denote the conjugate exponents of $p,q,$ and $s$ respectively, i.e., $\frac{1}{q^{\prime}} + \frac{1}{q} = 1$.
The Hausdorff-Young inequality says that if $1 \le p \le 2$
\begin{equation} \tag{1} \label{1} \|\widehat{f}\|_{p^{\prime}} \le \|f\|_{p} \quad \forall f \in L^{p}(\mathbb{R}^{n}). \end{equation}
I want to show that that if \eqref{1} holds for some $q$ instead of $p^{\prime}$, i.e.,
\begin{equation} \tag{2} \label{2} \| \widehat{f} \|_{q} \le \|f\|_{p} \quad \forall f \in L^{p}(\mathbb{R}^{n}) \end{equation}
then: (A) $q = q^{\prime}$ and (B) $1 \le p \le 2$.
To show (A), I have tried to use dilations. To that end, for $\lambda > 0$ define $g_{\lambda}(x) = \lambda^{-n} g(x \lambda^{-1}).$ Assume $g \in L^{s}(\mathbb{R}^{n})$. Then so is $g_{\lambda}$. Moreover, one can compute,
\begin{equation} \tag{3} \label{3} \| g_{\lambda} \|_{s} = \left( \int_{\mathbb{R}^{n}} \lambda^{-ns} |g(x \lambda^{-1})|^{s} dx \right)^{1/s} = \lambda^{-n} \left( \int \lambda^{n} |g(u)|^{s} du \right)^{1/s} = ( \lambda^{\frac{1}{s} - 1} )^{n} \| g \|_{s}. \end{equation}
So, my attempt at a proof of (A) goes as follows: Assume that \eqref{2} holds for some $p,q$. Then by applying \eqref{2} to some $f_{\lambda}$ with $f \in L^{p}$, we attain,
\begin{equation} \tag{4} \label{4} \| \widehat{f}_{\lambda} \|_{q} \le \| f_{\lambda} \|_{p}. \end{equation}
However, by \eqref{3}, we see that
\begin{equation} \tag{5} \label{5} \left( \lambda^{1/q - 1} \right)^{n} \| \widehat{f} \|_{q} \le \left( \lambda^{1/p - 1} \right)^{n} \| f \|_{p}. \end{equation}
Since $\frac{1}{q} - 1 = \frac{-1}{q^{\prime}}$ we see this is equivalent to
$$ \| \widehat{f} \|_{q} \le \left(\lambda^{\frac{1}{q^{\prime}} - \frac{1}{p^{\prime}}} \right)^{n} \| f \|_{p}. $$
Hence, depending on the sign of $(\frac{1}{q^{\prime}} - \frac{1}{p^{\prime}})$ you can send $\lambda$ to infinity or $0$, showing that $\| \widehat{f}\|_{q} = 0$. Thus, if \eqref{2} holds for all $f \in L^{p}$ we have that $q^{\prime} = p^{\prime}$. But, this means $q = p$, contradicting the Hausdorff-Young inequality. In fact, we want $q = p^{\prime}$... but I don't see where my argument went wrong.
Any corrections or alternative proof ideas are appreciated! Thanks.
I discovered my mistake. Since $\widehat{f}_{\lambda} (\xi) = \widehat{f}(\lambda \xi).$ This means that I cannot apply (3) on the left hand side of (4). Instead, I have to apply,
$$ \begin{equation} \tag{6} \label{6} \|\widehat{g}_{\lambda} \|_{s} = \lambda^{-\frac{n}{s}} \| \widehat{g} \|_{s}. \end{equation} $$
So that (5) becomes
$$ \lambda^{- \frac{n}{q}} \| \widehat{f} \|_{q} \le \left( \lambda^{1/p - 1} \right)^{n} \| f \|_{p}. $$
Hence,
$$ \| \widehat{f} \|_{q} \le \left( \lambda^{\frac{1}{p} + \frac{1}{q} - 1} \right)^{n} \| f \|_{p}, $$
resulting in $q = p^{\prime}$ as desired.
Let this serve as a warning to abuse of notation.