Suppose $1\leq p \leq 2$. The Hausdorff-Young inequality states that $$\|\hat{f}\|_{L^{p^{\prime}}(\mathbb{R}^n)}\leq \|f\|_{L^{p}(\mathbb{R}^n)}\quad (1)$$
where $p^{\prime}$ is the conjugate exponent of $p$.
Probably the easiest way to prove (1) is interpolation between the trivial estimate $$\|\hat{f}\|_{L^{\infty}(\mathbb{R}^n)}\leq \|f\|_{L^{1}(\mathbb{R}^n)},$$ and Plancherel's identity: $$\|\hat{f}\|_{L^{2}(\mathbb{R}^n)}\leq \|f\|_{L^{2}(\mathbb{R}^n)}$$.
My question is does the inequality (1) hold on bounded sets that contain the origin, i.e., is it true that
$$\|\hat{f}\|_{L^{p^{\prime}}(\Omega)}\leq
\|f\|_{L^{p}(\Omega)},$$
where $\Omega$ is a bounded measurable set with a finite Lebesgue measure such that $0\in \Omega$ ?
No. Take $n=1$ for simplicity. Let $\Omega=[-1,1]$. Let $f=\chi_{[2,3]}$. Then $f|_\Omega=0$, but $\hat f$ does not vanish on $\Omega$ (either calculate $\hat f$ explicitly or note that it's analytic, so it can't vanish on $[-1,1]$ without vanishing identically).