de Leeuw TheoremIf $m:\mathbb R^d\to \mathbb C$ is continuous and Fourier multiplier on $ L^{p}(\mathbb R^d) (1\leq p < \infty)$, i.e., $\|(m \hat{f})^{\vee}\|_{L^p} \leq C_1 \|f\|_{L^p}$, then $m|_{\mathbb Z^d},$ the restriction of $m$ to $\mathbb Z^d$, is a Fourier multiplier on $L^{p}(\mathbb T^d),$ i.e., $\|(m \hat{f})^{\vee}\|_{L^p (\mathbb T^d)} \leq C_2 \|f\|_{L^p(\mathbb T^d)}$.
My question is: Can we say: If $\|(m \hat{f})^{\vee}\|_{L^p} \leq C \|f\|_{L^p}$ then $\|(m \hat{f})^{\vee}\|_{L^p (\mathbb T^d)} \leq C \|f\|_{L^p(\mathbb T^d)}$. In other words, Can we say $C_1=C_2$? If not, any relation between $C_1$ and $C_2$?