Let $\dot{B}^s_{\infty,\infty}(\mathbb{R}^d)$ denote the homogeneous Besov space of order $s$ with second and third index $\infty$, i. e. the homogeneous Zygmund space. Let $\mathcal{S}(\mathbb{R}^d)$ denote the Schwartz space of rapidly decaying functions.
It is known that $\mathcal{S}(\mathbb{R}^d)$ is not dense in $\dot{B}^0_{\infty,\infty}(\mathbb{R}^d)$. It is also known that $\dot{B}^0_{\infty,\infty}(\mathbb{R}^d)$ is stricly larger than the Lebesgue space $L_\infty(\mathbb{R}^d)$.
My question is the following: Can we bound the $L_\infty$-norm of a Schwartz function $f$ by its $\dot{B}^0_{\infty,\infty}$-norm? In other words, is there a constant $C>0$ such that $$\|f\|_{L_\infty(\mathbb{R}^d)}\leq C\|f\|_{\dot{B}^0_{\infty,\infty}(\mathbb{R}^d)}$$ for every $f\in\mathcal{S}(\mathbb{R}^d)$? A somewhat weaker statement would be whether there is a constant $C>0$ and some arbitrarily large $s>0$ such that $$\|f\|_{L_\infty(\mathbb{R}^d)}\leq C\|f\|_{\dot{B}^0_{\infty,\infty}(\mathbb{R}^d)}+C\|f\|_{\dot{B}^s_{\infty,\infty}(\mathbb{R}^d)}$$ for every $f\in\mathcal{S}(\mathbb{R}^d)$.
Interpolation theory tells us that if one replaces $\dot{B}^0_{\infty,\infty}(\mathbb{R}^d)$ in the above statement by $\dot{B}^{-\varepsilon}_{\infty,\infty}(\mathbb{R}^d)$ for some $\varepsilon>0$ then the latter inequality is true for every $f\in\dot{B}^{-\varepsilon}_{\infty,\infty}(\mathbb{R}^d)\cap \dot{B}^{s}_{\infty,\infty}(\mathbb{R}^d)$. However, $\mathcal{S}(\mathbb{R}^d)$ is not dense in $\dot{B}^{-\varepsilon}_{\infty,\infty}(\mathbb{R}^d)\cap \dot{B}^{s}_{\infty,\infty}(\mathbb{R}^d)$ either, so there might still be some room for improvement. Also, if one replaces the homogeneous Besov spaces above by their inhomogeneous counterparts then at least the latter statement holds by the Sobolev embedding Theorem.