Leray projector in $L^{\infty}$ and negative order Besov spaces for the Navier-Stokes equations

I was reading the paper "Norm inflation for the generalized Navier-Stokes equations" which can be found here: https://arxiv.org/abs/1212.3801.

In Lemma 2.1, the authors said for any $$\phi \in L^{\infty}$$ and $$\alpha>0$$ $$\Vert \nabla e^{-t(-\Delta)^{\alpha}}\Vert_{L^\infty} \leq C t^{-\frac{1}{t^{2\alpha}}}\Vert \phi\Vert_{L^\infty} .$$

I am not sure why this is true since the Leray projector $$\mathbb P$$ is a bounded operator in $$L^{2}$$. Furthermore, if we want to replace the $$L^{\infty}$$ norm with some negative Besov norm $$\dot B^{-s}_{p,q}$$, is the above inequality still true? I appreciate if any reference could be provided.

Let $$K(x)$$ denote the Schwartz kernel of the Fourier multiplier $$\nabla e^{-t(-\Delta)^\alpha}$$. It's straightforward to check from Fourier inversion and Plancherel's theorem that for $$t>0$$, $$K(x) = (2\pi)^{-3}\int_{\mathbb{R}^3} i\xi e^{-t|\xi|^{2\alpha}}e^{i\xi\cdot x}d\xi,\qquad\forall x\in\mathbb{R}^3.$$ I claim that $$K\in L^1(\mathbb{R}^3)$$. To see this, observe that for any multi-index $$\beta=(\beta_1,\beta_2,\beta_3)$$ of order $$|\beta|=4$$ and $$x\in\mathbb{R}^3$$ such that $$x^\beta = x_1^{\beta_1}x_2^{\beta_2}x_3^{\beta_3}\neq0$$, we have that $$K(x)=\frac{1}{x^\beta (2\pi)^3}\int_{\mathbb{R}^3}i\xi e^{-t|\xi|^{2\alpha}} (\partial_{\xi}^\beta e^{i\xi\cdot x})d\xi.$$ Now integrate by parts in $$\xi$$. The most singular term in the application of the Leibnitz rule comes from repeatedly differentiating $$|\xi|^{2\alpha}$$. You can show that $$|K(x)|\lesssim_\alpha \frac{1}{|x^\beta|}\int_{\mathbb{R}^3}|\xi|^{2\alpha-3}e^{-t|\xi|^{2\alpha}}d\xi\lesssim \frac{t^{-1/2\alpha}}{|x^\beta|},$$ where the last expression follows from dilation invariance of Lebesgue measure. Since $$|\beta|>3$$, the desired conclusion follows.