Let's consider the propagator corresponding to the one-dimensional equation
$$
u\_t=\Lambda^\alpha u\_x,\; u(x,0)=f(x)
$$
where 
$$
\widehat{\Lambda^\alpha u}=|\xi|^\alpha\hat{u}(\xi),
$$
and $-1< \alpha\leq 1$. 


QUESTION 1) Do you know any good reference in the dispersive estimates for this operator? I mean inequalities of the following form
$$
\|u(t)\|\_{L^\infty(\mathbb{R}}\leq (1+t)^{-\theta}\|f\|\_{X},
$$
for certain $\theta$ and Banach space $X$.


QUESTION 2) Same for the propagator corresponding to
$$
u\_t=(1-\partial\_x^2)^\alpha u\_x,\; u(x,0)=f(x)
$$
and $-1\leq \alpha\leq 1$.



Dispersive estimate for linear semigroup
First note that when $\alpha = 0$ you have the linear advection equation with no decay whatsoever, in either case, as $L^\infty$ is conserved. So we focus on the situation where $\alpha \neq 0$. 


### Question 1


Since you have scaling homogeneity, it suffices to localize in frequency. Let $\chi\_\pm$ be bump functions supported near $\xi \approx \pm 1$. Your goal is to essentially estimate 
$$ \int e^{i t |\xi|^\alpha \xi + i x \xi} ~\mathrm{d}\xi \approx \sum\_{\lambda = -\infty}^\infty \int e^{it |\xi|^\alpha \xi + i x\xi} (\chi\_+(2^\lambda \xi) + \chi\_-(2^\lambda \xi)) ~\mathrm{d}\xi = \sum P^{-\lambda}\_+ + P^{-\lambda}\_- $$
And we have that
$$ P\_{\pm}^{-\lambda}(2^{\lambda(\alpha + 1)}t, 2^{\lambda} x) = \frac{1}{2^\lambda} P\_{\pm}^0 $$
The Van der Corput Lemmas imply that $P^0$ has a uniform bound by $1/\sqrt{t}$ as $\alpha(\alpha+1) > 0$ if $\alpha > -1$ and $\alpha \neq 0$. So this gives the bound in terms of $X$ being a Homogeneous Besov space 


$$ \| u\|\_{\infty} \lesssim \frac{1}{\sqrt{t}} \|f\|\_{B^{(1-\alpha)/2,1}\_1} $$


### Question 2


You don't have scaling homogeneity. But instead you can consider
$$P\_{\pm}^{\lambda} = 2^{\lambda} \int e^{i t \langle 2^{\lambda} \xi\rangle^{2\alpha} 2^{\lambda} \xi} e^{i 2^\lambda \xi x} \chi\_{\pm}(\xi) ~\mathrm{d}\xi $$
for the frequency $2^{\lambda}$ piece. You see that the phase function 
$\eta = 2^\lambda \xi \langle 2^\lambda \xi\rangle^{2\alpha} + 2^\lambda \xi (x / t)$ is such that 
$$ |\eta''| \gtrsim \alpha 2^{\lambda (2\alpha -1)} |\xi|^{2\alpha - 3} \left| 1 + (1+2\alpha) 2^{2\lambda} |\xi|^2 \right|. $$
We see that when $\alpha \neq 0$ and $\alpha > -\frac12$, we have that the lower bound is by $2^{\lambda(2\alpha + 1)}$, so by Van der Corput Lemma (applied to $\tilde{t} \tilde{\eta} = t\eta$ where $\tilde{\eta} = 2^{-\lambda(2\alpha + 1)} \eta$ and $\tilde{t} = 2^{\lambda(2\alpha + 1)} t$) we get exactly the same decay estimate as in Question 1. 


The regularization near $|\xi| = 0$ gives us however that, when $\alpha = -\frac12$, $|\eta''| \gtrsim 2^{-2\lambda}$. So in this case we have by Van der Corput Lemma that 
$$ \|u\|\_{\infty} \lesssim \frac{1}{\sqrt{t}} \| f\|\_{B^{2,1}\_1} $$


(a loss of one derivative compared to the $\alpha > -1/2$ case)




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For the small time estimate you can use $L^2$ conservation plus Sobolev to get the bound 


$$ \|u\|\_{\infty} \lesssim \|f\|\_{W^{1,2}}$$


