Question: Let $u_0\in L^1(\mathbb R)$ with $\text{supp}(u_0)\in \{\xi\in\mathbb R|\frac{1}{2}\le|\xi|\le 2\}$ and $u$ solves $$\begin{cases} \partial_t u +\partial_x^3u=0\\ u|_{t=0}=u_0. \end{cases}$$ I need to prove that there exists $C>0$ such that for all $t\neq0$, and $u_0$ satisfying the assumption, $$\Vert u(t,\cdot)\Vert_{L^\infty}\le \frac{C}{|t|^{1/2}}\Vert u_0\Vert_{L^1}.$$ I was hinted to use the following theorem which I managed to prove.
Theorem: Let $a\in\mathcal D(\mathbb R)$ and $\Phi\in C^2(\mathbb R)$ such that for all $\xi\in \text{supp}(a)$ it holds that $\Phi''(\xi)\ge c_0$ for $c_0>0$. Then for all $t\neq 0$, $$\bigg|\underbrace{\int_\mathbb R e^{it\Phi(\xi)}a(\xi)\,d\xi}_{:=I(t)}\,\bigg|\le \frac{C(c_0)}{|t|^{1/2}}\Vert a'\Vert_{L^1}.$$
My attempt: Taking Fourier transform on space variable, equation transforms into $$\begin{cases} \partial_t \hat u =i\xi^3\hat u\\ \hat u|_{t=0}=\hat u_0 \end{cases}$$ so we have $\hat u = e^{i\xi^3t}\hat u_0$. Let $$K_t:=\frac{1}{2\pi}\int_{\text{supp}(\hat u_0)}e^{i\xi x+i\xi^3t}d\xi=\frac{1}{\pi}\int_{1/2}^2e^{i\xi x+i\xi^3t}d\xi$$ so by Fourier inversion, $u=K_t\star u_0.$ Then we have the obvious bound from Young's inequality: \begin{equation}\Vert u(t,\cdot)\Vert_{L^\infty}\le \Vert K_t\Vert_{L^\infty}\Vert u_0\Vert_{L^1}.\tag{1}\end{equation} So it remains to show $\Vert K_t\Vert_{L^\infty}\le \frac{C}{|t|^{1/2}}$. In the theorem, set $a(\xi)=e^{i\xi x}\mathbb 1_{[1/2,2]}$, $\Phi(\xi)=\xi^3$ so on $[\frac{1}{2},2]$, $\Phi''(\xi)\ge 3$. Then $$|K_t(x)|=:|I(t,x)|\le \frac{C}{|t|^{1/2}}\Vert ixe^{i\xi x}\mathbb 1_{[1/2,2]}\Vert_{L_\xi^1}=\frac{C|x|}{|t|^{1/2}}$$ but I can't find a way to get rid of the $x$ dependence on this bound.
Thanks in advance for any help.
Updates: Just to check that $K_t$ is bounded function of $x$, I tried plotting it for fixed $t$ and (although this doesn't prove anything) it seems like $K_t$ is bounded. This indicates that taking Young's inequality to obtain (1) is the right approach so I really just need a bound on $K_t$.
