Consider the equation


$$\partial\_t u(t,x) = -\partial\_{xxx} u(t,x)$$
for $x \in \mathbb R.$


It is well-known that in general we have


$$\Vert u(t) \Vert\_{L^{\infty}} \le C t^{-1/3} \Vert u\_0 \Vert\_{L^1}.$$


I now read in some lecture notes, without proper argument, that for $u\_0$ such that $\operatorname{supp} (\hat u\_0) \subset [\tfrac{1}{2},2] \cup [-2,-\tfrac{1}{2}]$ we have the improved estimate


$$\Vert u(t) \Vert\_{L^{\infty}} \le C t^{-1/2} \Vert u\_0 \Vert\_{L^1}.$$


My question is: Why is that true?


Remark: Before that statement there is the remark that the integral operator


$$I(t)=\int\_{\mathbb R} e^{it \Phi(x)} a(x) \ dx$$ with $\Phi'' \ge c >0$


satisfies the estimate


$$I(t)=\int\_{\mathbb R} e^{it \Phi(x)} a(x) \ dx \le Ct^{-1/2} \Vert a' \Vert\_{L^1}.$$


However, I am not sure if one is supposed to follow from the other.



Improving dispersive estimates for linear KdV
Let $\chi$ be a smooth function with support on $[-3,-1/3] \cup [1/3,3]$ and equals 1 on $[-2,-1/2] \cup [1/2,2]$.


You can write
$$ u(t,x) = \int e^{it \xi^3}e^{ix\xi} \chi(\xi) \hat{u}\_0(\xi) ~d\xi $$
if $u\_0$ has Fourier support as you demanded.
If you write
$$ V(t,x) = \int e^{it\xi^3}e^{ix\xi} \chi(\xi) ~d\xi$$
this shows $u(t,x) = \int V(t,x-y) u\_0(y)~dy$ and so it suffices to show that
$ \|V(t,x)|\_{L^\infty\_x} \lesssim t^{-1/2}$.


For convenience write $\eta(\xi) = t\xi^3 + x \xi$.
We have that $\eta'(\xi\_0) = 0 \iff \xi\_0 = \pm\sqrt{ -x/t}$ (provided $t$ and $x$ are such that the quantity under the square root is positive).


Suppose for convenience $t > 0$ (the $t < 0$ case follows the same way).
And similarly we will only look at the integral over $[1/3,3]$, the other is similar.


On this domain, we have $\eta'' = 6 t \xi > 2t$ and so $\eta'$ is monotonic.


Let $\delta = \frac{1}{\sqrt{t}}$. Split the integral for $V$ into three pieces:


$$ \int\_{1/3}^{\xi\_0 - \delta} + \int\_{\xi\_0-\delta}^{\xi\_0+\delta} + \int\_{\xi\_0+\delta}^3 e^{i\eta}\chi ~d\xi $$


On the first and third piece we have that $|\eta'| > 2 \sqrt{t}$.
So the standard Van der Corput argument gives that the integral is bounded by


$$ \frac{1}{2\sqrt{t}} \left( \| \chi'\|\_{L^1} + 3\|\chi\|\_{L^\infty} \right) $$


In the middle term we are integrating over an interval of width $2\delta$ so we have the integral is bounded by


$$ \frac{2}{\sqrt{t}} \|\chi\|\_{L^\infty} $$


Since $\chi$ is a fixed function, we can just absorb them into constants. And have that $|V(t,x)| \lesssim t^{-1/2}$ as claimed for $t >0$.




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The key step is the lower bound on $\eta''$, which is available since we restricted away from $\xi = 0$. This ensures that if you go a distance $\delta$ away from the critical point you pick up a lower bound on $|\eta'|$ so the usual stationary phase argument gives you a bound by $\frac{1}{\inf |\eta'|}$.


Without this restriction $\eta'$ may stay small for a wider range of $\xi$. This would be the case when $x = 0$ (which in our case is not a problem because the corresponding $\xi\_0$ is outside of the domain of integration). In this case instead of $\delta = 1/\sqrt{t}$ you need to choose $\delta = 1/\sqrt[3]{t}$, which would guarantee (using that $\eta''' = 6t > 0$) that when you get $\delta$ away from $\xi\_0$, your $|\eta'| \geq 3 \sqrt[3]{t}$.


