I'm reading the proof of Lemma 2.11 of that book, for which Tao has an [errata](https://terrytao.wordpress.com/books/nonlinear-dispersive-equations-local-and-global-analysis) showing that the case $b=b'$ is not obvious. But I can't quite understand his explanation on how to show that case. Could anyone provide more details in that regard?


**Added**: An abridged online version of Tao's book can be found [here](http://www.math.ucla.edu/~tao/preprints/chapter.pdf).



Lemma 2.11 of Tao's Nonlinear Dispersive Equations
I have now figured out the question, so I'll record it here.


Let $L=ih(\nabla/i)$ be a constant coefficient differential operator, where $h$ is a polynomial. Recall the Bourgain norm is defined as


$$ \|u\|\_{X^{s,b}\_h} = \| \langle\xi\rangle^s\langle \tau-h(\xi) \rangle^b\hat u(\tau,\xi) \|\_{L^2}. $$


Let $\eta$ be a mollifier. We want to show, for any $s\in\mathbb R$, $-1/2<b<1/2$, and $0<T\le1$,


$$ \|\eta(t/T)u\|\_{X^{s,b}}\ll\_{s,b,\eta}\|u\|\_{X^{s,b}}. $$


First we make some reductions. Let $\Lambda$ be the operator such that $\widehat{\Lambda u}(\xi)=\langle\xi\rangle u(\xi)$. We replace $u$ by $\Lambda^s u$ so we can assume $s=0$.
Also, since


$$ \widehat{e^{-tL}u}(\tau,\xi)=\widehat{e^{-ith(\xi)}\hat u(t,\xi)}(\tau)=\hat u(\tau-h(\xi),\xi), $$


We replace $u$ by $e^{-tL}u$ so we can assume $h=0$. Now the inequality becomes


$$ \|\eta(t/T)u\|\_{L\_x^2H\_t^b}\ll\_b\|u\|\_{L\_x^2H\_t^b} $$


and it suffices to show the pointwise inequality


$$ \|\eta(t/T)u\|\_{H\_t^b}\ll\_b\|u\|\_{H\_t^b} $$


and integrate. By duality $(H\_t^b)^\*=H\_t^{-b}$ we can also assume $0\le b<1/2$.


To show this inequality, we use the Littlewood-Paley decomposition operator $P\_k$. We have


$$ \|\eta(t/T)u\|\_{H\_t^b}\ll\_b \sum\_k 2^{bk} \|P\_k(\eta(t/T)u)\|\_{L^2}
\ll\_b \sum\_k 2^{bk} (\|\eta(t/T)P\_{\ge k}u\|\_{L^2}+\|P\_{\ge k}\eta(t/T) P\_{<k}u\|\_{L^2}). $$


The first term is bounded by


$$ \|\eta(t/T)\|\_{L^\infty}\sum\_k 2^{bk}\sum\_{l\ge k}\|P\_lu\|\_{L^2}\ll\_b \|\eta(t/T)\|\_{L^\infty}\sum\_{k,l\ge k} 2^{b(k-l)} \|P\_l\Lambda^bu\|\_{L^2}\ll\_b \|\eta(t/T)\|\_{L^\infty}\|u\|\_{H^b}. $$


Similarly the second term is bounded by


$$ \sum\_k \|P\_{\ge k}\eta(t/T)\|\_{H^b}\|P\_{<k}u\|\_{L^\infty}. $$


Let $\phi\_{<k}$ be the convolution kernel of $P\_{<k}$. Then


$$P\_{<k}u=\phi\_{<k}\*u=\Lambda^{-b}\phi\_{<k}\*\Lambda^bu,$$


so


$$ \|P\_{<k}u\|\_{L^\infty}\le\|\Lambda^bu\|\_{L^2}\|\Lambda^{-b}\phi\_{<k}\|\_{L^2}\ll\_b 2^{(1/2-b)k}\|u\|\_{H^b}. $$


Thus the second term is bounded by


$$ \|u\|\_{H^b}\sum\_k 2^{(1/2-b)k}\|P\_{\ge k}\eta(t/T)\|\_{H^b}\ll\_b \|u\|\_{H^b}\sum\_{k,l\ge k} 2^{(1/2-b)(k-l)}\|P\_l\eta(t/T)\|\_{H^{1/2}}\ll\_b \|u\|\_{H^b}\|\eta(t/T)\|\_{H^{1/2}}. $$


Finally


$$ \|\eta(t/T)\|\_{H^{1/2}}^2=\int \langle\tau\rangle T^2|\hat\eta(T\tau)|^2d\tau\le\int \langle T\tau\rangle |\hat\eta(T\tau)|^2d(T\tau)=\|\eta\|\_{H^{1/2}}^2 $$


comletes the proof.


