I'm reading the proof of Lemma 2.11 of that book, for which Tao has an errata 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.
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.
