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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.

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    $\begingroup$ Since people who do not own that book may still be able to answer your question, it is perhaps better to included precisely what Lemma 2.11 is and what the number $b$ is in your question. $\endgroup$ Apr 7 '15 at 12:51
  • $\begingroup$ @WillieWong The same page I linked to has an abridged version of the book that contains all material in the first 3 chapters. $\endgroup$
    – Fan Zheng
    Apr 7 '15 at 18:56
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    $\begingroup$ It's still better, if possible, to try to make your question self-contained and not require people to read even an abridged version of the book. $\endgroup$
    – Deane Yang
    Apr 7 '15 at 21:50
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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.

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