In Appendix A of Tao's book Nonlinear Dispersive Equations, a Bernstein inequality has puzzled me a long time:
Suppose that $P_{\geq N}$ is smoothed out projection to the region $|\xi|\geq N$, where $N\in 2^{\mathbb{Z}}$ is a dyadic number, $\hat{\varphi}(\xi)$($|\hat{\varphi}(\xi)|\leq1$) is a real valued radially symmetric bump function that is equal to $1$ in ball $B(0;1)$ and supported in ball $B(0;2)$.
We define $\widehat{P_{\geq N}f}:=(1-\varphi(2\xi/N))\hat{f}(\xi)$.
Then Tao says that if $s\geq0$, we have $$\|P_{\geq N}f\|_{L^{p}(\mathbb{R}^{d})}\lesssim_{p,s,d}N^{-s}\||\nabla|^{s}P_{\geq N}f\|_{L^{p}(\mathbb{R}^{d})}.$$
How can we get this inequality?
My attempt is \begin{align} \widehat{P_{\geq N}f}=&(1-\varphi(2\xi/N))\hat{f}(\xi)\\ =&|\xi|^{-s}|\xi|^{s}(1-\varphi(2\xi/N))\hat{f}(\xi) \end{align} here $|\xi|^{-s}$ is well defined because the support of $(1-\varphi(2\xi/N))$ is away from $0$. Then what should we do to deal with the above?
We know that $|\xi|^{s}(1-\varphi(2\xi/N))\hat{f}(\xi)=\widehat{|\nabla|^{s}P_{\geq N}f}$ and $|\xi|^{-s}\leq 2^{s}N^{-s}$, and then how to get Tao's result? Maybe I can just get \begin{equation} |\widehat{P_{\geq N}f}|\leq 2^{s}N^{-s}|\widehat{|\nabla|^{s}P_{\geq N}f}|. \end{equation}
If we remove the hat, we should get \begin{align} P_{\geq N}f&=\mathscr{F}^{-1}\big(|\xi|^{-s}|\xi|^{s}(1-\varphi(2\xi/N))\hat{f}(\xi)\big)\\&=\mathscr{F}^{-1}\big(|\xi|^{-s}|(1-\varphi(3\xi/N))|\xi|^{s}(1-\varphi(2\xi/N))\hat{f}(\xi)\big)\\&=\mathscr{F}^{-1}\big(|\xi|^{-s}|(1-\varphi(3\xi/N))\big)*(|\nabla|^{s}P_{\geq N}f). \end{align}
Is $\mathscr{F}^{-1}\big(|\xi|^{-s}(1-\varphi(3\xi/N))\big)\in L^{1}(\mathbb{R}^{d})$?
Here is one approach. You can use Young's inequality and the principle of nonstationary phase to prove \begin{equation} \|P_M|\nabla|^{-s}g\|_{L^p}\lesssim M^{-s}\|g\|_{L^p}\tag{1}, \end{equation} where $P_M$ is the Fourier projection to frequencies $|\xi|\sim M$. If you've done this, then consider for $s > 0$ \begin{align*} \|P_{\ge N/2}|\nabla|^{-s}g\|_{L^p} \le \sum_{M>N/2\ \text{dyadic}}\|P_{M}|\nabla|^{-s}g\|_{L^p}\lesssim \sum_{M>N/2\ \text{dyadic}}M^{-s}\|g\|_{L^p} \sim N^{-s}\|g\|_{L^p}. \end{align*} Finally, we can set $g = |\nabla|^sP_{\ge N}f$ and use that $P_{\ge N/2}P_{\ge N} = P_{\ge N}$.
Here is a proof of $(1)$. By definition, $P_M|\nabla|^{-s}g = K\ast g$, where $K(x) = (\phi(\xi)|\xi|^{-s})^{\vee}(x)$ for an appropriate smooth $\phi$ supported in $|\xi|\sim M$. By Young's inequality, $$ \|K\ast g\|_{L^p}\le \|K\|_{L^1}\|g\|_{L^p}. $$ I claim that $\|K\|_{L^1}\lesssim M^{-s}$. First a heuristic and then a rigorous proof. Note that $$ K(x) = \int \phi(\xi)|\xi|^{-s}e^{2\pi ix\cdot\xi}\,d\xi $$ is an oscillatory integral. Moreover, $|K(x)| \lesssim M^{-s+d}$ by the definition of $\phi$, and by the uncertainty principle $|K(x)|$ is rapidly decaying if $|x|>C/M$. Therefore, $$ \int |K(x)|\,dx\lesssim \int_{|x|<C/M}M^{-s+d}\,dx + \int_{|x|>C/M}|K(x)|\,dx \sim M^{-s} $$ (the second term is minor by the uncertainty principle heuristic).
More rigorously, let $\Phi(\xi) = \phi(\xi)|\xi|^{-s}$, and consider $$ \int_{|x|>C/M}|\int \Phi(\xi)e^{2\pi ix\cdot\xi}\,d\xi|\,dx. $$ Let $\Delta_\xi = -\sum_{j=1}^d\partial_{\xi_j}^2$ be the Laplacian in the $\xi$ variables, and note that by the chain rule, $$ \frac{c_k}{|x|^{2k}}\Delta_\xi^ke^{2\pi ix\cdot\xi} = e^{2\pi ix\cdot\xi}. $$ Inserting this identity, using that $\Delta_\xi^k$ is self-adjoint (integrating by parts), and the triangle inequality, we get an upper bound of $$ \int_{|x|>C/M}\frac{dx}{|x|^{2k}}\cdot\int |c_k\Delta_\xi^{k}\Phi(\xi)|\,d\xi. $$ Since $\phi(\xi),|\xi|^{-s}$ are radial functions, so is $\Phi$, and we can use the following formula for the Laplacian of a radial function $\Phi(\xi) = f(r)$: $$ \Delta f = \frac{1}{r^{d-1}}\frac{d}{dr}(r^{d-1}f'). $$ Using this formula on $\Phi(r) = \phi(r)r^{-s}$, you can prove by induction on $k$ that $|\Delta^k\Phi|\lesssim M^{-s-2k}$ since $r\sim M$, and $|\phi^{(l)}(r)|\lesssim M^{-l}$. Therefore, the integral in $\xi$ is bounded by $M^{d-2k-s}$.
If $2k>d$, we can do the integral in $x$ by using polar coordinates: $$ \int_{|x|>C/M}\frac{dx}{|x|^{2k}}\sim\int_{C/M}^\infty r^{d-1-2k}\,dr \sim (\frac{C}{M})^{d-2k}=M^{2k-d}C^{d-2k}. $$ Overall we found $$ \int_{|x|>C/M}|\int \Phi(\xi)e^{2\pi ix\cdot\xi}\,d\xi|\,dx\lesssim_{s,d,k}M^{2k-d}C^{d-2k}\cdot M^{d-2k-s} = C^{d-2k}M^{-s}\sim_C M^{-s}, $$ as desired.