Equivalence of W^{m,p} and H^{s,p} norms I am reading Cazenave's Semilinear Schroedinger Euqation.
On page xi the two norms are defined:
$\| u \|_{W^{m,p}} = \sum\limits_{|\alpha|\leq m} \| D^{\alpha} u \|_{L^p}$,
$\| u \|_{H^{s,p}} = \left\| \mathcal{F}^{-1} \left[ \left( 1+|\xi|^2 \right)^{\frac{s}{2}} \mathcal{F}u \right] \right\|_{L^p}$,
where $\mathcal{F}$ and $\mathcal{F}^{-1}$ are the Fourier and inverse Fourier transforms, respectively.
On page 13 (Remark 1.4.1, (iv)) it is stated that from Mihlin's multiplier theorem and for non-negative integer $m$, the equivalence between the norms of $W^{m,p}(\mathbb R^N)$ and $H^{m,p}(\mathbb R^N)$ follows.
Mihlin's multiplier theorem claims that if $\sigma$ is a function defined in $\mathbb R^N\setminus\{0\}$ which has at least $[N/2]+1$ continuous derivatives satisfying
$|\partial^{\alpha}\sigma(\xi)|\leq C_{\alpha}|\xi|^{-|\alpha|}$, for all $|\alpha|\leq [N/2]+1$ ($[l]$ is the integer part of $l$), then
$\| T_{\sigma}(u)(x) \|_{L^p}\leq C \| u \|_{L^p}$ for functions $u\in \mathcal{S}(\mathbb R^N)$, the constant $C$ depends on $\sigma$, $N$ and $p$, where
$T_{\sigma}(u)(x)=\int\limits_{\mathbb R^n}\hat{u}(\xi)\sigma(\xi)e^{2\pi i x\cdot\xi}d\xi=\mathcal{F}^{-1}(\sigma\mathcal{F}(u))(x)$.
Unfortunately I don't understand how Mihlin's theorem assures the equivalence $\| u \|_{W^{m,p}}=\| u \|_{H^{m,p}}$, i.e. $\sum\limits_{|\alpha|\leq m} \| D^{\alpha} u \|_{L^p} = \left\| \mathcal{F}^{-1} \left[ \left( 1+|\xi|^2 \right)^{\frac{s}{2}} \mathcal{F}u \right] \right\|_{L^p}$.
Can someone please share how the proof is structured? Thank you!
 A: The two norms are not equal, but they are equivalent which suffices in showing that $H^{m,p} = W^{m,p}.$ There are many sources for this, but one reference I think is quite clear is Theorem 6.2.3 of the following text:
Bergh, Jöran; Löfström, Jörgen, Interpolation spaces. An introduction, Grundlehren der mathematischen Wissenschaften. 223. Berlin-Heidelberg-New York: Springer-Verlag. (1976). ZBL0344.46071.
I will reproduce the argument here. To simplify notation we denote the associated Bessel potential operators
$$ J^{s}u = \mathcal F^{-1} \left( (1+|\xi|^2)^{\frac{s}2} \mathcal Fu(\xi)\right).$$
Note we have $\lVert u \rVert_{H^{m,p}(\Bbb R^N)} = \lVert J^mu\rVert_{L^p(\Bbb R^N)}.$
Given $|\alpha| \leq m,$ we can check that
$$ \sigma_{\alpha}(\xi) = \xi^{\alpha} (1+|\xi|^2)^{-\frac{m}2} $$
is a Mihlin multiplier; indeed it is smooth and $|\partial^{\beta}\sigma_{\alpha}| \leq C\min\{1,|\xi|^{|\alpha|-m-|\beta|}\} \leq C |\xi|^{-\beta}.$ Hence by the Mihlin multiplier theorem we have
$$ \lVert \partial^{\alpha}u \rVert_{L^p(\Bbb R^N)} = \lVert \mathcal F^{-1}\left(\sigma_{\alpha}(\xi) \mathcal F(J^mu)(\xi)\right)\rVert_{L^p(\Bbb R^N)} \leq C \lVert J^mu\rVert_{L^p(\Bbb R^N)}, $$
and so summing over all $|\alpha| \leq k$ gives
$$ \lVert u \rVert_{W^{m,p}(\Bbb R^N)} \leq C \lVert J^ku\rVert_{L^p(\Bbb R^N)} = C \lVert u \rVert_{H^{m,p}(\Bbb R^N)},$$
which proves one direction. For the converse let $\chi \in C^{\infty}(\Bbb R)$ be a cutoff such that $1_{(-1,1)} \leq 1-\chi \leq 1_{(-2,2)}$ and consider
$$\sigma_0(\xi) = (1+|\xi|^2)^{\frac{m}2} \left(1+ \sum_{j=1}^N \chi(\xi_j) |\xi_j|^m \right)^{-1}, \text{ and } \sigma_j(\xi) = \chi(\xi_j) |\xi_j|^m \xi_j^{-m}$$
for $1 \leq j \leq N.$ One can verify these are Mihlin multipliers, and that
$$ \left(1+\sum_{j=1}^N \sigma_j(\xi)\xi_j^m\right) \sigma_0(\xi) = (1+|\xi|^2)^{\frac m2}, $$
so we can compute
\begin{align*}
  \lVert J^mu \rVert_L^p(\Bbb R^N) &\leq C \left\lVert \mathcal F^{-1}\left(\left(1+\sum_{j=1}^N \sigma_j(\xi)\xi_j^m\right)\mathcal Fu\right)\right\rVert_{L^p(\Bbb R^N)} \tag{with $\sigma_0$} \\
&\leq C \left( \lVert \mathcal F^{-1}(\mathcal Fu)\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \mathcal F^{-1}(\sigma_j(\xi)\xi_j^m\mathcal Fu)\rVert_{L^p(\Bbb R^N)} \right) \tag{$\Delta$-inequality}\\
&= C \left( \lVert u\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \mathcal F^{-1}(\sigma_j(\xi)\mathcal F(\partial_j^mu))\rVert_{L^p(\Bbb R^N)} \right) \tag{Properties of $\mathcal F$}\\
&= C \left( \lVert u\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \partial_j^mu\rVert_{L^p(\Bbb R^N)} \right) \tag{with $\sigma_j$}\\
&\leq C \lVert u \rVert_{W^{m,p}(\Bbb R^N)},
\end{align*}
as required.

Added later: To prove the symbol estimates, the idea is not to explicitly compute the derivatives, but to separate the terms using the Leibniz rule and estimate the individual terms by an inductive argument. I will sketch the details to illustrate this for $\sigma_{\alpha};$ by the Leibniz rule we have
$$ \partial^{\beta}\sigma_{\alpha}(\xi) = \sum_{\gamma \leq \beta} {\beta \choose \gamma} \partial^{\beta-\gamma}\xi^{\alpha} \partial^{\gamma}\left( (1+|\xi|^2)^{-\frac m2} \right), $$
and we claim all the terms in the sum decay as $O(|\xi|^{-m+|\alpha|-|\beta|}).$ It is easy to see that $|\partial^{\beta-\gamma}\xi^{\alpha}| \leq C |\xi|^{|\alpha|-|\beta|+|\gamma|},$ noting that each partial derivative will decrease of the order of the polynomial if it is not already zero. Moreover noting the term vanishes if $|\alpha|-|\beta|+|\gamma|<0,$ we can also bound the term as
$$ |\partial^{\beta-\gamma}\xi^{\alpha}| \leq C(1+|\xi|^2)^{\frac{|\alpha|-|\beta|+|\gamma|}2}.$$
For the second term we claim by induction on $|\gamma|$ that for all $k$ we have
$$ \partial^{\gamma}\left( (1+|\xi|^2)^{-\frac k2} \right) \leq C(1+|\xi|^2)^{-\frac{k+|\gamma|}2}. $$
Here we write $\gamma = \gamma' + e_i,$ applying the inductive hypothesis for $\gamma'$ and $k+1.$
Putting everything together we arrive at the estimate
$$ \partial^{\beta}\sigma_{\alpha} \leq C (1+|\xi|^2)^{\frac{-m+|\alpha|-|\beta|}2} \leq C (1+|\xi|^2)^{-\frac{|\beta|}2}, $$
where we have used the fact that $|\alpha| \leq m$ in the last inequality.
For the other two terms we need to distinguish when each $|\xi_j| \leq 1$ or not; if $|\xi_j|<1$ then $\sigma_j(\xi) = 0$ and we can also since $\chi_j(\xi)=0$ the expression for $\sigma_0$ simplifies. If $|\xi_j|>1$ then $\chi_j(\xi)$ is just a smooth function whose derivatives can be bounded. One needs a similar induction argument to estimate the derivatives of $(1+\sum_{j=1}^N \chi_j(\xi)|\xi|^m)^{-1}$ (considering cases at each step), but I hope you can see how to fill in the details.
