An inequality of J. Necas 
The above inequality belongs to J. Necas. Sur les normes equivalentes dans $W^k_p$ et sur la coercivite des formellement positives. Sem. math. sup. Universite Montreal, 102-128 (1966). But I can't get this article. Can any one help me on proving this? Or are there other references?
I find the proof of the case $k=1,p=2$ in the book Mathematical tools for the study of the imcompressible Navier-Stokes equations and related models. And there gives a hint on the case of $k=1,p\neq2,\Omega=\mathbb{R}^d$:characterise the Fourier Multipliers that continuously map $L^p(\mathbb{R}^d)$ into itself. But I'm nor familiar with Fourier Multiplier.
 A: J. Ne&ccaron;as was not the first to establish this inequality
in the special case $k=1$.  Five years earlier, Lamberto Cattabriga, assisted by Giovanni Prodi, proved an equivalent inequality
$$
\|u\|_{0,p}\leqslant C\Bigl(\Bigl|\int\limits_{\Omega}u\,dx\Bigr|
+\sum\limits_{j=1}^d\|\partial_j u\|_{-1,p}\Bigr) \tag{$\ast$}
$$
for a bounded smooth $\,\Omega\,$ in the article "Su un problema al contorno relativo al sistema di equazioni di Stokes" in  Rendiconti del Seminario Matematico della Università di Padova,  31(1961), p. 308-340 (see p. 312-313  therein).  The inequality $(\ast)$, referred to as a generalization of Poincaré's inequality,  can as well be found along with its rather detailed proof in  G.P. Galdi's fundamental monograph 

Galdi, G. P. An introduction to the mathematical theory of the Navier-Stokes equations. Steady-state problems. Second edition. Springer Monographs in Mathematics. Springer, New York, 2011. xiv+1018 pp. ISBN: 978-0-387-09619-3 MR2808162

(see Exercise III.3.4 on p. 175 and Exercise III.3.10 on p. 191  therein).  For bounded $\,\Omega\,$ satisfying the cone condition,  Galdi's approach to the case $\,k=1$,$\,1<p<\infty\,$ is based on applying the Bogovskii explicit representation formula for the right inverse of the divergence operator with homogeneous boundary data on $\,\partial\Omega$, being the adjoint to the gradient operator $\,\nabla\,\colon L^p(\Omega)\to W^{-1,p}(\Omega)\,$ whose range is closed in $W^{-1,p}(\Omega)\,$ iff  the inequality $\,(\ast)\,$ does hold for all $\,u\in L^p(\Omega)\,$ with some constant $\,C>0\,$ independent of $u$. 
A relatively trivial case of $\,\Omega=\mathbb{R}^d\,$ is readily solved by employing 
the so-called Bessel potentials along with $L^p$-estimates for the Fourier transform
multipliers.  Namely, let 
$$
\widehat{f}(\xi)=F[f]\overset{\rm def}{=}\int\limits_{\mathbb{R}^d}
f(x)e^{-i(\xi,x)}dx.
$$
By definition, for any integer $m\in\mathbb{Z}$, the space
$$
W^{m,p}(\mathbb{R}^d)=\{w\in \mathcal{S}'(\mathbb{R}^d)\,\colon 
F^{-1}\bigl[(1+|\xi|^2)^{m/2}\widehat{w}(\xi)\bigr]\in  L^p(\mathbb{R}^d)\}
$$
with the norm
$$
\|w\|_{m,p}=\bigl\|F^{-1}\bigl[(1+|\xi|^2)^{m/2}
\widehat{w}(\xi)\bigr]\bigr\|_{L^p(\mathbb{R}^d)}\,,\quad -
\infty<m<+\infty,\;1<p<\infty,
$$
is the space of Bessel potentials
$$
w(x)=B_m f\overset{\rm def}{=}F^{-1}\bigl[(1+|\xi|^2)^{-m/2}\widehat{f}(\xi)\bigr]
$$
of all functions $f\in L^p(\mathbb{R}^d)$.  Of course, operators $B_m$ will be potentials in a direct sense if only $m\geqslant 0$. When $\,m<0\,$, the potentials $\,B_m\,$ are to be treated as pseudo-differential operators of the order "$-m$".  The Bessel potentials space $W^{m,p}(\mathbb{R}^d)$ does coincide with the habitual Sobolev space $W^{m,p}(\mathbb{R}^d)$ for any integer $m\geqslant 1$, and with 
$L^p(\mathbb{R}^d)\,$ when $m=0$.  For any integer $k\geqslant 1$ and conjugate exponent $p'=p/(p-1)$, $\,1<p<\infty$,  the dual of the Sobolev space $W^{k,p'}(\mathbb{R}^d)$ is identified with the Bessel potentials space 
$W^{-k,p}(\mathbb{R}^d)$.
It is clear that the Schwartz space $\mathcal{S}(\mathbb{R}^d)$ will be a dense subspace in $W^{-k,p}(\mathbb{R}^d)$ for all integer $k\geqslant 1$ and $\,1<p<\infty$. Hence, 
to prove the lemma in question it just suffices to establish the inequality
$$
\|w\|_{-k+1,p}\leqslant \|w\|_{-k,p}+C\!\sum\limits_{j=1}^d 
\|\partial_j w\|_{-k,p}\quad \forall\,w\in \mathcal{S}(\mathbb{R}^d)
\tag{$\ast\ast$}
$$
with some constant $C>0$ depending only on $k,d,p$. Notice that
\begin{align*}
(1+|\xi|^2)^{(-k+1)/2}=(1+|\xi|^2)^{-k/2}+(1+|\xi|^2)^{-k/2}\bigl[(1+|\xi|^2)^{1/2}-1\bigl]\\
=(1+|\xi|^2)^{-k/2}+(1+|\xi|^2)^{-k/2}\frac{|\xi|^2}
{1+\sqrt{1+|\xi|^2}}
\end{align*}
whence follows
$$
(1+|\xi|^2)^{(-k+1)/2}\widehat{w}(\xi)
=(1+|\xi|^2)^{-k/2}\widehat{w}(\xi)-(1+|\xi|^2)^{-k/2}
\sum\limits_{j=1}^d m_j(\xi)F\bigl[{\partial_j}w\bigr]
$$
with functions
$$
m_j(\xi)\overset{\rm def}{=}\frac{i\xi_j}{1+\sqrt{1+|\xi|^2}}\,,\quad j=1,\dots,d,
$$
satisfying the condition
$$
\sup_{\xi\in\mathbb{R}^d}|\xi^{\alpha}\partial^{\alpha}_{\xi}m_j(\xi)|<\infty\,,\quad j=1,\dots,d,
$$
for all multi-indices $\alpha$ such that $|\alpha|\leqslant 1+d/2$,  which is sufficient for $m_j$ 
to be the Fourier transform $L^p(\mathbb{R}^d)$-multipliers, i.e.,
$$
\|F^{-1}[m_j\widehat{\varphi}]\|_{L^p(\mathbb{R}^d)}\leqslant 
C_0\|\varphi\|_{L^p(\mathbb{R}^d)}
\quad \forall\,\varphi\in \mathcal{S}(\mathbb{R}^d),\;j=1,\dots,d,
$$
with some constant $C_0>0$ depending only on $k,d,p$. 
By definition of the norm in $W^{-k,p}(\mathbb{R}^d)$ now follows $(\ast\ast)$, 
and hence lemma 2.6.
A: I've been looking for that article for weeks, without success, nevertheless concerning the inequality you write, you can find it for $p=2$ in
Necas, J. Direct Methods in the Theory of Elliptic Equations, Springer, 2012, p.186-190.
For $1<p<\infty$ I suppose you can use Marcinkiewicz interpolation, but I'm looking for that article, because I think there is a better method, without using Fouries multipliers directly, that's why I'm interested in this article.
If someone there has the paper, I'd be in debt.
Thanks a lot.
A: I found a complete proof for the case $p=2$.A reference is provided for other values of $p$. The reference in question is 
Pierre Fabrie, Franck Boyer, Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models, 2012, Springer. 
Section IV.1 is devoted to the Nečas inequality.
This question has been asked and answered a long time ago. However, I hope it will help future visitors.
