Equivalent Norms on Sobolev Spaces When $k$ is a positive integer and $1<p<\infty$, we know that there is some
$C>0$ such that for all $u\in W^{k,p}\left(\mathbb{R}^{N}\right)  :$
$$
\left\Vert \left(  -I+\Delta\right)  ^{\frac{k}{2}}u\right\Vert _{p}\leq
C\left(  \left\Vert \left(  -\Delta\right)  ^{\frac{k}{2}}u\right\Vert
_{p}+\left\Vert u\right\Vert _{p}\right)  .
$$
(Stein, Singular Integrals and Differentiability Properties of Functions; etc)
The question is that can we have the following estimate: for $\varepsilon>0$,
there exists $C\left(  \varepsilon\right)  >0$ such that for all $u\in
W^{k,p}\left(\mathbb{R}^{N}\right)  :$
$$
\left\Vert \left(  -I+\Delta\right)  ^{\frac{k}{2}}u\right\Vert _{p}%
\leq\left(  1+\varepsilon\right)  \left\Vert \left(  -\Delta\right)
^{\frac{k}{2}}u\right\Vert _{p}+C\left(  \varepsilon\right)  \left\Vert
u\right\Vert _{p}?
$$
Note that when $p=2$, the answer is yes via Fourier transform. But I have no idea in the general case.
 A: It is at least true for $k = 2\ell$ and $p \in (1,2]$. The restriction on $k$ is just to make the computation simpler, and shouldn't be too hard to extend to $k$ being odd. The restriction to $p\in (1,2]$ is essential in the method of proof below. 
By triangle inequality it suffices to prove that for every $0 < j < \ell$ you can interpolate
$$ \| \triangle^ju \|_p \leq \epsilon \|\triangle^\ell u\|_p + C(\epsilon,j) \|u\|_p $$
Let $P_n$ denote the standard Littlewood Paley projectors with $n \in \mathbb{N}$, then we have (the constant $C$ differs from line to line, but are "universal")
$$ \begin{align}
\| \triangle^j u\|_p &= \| \sum_n P_n \triangle^j u\|_p \\
& \leq C \sum_{n} 2^{2nj} \| P_n u\|_p \\
& \leq 2^{2n^*j}C \sum_{n = 0}^{n^*} \|P_n u\|_p + C \sum_{n = n^* + 1}^\infty 2^{2nj} \|P_n u\|_p \end{align} $$
where $n^*$ is to be determined.  The first term can be bounded (very roughly) by 
$$ (n^* + 1) 2^{n^* j} C \|u\|_p; $$
the second term we control with (throughout $C$ is independent of $n^*$)
$$ 
C\sum_{n = n^* + 1}^{\infty} 2^{2n(j-\ell)} 2^{2n\ell} \|P_n u\|_p \leq
C\left( \sum_{n > n^*} 2^{4n(j-\ell)} \right)^\frac12 \left( \sum_{n > n^*} \|P_n \triangle^\ell u\|_p^2\right)^\frac12 $$
by Cauchy-Schwarz. Using that $j-\ell < 0$ the first factor converges and can be bounded by $2^{2n^*(j-\ell)}C $ for some $C$ independent of $n^*$. For the second factor we can use the fact that $p \leq 2$, which implies via Minkowski's inequality that
$$ \left( \sum_{n > n^*} \|P_n \triangle^\ell u\|_p^2\right)^\frac12 \leq \| \left( \sum_{n > n^*} |P_n \triangle^\ell u|\right)^\frac12 \|_p $$
the right hand side now is dominated by the square function and for $p > 1$ we can use the square function estimate to conclude that it is bounded by $C \|\triangle^\ell u\|_p$. Putting everything together we have that for some universal constant $C$, independent of $n^*$, we have
$$ \|\triangle^j u\|_p \leq (n^* + 1) 2^{2n^*j} C \|u\|_p + 2^{2n^*(j-\ell)} C \|\triangle^\ell u \|_p $$
Now it suffices to take $n^*$ sufficiently large so that the coefficient in front of $\|\triangle^\ell u\|_p$ is less than $\epsilon$. 
A: If $k \in (0, 2]$, we define the multiplier 
$$
  m (\xi) =  (1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k.
$$
We observe that if $\vert \xi \vert \ge 2$, then by differentiability
$$
  \big \vert (1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k \big\vert
= \vert \xi \vert^k \Big \vert \Big(1 + \frac{1}{\vert \xi \vert^2}\Big)^\frac{k}{2} - 1 \Big\vert
\le C \vert \xi \vert^{k - 2},
$$
so that $m$ is bounded on $\mathbb{R}^N$.
Similarly,
$$
\vert   D^\ell m (\xi)\vert \le \frac{C_\ell}{\vert \xi \vert^\ell}.
$$
By the classical Mikhlin multiplier theorem, this implies that 
$$
 \big\Vert(-\Delta + I)^\frac{k}{2}u - (-\Delta)^\frac{k}{2}u \big\Vert_p
\le C \Vert u \Vert_p,
$$
from which the estimate follows.
If $k \in (2, \infty)$, we define the multiplier
$$
  m (\xi) =  \frac{(1 + \vert \xi \vert^2)^\frac{k}{2} - \vert \xi \vert^k}{1 + \vert \xi \vert^{k - 2}}.
$$
It can be checked that the multiplier satisfies also the conditions of the Mikhlin multiplier theorem. Therefore,
$$
\big \Vert \big((- \Delta + I)^\frac{k}{2} - (-\Delta)^\frac{k}{2}\bigr)\big((-\Delta)^{\frac{k}{2} - 1} + I\big)^{-1} v\Vert_p
\le C\Vert v \Vert_p.
$$
If we set $v = (-\Delta)^{\frac{k}{2} - 1}u + u$,
then 
$$\big \Vert \big((- \Delta + I)^\frac{k}{2} - (-\Delta)^\frac{k}{2}\bigr)u\Vert_p
\le \Vert (-\Delta)^{\frac{k}{2} - 1}u + u \Vert_p
\le \Vert (-\Delta)^{\frac{k}{2} - 1}u\Vert_p + \Vert u \Vert_p
$$
By a classical interpolation
$$ 
\Vert (-\Delta)^{\frac{k}{2} - 1}u\Vert_p
\le C \Vert (-\Delta)^\frac{k}{2}u\Vert_p^\frac{k - 2}{k}
\Vert u \Vert_p^\frac{2}{k}.
$$
The requested inequality follows then from Young’s inequality.
