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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.

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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 $\epsilon$.

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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.

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