$L^p$ interpolation absorption of first derivatives by second derivatives

The multiplier symbol. Fix an index $i$ and, for $\epsilon >0$, define $m_{\epsilon }(\xi ):={\displaystyle \frac{2\pi i{\xi }_i}{4{\pi }^2\epsilon \mathrm{\mid }\xi {\mathrm{\mid }}^2+1}}$ for $\xi \in {\mathbb{R}}^n$. Writing $\eta :=2\pi \sqrt{\epsilon }\xi$ and $g(\eta ):={\eta }_i\mathrm{/}(\mathrm{\mid }\eta {\mathrm{\mid }}^2+1)$, we have $m_{\epsilon }(\xi )=i{\epsilon }^{-1\mathrm{/}2}g(\eta )$ and hence ${\mathrm{\partial }}_{\xi }^{\alpha }{m}_{\epsilon }(\xi )=i{\epsilon }^{-1\mathrm{/}2}(2\pi \sqrt{\epsilon }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}({\mathrm{\partial }}^{\alpha }g)(\eta )$; therefore $\mathrm{\mid }\xi {\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }{\mathrm{\partial }}_{\xi }^{\alpha }{m}_{\epsilon }(\xi )\mathrm{\mid }={\epsilon }^{-1\mathrm{/}2}\mathrm{\mid }\eta {\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }({\mathrm{\partial }}^{\alpha }g)(\eta )\mathrm{\mid }\le C_{\alpha }{\epsilon }^{-1\mathrm{/}2}$ and $\mathrm{\parallel }{m}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}={\epsilon }^{-1\mathrm{/}2}{\sup }_{\eta }\mathrm{\mid }\eta \mathrm{\mid }\mathrm{/}(1+\mathrm{\mid }\eta {\mathrm{\mid }}^2)=\frac{1}{2}{\epsilon }^{-1\mathrm{/}2}$, because $g$ and all of its derivatives are bounded on ${\mathbb{R}}^n$ and $\mathrm{\mid }\eta {\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }{\mathrm{\partial }}^{\alpha }g\mathrm{\mid }$ is bounded (near zero the derivatives of $g$ are bounded, while for $\mathrm{\mid }\eta \mathrm{\mid }\ge 1$ the quotient rule gives $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }g(\eta )\mathrm{\mid }\le C_{\alpha }\mathrm{\mid }\eta {\mathrm{\mid }}^{-1-\mathrm{\mid }\alpha \mathrm{\mid }}$). Thus $m_{\epsilon }$ is a Mihlin symbol with constants $C_{\alpha }{\epsilon }^{-1\mathrm{/}2}$, and [F1] gives the $L^p$ bound $\mathrm{\parallel }{T}_{m_{\epsilon }}h{\mathrm{\parallel }}_{L^p}\le C_{n,p}{\epsilon }^{-1\mathrm{/}2}\mathrm{\parallel }h{\mathrm{\parallel }}_{L^p}$ for every $h\in L^p$, and $T_{m_{\epsilon }}(\epsilon (-\mathrm{\Delta }u)+u)={\mathrm{\partial }}_{i}u$ for $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$.

One-dimensional identity along a coordinate line. Let $v\in C^2({\mathbb{R}}^n)$, $y\in {\mathbb{R}}^n$, $s>0$ and $i\in \{1,\dots ,n\}$, and put $g(t):=v(y+t{e}_i)$ for $t\in [-s,s]$. Since $g^{\mathrm{\prime }}(0)={\mathrm{\partial }}_{i}v(y)$, $g^{\mathrm{\prime }}(t)={\mathrm{\partial }}_{i}v(y+t{e}_i)$ and $g^{\mathrm{\prime }\mathrm{\prime }}(t)={\mathrm{\partial }}_i^{2}v(y+t{e}_i)$, the first identity of [F5] applied on $[-s,s]$ gives $\frac{1}{2s}{\int }_{-s}^s{g}^{\mathrm{\prime }}(t)dt=\frac{g(s)-g(-s)}{2s}$, while the weighted second identity of [F5] gives $\mathrm{\mid }{g}^{\mathrm{\prime }}(0)-\frac{g(s)-g(-s)}{2s}\mathrm{\mid }=\mid \frac{1}{2s}{\int }_{-s}^s\mathrm{sign}(t)(s-\mathrm{\mid }t\mathrm{\mid })g^{\mathrm{\prime }\mathrm{\prime }}(t)dt\mid \le \frac{1}{2}{\int }_{-s}^s\mathrm{\mid }{g}^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{\mid }dt$. Hence $\mathrm{\mid }{\mathrm{\partial }}_{i}v(y)\mathrm{\mid }\le \frac{1}{2s}\mid v(y+s{e}_i)-v(y-s{e}_i)\mid +\frac{1}{2}{\int }_{-s}^s\mid {\mathrm{\partial }}_i^{2}v(y+t{e}_i)\mid dt$.

Global form for smooth compact data. Let $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. On the Fourier side $\mathcal{F}(\epsilon (-\mathrm{\Delta }u)+u)=(4{\pi }^2\epsilon \mathrm{\mid }\xi {\mathrm{\mid }}^2+1)\widehat{u}$ by [F2], so $m_{\epsilon }\cdot \mathcal{F}(\epsilon (-\mathrm{\Delta }u)+u)=2\pi i{\xi }_i\widehat{u}=\mathcal{F}({\mathrm{\partial }}_{i}u)$; two tempered distributions with the same Fourier transform are equal by [F2], hence $T_{m_{\epsilon }}(\epsilon (-\mathrm{\Delta }u)+u)={\mathrm{\partial }}_{i}u$. Step 1.1 and $\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_{L^p}\le n\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p}$ therefore give $\mathrm{\parallel }{\mathrm{\partial }}_{i}u{\mathrm{\parallel }}_{L^p}\le C_{n,p}{\epsilon }^{-1\mathrm{/}2}(\epsilon n\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p})$, that is $\mathrm{\parallel }{\mathrm{\partial }}_{i}u{\mathrm{\parallel }}_{L^p}\le C_{n,p}n\sqrt{\epsilon }\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p}+C_{n,p}{\epsilon }^{-1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$; replacing $\sqrt{\epsilon }{C}_{n,p}n$ by a new $\epsilon >0$ yields the global form $\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p}\le \epsilon \mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p}+C_{n,p}{\epsilon }^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$ for every smooth compactly supported $u$ and every $\epsilon >0$.

Local estimate for smooth functions. Fix a ball $B_R(x_0)$, let $0<s\le R$ and $v\in C^{\mathrm{\infty }}(B_{2R}(x_0))$; the points $y\pm s{e}_i$ and $y+t{e}_i$, $\mathrm{\mid }t\mathrm{\mid }\le s$, lie in $B_{2R}(x_0)$ whenever $y\in B_R(x_0)$. Raise the inequality of step 1.2 to the power $p$, use $(a+b{)}^p\le 2^p(a^p+b^p)$, integrate over $y\in B_R(x_0)$ and apply [F6] to the inner integral over $t\in [-s,s]$: ${\int }_{B_R}\mid {\mathrm{\partial }}_{i}v{\mid }^{p}dy\le 2^{2p}[(2s{)}^{-p}{\int }_{B_R}(\mathrm{\mid }v(y+s{e}_i){\mathrm{\mid }}^p+\mathrm{\mid }v(y-s{e}_i){\mathrm{\mid }}^p)dy+2^{-p}(2s{)}^{p-1}{\int }_{B_R}{\int }_{-s}^s\mid {\mathrm{\partial }}_i^{2}v(y+t{e}_i){\mid }^{p}dtdy]\mathrm{.}$ By translation invariance [F5] the first integral is at most $2\mathrm{\parallel }v{\mathrm{\parallel }}_{L^p(B_{2R})}^p$ and the second at most $2s\mathrm{\parallel }{D}^{2}v{\mathrm{\parallel }}_{L^p(B_{2R})}^p$ (the inner $y$-integrals are integrals over translate balls contained in $B_{2R}(x_0)$). Taking the $p$-th root and the maximum over $i$ gives $\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^p(B_R)}\le C_{n,p}(s^{-1}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^p(B_{2R})}+s\mathrm{\parallel }{D}^{2}v{\mathrm{\parallel }}_{L^p(B_{2R})})$ for every $0<s\le R$.

Density. Let $u\in W^{2,p}({\mathbb{R}}^n)$ and let $u_k\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ satisfy $u_k\to u$ in $W^{2,p}({\mathbb{R}}^n)$, which exists by [F4]. Applying step 2.1 to $u_k-u_l$ and to $u_k$ and letting $k\to \mathrm{\infty }$ gives, in the limit, $\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p}\le \epsilon \mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{L^p}+C_{n,p}{\epsilon }^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$: all three norms converge along the sequence and the constant is unchanged. This is the global form of the statement for every $W^{2,p}$ class.

Choosing the scale and passing to $W^{2,p}$ on the ball. In step 2.2 put $s:=\min \{R,\epsilon R\mathrm{/}(C_{n,p})\}$; then $s\le R$ and $s^{-1}\le \max \{R^{-1},C_{n,p}{\epsilon }^{-1}{R}^{-1}\}\le C_{n,p}{\epsilon }^{-1}{R}^{-1}$ for $0<\epsilon \le C_{n,p}$, while for $\epsilon >C_{n,p}$ the inequality is implied by the case $\epsilon =C_{n,p}$ (the right-hand side is increasing in $\epsilon$); hence for every $\epsilon >0$ there is $C(n,p,\epsilon )<\mathrm{\infty }$ with $\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^p(B_R)}\le \epsilon R\mathrm{\parallel }{D}^{2}v{\mathrm{\parallel }}_{L^p(B_{2R})}+C{R}^{-1}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^p(B_{2R})}$ for every smooth $v$ on $B_{2R}(x_0)$. Finally let $u\in W^{2,p}(B_{2R}(x_0))$ and approximate it in $W^{2,p}(B_{2R}(x_0))$ by smooth functions on that ball, which exist by [F4]; the estimate is stable under this convergence, so it holds for $u$ as well. This is the scaled form of the statement, uniformly in $x_0$ and $R$.

Conclusion. The global form is step 3.1 and the scaled form is step 3.2. Both were derived using only the Countable Choice instances recorded in [A1], namely those of the Fourier, multiplier, Sobolev-density and measure-translation interfaces; no extension operator and no full Axiom of Choice is used, which is why the scaled form is stated with the doubled ball on the right. The strict range $1<p<\mathrm{\infty }$ is used in the Mihlin theorem and nowhere else; the first-order identity of step 1.2 and the absorption of step 2.2 are elementary.