2.9.9 Second derivatives and singular integrals4D259A5A576A

Proof. The multiplier is real and bounded by one in absolute value, so Plancherel's theorem shows that $T_{ab}$ is self-adjoint with norm at most one. For $v\in C_ c^\infty $, the first rule in (2.9.3) (lemma 2.9.2), applied twice, gives $\widehat{-\Delta v}(\xi )=4\pi ^2|\xi |^2\widehat v(\xi )$ and $\widehat{\partial _ a\partial _ bv}(\xi )=-4\pi ^2\xi _ a\xi _ b\widehat v(\xi )$. Multiplying the first identity by the multiplier of $T_{ab}$ gives the second, which proves (3).

To construct the kernel we use the heat kernel $H_ t(z)=(4\pi t)^{-d/2}\mathrm e^{-|z|^2/(4t)}$, $t>0$. By lemma 2.9.2, $\widehat H_ t(\xi )=\mathrm e^{-4\pi ^2t|\xi |^2}$, and hence $\widehat{\partial _ a\partial _ bH_ t}(\xi )=-4\pi ^2\xi _ a\xi _ b\mathrm e^{-4\pi ^2t|\xi |^2}$. For $0<\varepsilon <R<\infty $ put

Since $\partial _ a\partial _ bH_ t(z)=\bigl (z_ az_ b/(4t^2)-\delta _{ab}/(2t)\bigr )H_ t(z)$, where $\delta _{ab}$ is the Kronecker symbol, we have

Hence $\sup _ z|\partial _ a\partial _ bH_ t(z)|\leq C_ dt^{-d/2-1}$, and the substitution $z=\sqrt t\, w$ gives $\int |\partial _ a\partial _ bH_ t(z)|\, \mathrm d z\leq C_ dt^{-1}$. Therefore the kernel $k_{\varepsilon ,R}$ is bounded and integrable, and Fubini's theorem, applied with the integrable majorant $|\partial _ a\partial _ bH_ t(z)|$ on $\mathbb R^ d\times [\varepsilon ,R]$, gives its Fourier transform:

Let $f\in L^1\cap L^2$. Then $k_{\varepsilon ,R}*f$ is bounded and integrable, hence in $L^2$, and by Fubini's theorem and the last assertion of theorem 2.9.4 its Fourier transform is $\widehat{k_{\varepsilon ,R}}\widehat f$. As $\varepsilon \downarrow 0$ and $R\uparrow \infty $, these multipliers are bounded by one and converge to $-\xi _ a\xi _ b/|\xi |^2$ for $\xi \neq 0$. By dominated convergence on the Fourier side and Plancherel's theorem, $k_{\varepsilon ,R}*f\to T_{ab}f$ in $L^2$.

For $z\neq 0$ define

Since $H_ t(z)=t^{-d/2}H_1(z/\sqrt t)$, every derivative satisfies

and induction on $|\gamma |$ shows that $\partial ^\gamma H_1(w)=P_\gamma (w)\mathrm e^{-|w|^2/4}$ for a polynomial $P_\gamma $, of degree $m_\gamma $ say. Fix $0<c\leq c'$, and let $c\leq |z|\leq c'$ and $|\gamma |\geq 2$. Since $|P_\gamma (w)|\leq C_\gamma (1+|w|)^{m_\gamma }$ and $\mathrm e^{-|z|^2/(4t)}\leq \mathrm e^{-c^2/(4t)}$, we obtain $|\partial ^\gamma H_ t(z)|\leq g_\gamma (t)$, where

This function is integrable on $(0,\infty )$ and does not depend on $z$: near $t=0$ the factor $\mathrm e^{-c^2/(4t)}$ dominates every power of $t^{-1}$, and for $t\geq 1$ the exponent satisfies $(d+|\gamma |)/2\geq 3/2>1$. By the mean value theorem and dominated convergence, we may therefore differentiate under the integral sign any number of times. Hence $K_{ab}$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, and its derivatives are the integrals of the corresponding derivatives of $\partial _ a\partial _ bH_ t$. Moreover, the integrals $\int _0^\varepsilon $ and $\int _ R^\infty $ of $|\partial _ a\partial _ bH_ t(z)|$ tend to zero as $\varepsilon \downarrow 0$ and $R\uparrow \infty $, uniformly for $c\leq |z|\leq c'$.

Differentiating the identity $H_{r^2s}(rz)=r^{-d}H_ s(z)$ twice in $z$ and substituting $t=r^2s$, we find that $K_{ab}(rz)=r^{-d}K_{ab}(z)$ for $r>0$. Thus $K_{ab}$ is homogeneous of degree $-d$ and $\nabla K_{ab}$ is homogeneous of degree $-d-1$, and the bounds in (1) follow from the boundedness of the continuous functions $K_{ab}$ and $\nabla K_{ab}$ on the unit sphere. The same scaling shows that the function $z\mapsto \int _0^\infty |\partial _ a\partial _ bH_ t(z)|\, \mathrm d t$ is homogeneous of degree $-d$; it is bounded on the unit sphere by $\int _0^\infty g_\gamma (t)\, \mathrm d t$, where $\partial ^\gamma =\partial _ a\partial _ b$ and $c=c'=1$, so it is at most $C_ d|z|^{-d}$.

To prove (2), let $f$ and $F$ be as there. By the Cauchy–Schwarz inequality on $F$, the function $f$ is integrable, so $f\in L^1\cap L^2$. Let $L$ be a compact set disjoint from $F$. For $x\in L$,

where the outer integral may be taken over $F$. The differences $x-y$, with $x\in L$ and $y\in F$, form a compact set not containing the origin, on which $c\leq |x-y|\leq c'$ for some $0<c\leq c'$. By the uniform estimates just proved, the inner integral converges to $K_{ab}(x-y)$ uniformly on this set, and it is bounded by $\int _0^\infty |\partial _ a\partial _ bH_ t(x-y)|\, \mathrm d t\leq C_ dc^{-d}$. Hence, since $f$ is integrable, $k_{\varepsilon ,R}*f$ converges uniformly on $L$ to $\int K_{ab}(x-y)f(y)\, \mathrm d y$. Since it also converges to $T_{ab}f$ in $L^2$, a subsequence converges almost everywhere to $T_{ab}f$, and the two limits agree almost everywhere on $L$. Covering the open set $\mathbb R^ d\setminus F$ by countably many such compact sets proves (2). For $d=1$ the kernel vanishes, in accordance with $T_{11}=-1$: a direct differentiation gives $\partial _ z^2H_ t=\bigl (z^2/(4t^2)-1/(2t)\bigr )H_ t=\partial _ tH_ t$, so $K_{11}(z)=\lim _{t\uparrow \infty }H_ t(z)-\lim _{t\downarrow 0}H_ t(z)=0$ for $z\neq 0$. $\square$

Proof. We use the Calderón–Zygmund decomposition of $f$ at height $\lambda $ into a bounded part $g$, handled by the $L^2$ bound for $T_{ab}$, and parts $b_ Q$ of mean zero on disjoint cubes $Q$, handled by lemma 2.9.10. The dyadic cubes are the cubes $2^ k(m+[0,1)^ d)$ with $k\in \mathbb Z$ and $m\in \mathbb Z^ d$. Two dyadic cubes are either disjoint or nested, and each dyadic cube $Q$ lies in a unique dyadic cube of twice its side length, its parent, which has volume $2^ d|Q|$. Call a dyadic cube $Q$ selected if the average of $|f|$ over $Q$ exceeds $\lambda $, while the average over every strictly larger dyadic cube containing $Q$ does not. Every dyadic cube on which the average of $|f|$ exceeds $\lambda $ is contained in a selected cube, because the averages over its successive ancestors are at most $\| f\| _1$ divided by their volumes and therefore tend to zero. The selected cubes are pairwise disjoint. For each of them, the average over its parent is at most $\lambda $, and hence

Moreover, $|f|\leq \lambda $ almost everywhere outside the union of the selected cubes. Indeed, let $x$ be a point outside this union at which the conclusion of theorem 2.8.3 holds. Every dyadic cube containing $x$ has average of $|f|$ at most $\lambda $, since otherwise it would lie in a selected cube. The dyadic cube of side $2^ k$ containing $x$ lies in the cube $x+(-2^ k,2^ k)^ d$, whose volume is $2^ d$ times larger. Hence the average of $|f-f(x)|$ over it tends to zero as $k\to -\infty $, so its average of $|f|$ tends to $|f(x)|$, and therefore $|f(x)|\leq \lambda $.

For a selected cube $Q$, let $f_ Q=|Q|^{-1}\int _ Qf$ be the average of $f$. Define $g=f$ outside the selected cubes and $g=f_ Q$ on each selected cube $Q$, and put $b_ Q=(f-f_ Q)\mathbf1_ Q$. Each $b_ Q$ has integral zero, and

By the Cauchy–Schwarz inequality, $|Q||f_ Q|^2\leq \int _ Q|f|^2$, so $\| b_ Q\| _2\leq 2\| f\mathbf1_ Q\| _2$. Since the $b_ Q$ have disjoint supports, the series $\sum _ Qb_ Q$ converges to $f-g$ in $L^2$.

Let $c_ Q$ be the center of $Q$, and let $Q^*$ be the cube with center $c_ Q$ and $4\sqrt d$ times the side length of $Q$. If $x\notin Q^*$ and $y\in Q$, then $|x-c_ Q|\geq 2\operatorname {diam}Q$ and $|y-c_ Q|\leq \operatorname {diam}Q/2$, so every point of the segment from $x-c_ Q$ to $x-y$ has norm at least $|x-c_ Q|/2$. The function $b_ Q\in L^2$ vanishes outside the compact set $\overline Q$, which does not meet $\mathbb R^ d\setminus Q^*$, and it has integral zero. Therefore parts (1) and (2) of lemma 2.9.10 and the mean value theorem give, for almost every $x\notin Q^*$,

Integrating in polar coordinates about $c_ Q$ over the region $|x-c_ Q|\geq 2\operatorname {diam}Q$, which contains $\mathbb R^ d\setminus Q^*$, we obtain

Put $B^*=\bigcup _ QQ^*$. Then $|B^*|\leq (4\sqrt d)^ d\sum _ Q|Q|\leq C_ d\lambda ^{-1}\| f\| _1$. Enumerate the selected cubes, and let $S_ N$ be the partial sums of $\sum _ Qb_ Q$. Since $S_ N\to f-g$ in $L^2$, also $T_{ab}S_ N\to T_{ab}(f-g)$ in $L^2$, and a subsequence converges almost everywhere. Moreover $\int _{\mathbb R^ d\setminus B^*}|T_{ab}S_ N|\leq \sum _ Q\int _{\mathbb R^ d\setminus Q^*}|T_{ab}b_ Q|\leq C_ d\| f\| _1$, so Fatou's lemma gives

Finally, $T_{ab}f=T_{ab}g+T_{ab}(f-g)$, and the set where $|T_{ab}f|>\lambda $ is contained in the union of $B^*$, the set where $|T_{ab}g|>\lambda /2$, and the set of $x\notin B^*$ with $|T_{ab}(f-g)(x)|>\lambda /2$. By Chebyshev's inequality and the $L^2$ bound for $T_{ab}$, the second set has measure at most $4\lambda ^{-2}\| g\| _2^2\leq 2^{d+2}\lambda ^{-1}\| f\| _1$, and the third has measure at most $2\lambda ^{-1}\int _{\mathbb R^ d\setminus B^*}|T_{ab}(f-g)|\leq C_ d\lambda ^{-1}\| f\| _1$. Together with the bound for $|B^*|$, this proves the lemma. $\square$

Proof. Assume first that $1<p<2$, and let $f\in L^ p\cap L^2$. Fix $\lambda >0$ and split $f=f_1+f_2$ with $f_1=f\mathbf1_{\{ |f|>\lambda \} }$ and $f_2=f\mathbf1_{\{ |f|\leq \lambda \} }$. Since $|f_1|\leq \lambda ^{1-p}|f|^ p$, we have $f_1\in L^1\cap L^2$, and $f_2\in L^2$. Apply lemma 2.9.11 to $f_1$, and Chebyshev's inequality with the $L^2$ bound to $f_2$, both with threshold $\lambda /2$. This gives

For a measurable function $h$, Tonelli's theorem applied to $|h(x)|^ p=\int _0^{|h(x)|}p\lambda ^{p-1}\, \mathrm d\lambda $ expresses $\| h\| _ p^ p$ as the integral of $p\lambda ^{p-1}$ times the distribution function $\lambda \mapsto |\{ |h|>\lambda \} |$, also when the norm is infinite. We apply this to $h=T_{ab}f$: multiply the last inequality by $p\lambda ^{p-1}$ and integrate in $\lambda $. By Tonelli's theorem again, the first term on the right contributes $C_ d\int |f|\int _0^{|f|}p\lambda ^{p-2}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(p-1)$, and the second contributes $C_ d\int |f|^2\int _{|f|}^\infty p\lambda ^{p-3}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(2-p)$; the inner integrals converge precisely because $p>1$ and $p<2$. Therefore

For $p=2$ the bound is Plancherel's theorem.

Now let $2<p<\infty $, put $p'=p/(p-1)\in (1,2)$, let $f\in L^ p\cap L^2$, and write $h=T_{ab}f\in L^2$. For integers $N\geq 1$, put $E_ N=B(0,N)\cap \{ |h|\leq N\} $ and $g_ N=\mathbf1_{E_ N}h|h|^{p-2}$. Then $|g_ N|\leq N^{p-1}\mathbf1_{E_ N}$, so $g_ N\in L^2\cap L^{p'}$, and $|g_ N|^{p'}=\mathbf1_{E_ N}|h|^ p$. Since $T_{ab}$ is self-adjoint on $L^2$, the case $p'$ already proved and Hölder's inequality give

The integral on the left is finite. If it is nonzero, dividing by its $(1/p')$th power gives $(\int _{E_ N}|h|^ p)^{1/p}\leq C_{d,p'}\| f\| _ p$; if it is zero, this bound is trivial. The sets $E_ N$ increase to $\{ |h|<\infty \} $, which has full measure since $h\in L^2$, so monotone convergence gives $\| T_{ab}f\| _ p\leq C_{d,p'}\| f\| _ p$.

By lemma 2.5.1(5), $C_ c^\infty \subset L^ p\cap L^2$ is dense in $L^ p$, and $L^ p$ is complete, so $T_{ab}$ has a unique bounded extension from $L^ p\cap L^2$ to $L^ p$. Finally, for $v\in C_ c^\infty $, lemma 2.9.10(3) gives $\| \partial _ a\partial _ bv\| _ p=\| T_{ab}\Delta v\| _ p\leq C_{d,p}\| \Delta v\| _ p$, and summing over $a$ and $b$ proves (2.9.13). $\square$

Proof. (1) For $k=0$ the first assertion holds because $\chi $ is bounded on compact sets. Let $k\geq 1$ and $1\leq j\leq d$. For a test function $\phi $ on $U$, the function $\chi \phi $ is a test function with $\partial _ j(\chi \phi )=\chi \partial _ j\phi +\phi \partial _ j\chi $, so

Thus $\partial _ j(\chi v)=\chi \partial _ jv+(\partial _ j\chi )v$. The functions $\partial _ jv$ and $v$ belong to $W^{k-1,p}_{\mathrm{loc}}(U)$, so by induction on $k$ both terms belong to $W^{k-1,p}_{\mathrm{loc}}(U)$; hence $\chi v\in W^{k,p}_{\mathrm{loc}}(U)$, and iterating the formula for first derivatives gives the Leibniz formula, exactly as for smooth functions.

Now let $\operatorname{Supp}\chi $ be a compact subset of $U$, and let $W$ be as in the statement. By the Leibniz formula, $\chi v$ and its derivatives of order at most $k$ vanish almost everywhere outside $\operatorname{Supp}\chi $, and there they are bounded by $C\sum _{|\beta |\leq k}|\partial ^\beta v|$, where $C$ depends on $d$, $k$ and the derivatives of $\chi $ of order at most $k$. Hence their $L^ p$ norms are at most $C\| v\| _{W^{k,p}(W)}$, which is finite when $W\Subset U$. To identify the derivatives of the extension by zero, let $\phi $ be a test function on $\mathbb R^ d$, and choose a smooth cutoff function $\zeta $ for $\operatorname{Supp}\chi $ in $U$ which equals $1$ on a neighborhood $N$ of $\operatorname{Supp}\chi $ (lemma 2.5.1(4)). Then $\zeta \phi $ is a test function on $U$, and $\partial ^\alpha (\zeta \phi )=\partial ^\alpha \phi $ on $N$. Since $\chi v$ and $\partial ^\alpha (\chi v)$ vanish almost everywhere outside $\operatorname{Supp}\chi $,

This proves (1).

(2) The composite $v\circ \Phi $ is defined almost everywhere: if $Z\subset U$ is a Borel null set, the change of variables formula for the diffeomorphism $\Phi ^{-1}$ gives $|\Phi ^{-1}(Z)|=\int _ Z|\det D\Phi ^{-1}(x)|\, \mathrm d x=0$. For a smooth function $w$ on $U$ and $1\leq |\alpha |\leq k$, the classical chain rule gives

with smooth functions $c_{\alpha \beta }$ on $V$ that depend only on $\Phi $: for $|\alpha |=1$ this is $\partial _ j(w\circ \Phi )=\sum _ i\partial _ j\Phi _ i\, (\partial _ iw)\circ \Phi $, and differentiating the formula once more, with the product rule and the chain rule applied to each $(\partial ^\beta w)\circ \Phi $, gives the induction step.

Assume first that $v$ vanishes almost everywhere outside the compact set $K$. By (1), applied with a smooth cutoff function equal to $1$ near $K$, the extension of $v$ by zero belongs to $W^{k,p}(\mathbb R^ d)$. Its derivatives of order at most $k$ vanish almost everywhere outside $K$, because $v$ vanishes almost everywhere on the open set $\mathbb R^ d\setminus K$ and hence so do its distributional derivatives. Choose $\delta >0$ such that the compact set $K'=\{ x:\operatorname {dist}(x,K)\leq \delta \} $ lies in $U$. For $\varepsilon <\delta $, the mollification $v_\varepsilon =v*\rho _\varepsilon $ is smooth and vanishes outside $K'$, and $\partial ^\beta v_\varepsilon =(\partial ^\beta v)*\rho _\varepsilon \to \partial ^\beta v$ in $L^ p(\mathbb R^ d)$ for $|\beta |\leq k$, by the remark preceding the lemma, applied on $\mathbb R^ d$ to a bounded open set containing $K'$, outside of which both sides vanish.

Let $J$ be the maximum of $|\det D\Phi ^{-1}|$ on $K'$, and let $M\geq 1$ bound the functions $|c_{\alpha \beta }|$ on the compact set $\Phi ^{-1}(K')$. By the change of variables formula,

for every measurable function $g$ on $U$ that vanishes outside $K'$. The smooth functions $w_\varepsilon =v_\varepsilon \circ \Phi $ vanish outside $\Phi ^{-1}(K')$. Applying this inequality to $v_\varepsilon -v$ and its derivatives, we find that $w_\varepsilon \to v\circ \Phi $ and, for $1\leq |\alpha |\leq k$,

with $\| G_\alpha \| _ p\leq MJ^{1/p}\sum _{1\leq |\beta |\leq |\alpha |}\| \partial ^\beta v\| _ p$. For a test function $\phi $ on $V$, we obtain

Hence $\partial ^\alpha (v\circ \Phi )=G_\alpha \in L^ p(V)$. These functions, and $v\circ \Phi $, vanish almost everywhere outside $\Phi ^{-1}(K)$, and summing the bounds over $|\alpha |\leq k$ gives $\| v\circ \Phi \| _{W^{k,p}(V)}\leq C\| v\| _{W^{k,p}(U)}$, with $C$ depending only on $k$, $p$, $J$ and $M$, that is, on $k$, $p$, $\Phi $ and $K$ once $\delta $ has been fixed in terms of $K$ and $U$.

In general, let $V'\Subset V$ be open. Choose a smooth cutoff function $\chi $ for the compact set $\Phi (\overline{V'})$ in $U$ (lemma 2.5.1(4)). By (1), $\chi v\in W^{k,p}(U)$, and it vanishes outside $\operatorname{Supp}\chi $; by the case just treated, $(\chi v)\circ \Phi \in W^{k,p}(V)$. Since $(\chi v)\circ \Phi =v\circ \Phi $ on $V'$, we obtain $v\circ \Phi \in W^{k,p}(V')$. Hence $v\circ \Phi \in W^{k,p}_{\mathrm{loc}}(V)$. $\square$

Proof. Let $e_ a$ be the $a$th coordinate vector and $h>0$. Taylor's formula with integral remainder gives

The $L^ p$ norm is invariant under translation. Moreover, $\omega (t)\, \mathrm d t=2h^{-2}(h-t)\, \mathrm d t$ is a probability measure on $[0,h]$, so Hölder's inequality gives $|\int _0^ hg\, \omega \, \mathrm d t|^ p\leq \int _0^ h|g|^ p\omega \, \mathrm d t$ for every measurable function $g$ on $[0,h]$. Applying this to $g(t)=\partial _ a^2v(x+te_ a)$, integrating in $x$ and using Tonelli's theorem, we obtain

Hence $h\| \partial _ av\| _ p\leq 2\| v\| _ p+(h^2/2)\| \partial _ a^2v\| _ p$. Since $|\nabla v|\leq \sum _ a|\partial _ av|$ and $|\partial _ a^2v|\leq |D^2v|$, summing over $a$ yields $\| \nabla v\| _ p\leq 2dh^{-1}\| v\| _ p+(dh/2)\| D^2v\| _ p$. The choice $h=2\varepsilon /d$ proves the lemma with $C_{d,\varepsilon }=d^2/\varepsilon $. $\square$

Proof. We prove the estimate for smooth $v$, pass to general $v$ by mollification, and then apply the result to first derivatives. Assume first that $v$ is smooth on $B$. Let $Q\subset B$ be a ball of radius $r$, let $v_ Q=|Q|^{-1}\int _ Qv$ be the average of $v$ over $Q$, and let $x\in Q$. We claim that

For $z\in Q$, write $z=x+s\theta $ with $s=|z-x|$ and $\theta \in S^{d-1}$; then $|v(z)-v(x)|\leq \int _0^ s|\nabla v(x+t\theta )|\, \mathrm d t$, and $|v(x)-v_ Q|$ is at most the average over $z\in Q$ of these segment integrals. Since $Q$ is convex and open and contains $x$, for each $\theta $ the radii $s>0$ with $x+s\theta \in Q$ form an interval $0<s<R(\theta )$, and $R(\theta )\leq 2r$. In polar coordinates about $x$ ((3.1.2) on $B(x,2r)\supset Q$), the average of the segment integrals is

Interchanging the integrals in $s$ and $t$ produces the factor $|Q|^{-1}\int _ t^{R(\theta )}s^{d-1}\, \mathrm d s$, which is at most $(2r)^ d/(d|Q|)$ and hence bounded by a constant depending only on $d$. The remaining integral $\int _{S^{d-1}}\int _0^{R(\theta )}|\nabla v(x+t\theta )|\, \mathrm d t\, \mathrm d S(\theta )$ is the integral on the right side of the claim, written in polar coordinates. This proves the claim.

By Hölder's inequality with $p'=p/(p-1)$, the right side of the claim is at most $C_{d,p}r^{1-d/p}\| \nabla v\| _{L^ p(Q)}$, because

and this integral is finite exactly when $(1-d)p'>-d$, that is, when $p>d$. (For $d=1$ the exponent $(1-d)p'$ is zero.)

Let $\delta >0$ be the distance from $B'$ to $\mathbb R^ d\setminus B$. If $x,y\in B'$ and $0<|x-y|<\delta /2$, the ball $Q=B(x,2|x-y|)$ lies in $B$ and contains $x$ and $y$. Applying the preceding bound at both points gives

For $x\in B'$, the same bound on the ball $Q=B(x,\delta /2)\subset B$, together with the estimate $|v_ Q|\leq |Q|^{-1/p}\| v\| _{L^ p(B)}$ from Hölder's inequality, gives $\sup _{B'}|v|\leq C\| v\| _{W^{1,p}(B)}$ with $C$ depending on $d,p,\delta $. If $|x-y|\geq \delta /2$, then $|v(x)-v(y)|/|x-y|^\alpha \leq 2(\delta /2)^{-\alpha }\sup _{B'}|v|$. These bounds prove the estimate for smooth $v$ on $B'$, and hence on $\overline{B'}$, since $v$ is continuous on $B\supset \overline{B'}$.

For general $v\in W^{1,p}(B)$, choose a ball $B''$, depending only on $B'$ and $B$, with $B'\Subset B''\Subset B$, and let $v_\varepsilon =v*\rho _\varepsilon $, which is smooth on $B''$ for small $\varepsilon $. By the remark preceding lemma 2.9.14, $\partial _ av_\varepsilon =(\partial _ av)_\varepsilon $, and $v_\varepsilon \to v$ and $\partial _ av_\varepsilon \to \partial _ av$ in $L^ p(B'')$. The estimate for smooth functions, applied on $B'\Subset B''$ to the differences of the $v_\varepsilon $, shows that they form a Cauchy family in the norm of $C^{0,\alpha }(\overline{B'})$. Their uniform limit is continuous on $\overline{B'}$, agrees almost everywhere with $v$, and satisfies the asserted estimate, because $\| v\| _{W^{1,p}(B'')}\leq \| v\| _{W^{1,p}(B)}$. Two continuous functions that agree almost everywhere on an open set agree everywhere on it. Hence the continuous representatives obtained from different balls $B'\Subset B$ agree on their overlaps and define a single continuous representative on $B$.

Finally, let $v\in W^{2,p}_{\mathrm{loc}}(U)$. On every ball $B\Subset U$, both $v$ and its first derivatives $\partial _ av$ belong to $W^{1,p}(B)$. By the preceding paragraph, on each smaller ball $B'\Subset B$ the mollifications $v_\varepsilon $ and $\partial _ av_\varepsilon =(\partial _ av)_\varepsilon $ converge uniformly to the continuous representatives $\tilde v$ of $v$ and $\tilde w_ a$ of $\partial _ av$. Passing to the limit in

shows that $\tilde v$ has the continuous partial derivatives $\partial _ a\tilde v=\tilde w_ a$. Hence $\tilde v$ is continuously differentiable, and by the first part its first derivatives are Hölder continuous with exponent $\alpha $ on each $\overline{B'}$. Every compact subset of $U$ is covered by finitely many such balls $B'$; pairs of points closer than a Lebesgue number of this cover lie in a common ball, and for the remaining pairs the supremum bound suffices. Hence $\tilde v\in C^{1,\alpha }_{\mathrm{loc}}(U)$. $\square$