Smooth elliptic Dirichlet solvability from a boundary Sobolev bootstrap

First establish the regularity fact needed to test the homogeneous kernel: a zero-trace weak solution with smooth coefficients and smooth forcing on a smooth bounded domain lies in $H^k(V)$ for every integer $k$. Write the negative operator as $P=-L=-{\mathrm{\partial }}_i(A_{ij}{\mathrm{\partial }}_j)+({\mathrm{\partial }}_i{A}_{ij}-{\beta }_j){\mathrm{\partial }}_j-c\mathrm{.}$ Its principal coefficients are Lipschitz and uniformly elliptic, and its lower coefficients are bounded. Global $H^2$ regularity gives the base $u\in H^2(V)$.

Here is the full higher-order induction, rather than an appeal to boundary smoothness. Assume $u\in H^{m+1}(V)$ for some $m\ge 1$. On an interior patch, the local higher-order Sobolev theorem gives $H^{m+2}$ on a smaller patch. On a boundary patch, use a smooth graph chart and nested smooth cutoffs as in the actual global $H^2$ proof. In the flattened coordinates the cutoff solution $v$ has compact lateral support in $\bar{H}$, belongs to $H^{m+1}(H)\cap H_0^1(H)$, and satisfies $Pv=-a^{ij}{D}_i{D}_{j}v+B^i{D}_{i}v+Cv=F_0\mathrm{.}$ The transformed coefficients are smooth with $a^{nn}\ge {\lambda }^{\mathrm{\prime }}>0$, and $F_0\in H^m$ on the patch after localization; the cutoff commutators have order at most one and thus belong to $H^m$ by the induction hypothesis. A further cutoff identically one on the desired smaller patch avoids all artificial lateral boundaries.

For every tangential multi-index $\gamma$ of order $m$, put $w=D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma }v$. It belongs to $H_0^1(H)$ directly from the closure definition. Indeed every iterated tangential difference quotient of $v$ lies in $H_0^1(H)$ by [F6]. For $v\in H^{m+1}(H)$ these quotients converge to $w$ in $H^1$: weak derivatives commute with the quotients, and repeated application of their strong $L^2$ convergence and the bound $\mathrm{\parallel }{\delta }_h^{j}z{\mathrm{\parallel }}_2\le \mathrm{\parallel }{D}_{j}z{\mathrm{\parallel }}_2$ proves convergence for $v$ and each first derivative. Closedness of $H_0^1(H)$ gives the assertion without any half-space trace-kernel theorem. Apply the product rule in distributions to obtain $Pw=D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma }{F}_0+{\sum }_{0<\eta \le \gamma }\left(\genfrac{}{}{0px}{}{\gamma }{\eta }\right)(D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\eta }{a}^{ij})D_i{D}_j{D}_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma -\eta }v-{\sum }_{0<\eta \le \gamma }\left(\genfrac{}{}{0px}{}{\gamma }{\eta }\right)(D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\eta }{B}^i)D_i{D}_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma -\eta }v-{\sum }_{0<\eta \le \gamma }\left(\genfrac{}{}{0px}{}{\gamma }{\eta }\right)(D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\eta }C)D_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma -\eta }v\mathrm{.}$ Every term is $L^2$: the largest derivative of $v$ in a commutator has order $m+1$, since $\mathrm{\mid }\eta \mathrm{\mid }\ge 1$. Multiplying $w$ by the further cutoff adds only $L^2$ commutators, since $w\in H^1$. Convert $P$ back to divergence form and apply precisely the flat-boundary tangential $H^2$ lemma plus normal recovery (and the interior $H^2$ input) used in the global $H^2$ proof. This gives $w\in H^2$ on the smaller half-patch. Therefore all derivatives of $v$ of total order $m+2$ with at most two normal differentiations are $L^2$ there.

Recover the remaining order-$m+2$ derivatives by induction on their number $q\ge 3$ of normal differentiations. In distributions differentiate the equation by $D_n^{q-2}{D}_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma }$, where $\mathrm{\mid }\gamma \mathrm{\mid }=m+2-q$, and isolate the term $-a^{nn}{D}_n^q{D}_{\mathrm{t}\mathrm{a}\mathrm{n}}^{\gamma }v\mathrm{.}$ When all $m$ derivatives fall on $v$, every other principal term has at most $q-1$ normal differentiations and is already $L^2$ by this normal-count induction. If any derivative falls on a coefficient, the derivative of $v$ has order at most $m+1$, hence is $L^2$ by the original Sobolev induction. Differentiated lower-order terms also have order at most $m+1$, and the differentiated $F_0$ is $L^2$. The reciprocal $1\mathrm{/}{a}^{nn}$ is smooth and bounded on the patch. Division proves that the missing distributional derivative is represented by an $L^2$ function. There is no circular assumption of its boundary integrability: the distribution identity itself supplies the derivative, exactly as in the normal-recovery lemma. This proves $H^{m+2}$ on every smaller boundary patch. A finite cover proves $u\in H^{m+2}(V)$, completing the induction.

For any fixed $r$, choose $k$ with $k-n\mathrm{/}2>r+1\mathrm{/}2$. Smooth-domain extension and the higher-order Sobolev embedding give a $C^r(\bar{V})$ representative. Representatives at different $r$ coincide, since continuous representatives equal almost everywhere must agree everywhere. Thus $u\in C^{\mathrm{\infty }}(\bar{V})$. Its zero trace equals its zero classical boundary value.

Now apply the weak Fredholm theorem to $P$ on $H_0^1(V)$. Every element of the homogeneous kernel has the just-proved smoothness up to the boundary. The residual estimate for $L$ with $c\le 0$ makes that kernel zero, since both its forcing norm and boundary norm vanish. The first Fredholm alternative therefore gives a unique weak solution for every smooth forcing. For the specified boundary value, apply it to $v=u-G$ with forcing $-F-PG\in C^{\mathrm{\infty }}(\bar{V})$. The regularity argument makes $v$ smooth to the boundary, and $u=v+G$ solves the required problem. Uniqueness follows from the same residual estimate. This proves the lemma without a boundary Schauder theorem or continuity-method premise.