[proofplan]
We proceed by induction on $m$. The base case $m = 0$ is the [Elliptic Interior Regularity](/theorems/95) theorem. For the inductive step, we differentiate the PDE $Lu = f$ to obtain a new elliptic equation $L\tilde{u} = \tilde{f}$ for $\tilde{u} = \partial_{x_k} u$. The new source $\tilde{f}$ involves derivatives of $f$ and products of coefficient derivatives with derivatives of $u$; a regularity analysis shows $\tilde{f} \in H^m$. The inductive hypothesis applied to $\tilde{u}$ gives $\tilde{u} \in H^{m+2}_{\text{loc}}$, and since this holds for every first derivative, $u \in H^{m+3}_{\text{loc}}$.
[/proofplan]
[step:Establish the base case $m = 0$ via interior regularity]
For $m = 0$, the hypotheses require $a_{ij}, b_i, c \in C^1(U)$ and $f \in L^2(U)$, and the conclusion is $u \in H^2(V)$ for every $V \subset\subset U$. This is exactly the [Elliptic Interior Regularity](/theorems/95) theorem.
[/step]
[step:State the inductive hypothesis and set up the inductive step]
Assume the theorem holds for a non-negative integer $m$: if the coefficients are in $C^{m+1}(U)$ and the source is in $H^m(U)$, then $u \in H^{m+2}_{\text{loc}}(U)$.
For the inductive step, assume:
\begin{align*}
a_{ij}, b_i, c \in C^{m+2}(U), \quad f \in H^{m+1}(U), \quad u \in H^1(U), \quad Lu = f.
\end{align*}
Since $C^{m+2}(U) \subset C^{m+1}(U)$ and $H^{m+1}(U) \subset H^m(U)$, the inductive hypothesis applies and gives $u \in H^{m+2}_{\text{loc}}(U)$. The goal is to show $u \in H^{m+3}_{\text{loc}}(U)$.
[/step]
[step:Differentiate the PDE to obtain an elliptic equation for $\tilde{u} = \partial_{x_k} u$]
Let $k \in \{1, \dots, n\}$. Since $u \in H^{m+2}_{\text{loc}}(U)$, the derivative $\tilde{u} := \partial_{x_k} u \in H^{m+1}_{\text{loc}}(U) \subset H^1_{\text{loc}}(U)$. Apply $\partial_{x_k}$ to $Lu = f$. Using the product rule on terms of the form $\partial_{x_j}(a_{ij}\, \partial_{x_i} u)$:
\begin{align*}
L\tilde{u} = \tilde{f},
\end{align*}
where the new source term is:
\begin{align*}
\tilde{f} := \partial_{x_k} f + \sum_{i,j=1}^n \partial_{x_j}\bigl((\partial_{x_k} a_{ij})\, \partial_{x_i} u\bigr) - \partial_{x_k}\left(\sum_{i=1}^n b_i\, \partial_{x_i} u + c\, u\right).
\end{align*}
The operator $L$ acting on $\tilde{u}$ has the same principal part as the original, so uniform ellipticity is preserved.
[guided]
Why does differentiating the PDE work? The idea is to bootstrap regularity: if we know $u \in H^{m+2}_{\text{loc}}$, we want to show $u \in H^{m+3}_{\text{loc}}$. Instead of proving this directly, we show that every first derivative $\tilde{u} = \partial_{x_k} u$ belongs to $H^{m+2}_{\text{loc}}$. Since $\tilde{u}$ satisfies its own elliptic equation $L\tilde{u} = \tilde{f}$, we can apply the inductive hypothesis to $\tilde{u}$ — provided $\tilde{f} \in H^m$.
The differentiation produces commutator terms from the product rule. Explicitly, applying $\partial_{x_k}$ to $-\partial_{x_j}(a_{ij} \partial_{x_i} u)$ gives:
\begin{align*}
-\partial_{x_j}(a_{ij} \partial_{x_i x_k} u) - \partial_{x_j}((\partial_{x_k} a_{ij}) \partial_{x_i} u).
\end{align*}
The first term is exactly $L$ applied to $\tilde{u} = \partial_{x_k} u$ (up to lower-order terms), while the second term is a commutator contribution involving $\partial_{x_k} a_{ij}$ — a derivative of the coefficients multiplied by a derivative of $u$. These commutator terms land in $\tilde{f}$.
The smoothness assumption on the coefficients ($C^{m+2}$) ensures that $\partial_{x_k} a_{ij} \in C^{m+1}$, and the known regularity $u \in H^{m+2}_{\text{loc}}$ ensures that $\partial_{x_i} u \in H^{m+1}_{\text{loc}}$. Products of $C^{m+1}$ functions with $H^{m+1}$ functions lie in $H^{m+1}$, and after the divergence $\partial_{x_j}$ these terms drop to $H^m$ — exactly the regularity needed for $\tilde{f}$ so that the inductive hypothesis applies.
[/guided]
[/step]
[step:Verify that $\tilde{f} \in H^m(V)$ for any $V \subset\subset U$]
Fix $V \subset\subset U$. We check each term in $\tilde{f}$:
- $\partial_{x_k} f \in H^m(V)$, since $f \in H^{m+1}(U)$.
- $\partial_{x_k} a_{ij} \in C^{m+1}(V)$ and $\partial_{x_i} u \in H^{m+1}(V)$ (since $u \in H^{m+2}(V)$). Products of $C^{m+1}$ functions with $H^{m+1}$ functions lie in $H^{m+1}(V)$, so $(\partial_{x_k} a_{ij})\, \partial_{x_i} u \in H^{m+1}(V)$, and the divergence $\partial_{x_j}\bigl((\partial_{x_k} a_{ij})\, \partial_{x_i} u\bigr) \in H^m(V)$.
- Similarly, $\partial_{x_k}(b_i\, \partial_{x_i} u) \in H^m(V)$ and $\partial_{x_k}(c\, u) \in H^m(V)$.
Therefore $\tilde{f} \in H^m(V)$.
[/step]
[step:Apply the inductive hypothesis to $\tilde{u}$ and conclude $u \in H^{m+3}_{\text{loc}}(U)$]
The function $\tilde{u} = \partial_{x_k} u \in H^1_{\text{loc}}(U)$ is a weak solution of $L\tilde{u} = \tilde{f}$ on $V$. The coefficients of $L$ satisfy $a_{ij}, b_i, c \in C^{m+2}(V) \subset C^{m+1}(V)$, and $\tilde{f} \in H^m(V)$. The inductive hypothesis applied to $\tilde{u}$ on any $W \subset\subset V$ gives:
\begin{align*}
\tilde{u} \in H^{m+2}(W).
\end{align*}
Since $\tilde{u} = \partial_{x_k} u$ and this holds for every $k \in \{1, \dots, n\}$, every first derivative of $u$ belongs to $H^{m+2}(W)$. Therefore $u \in H^{m+3}(W)$. Since $W \subset\subset V \subset\subset U$ was arbitrary, $u \in H^{m+3}_{\text{loc}}(U)$, completing the induction.
The estimate $\|u\|_{H^{m+3}(W)} \le C(\|f\|_{H^{m+1}(U)} + \|u\|_{L^2(U)})$ follows by chaining the inductive estimates.
[/step]
