# $L^p$ estimates for elliptic PDE in nondivergence form

I'm reading Chen and Wu's book, Second order elliptic equations and elliptic systems. I'm having some truble in understanding a lemma, precisely it is lemma 4.1, chapter 3, page 46.

To give some context, we assume that $u \in W^{2,p}(\Omega)$ with $1 < p < \infty$, satisfies $Lu\ =\ -a^{ij}D_{ij}u + b^i D_i u + cu = f$ almost everywhere ( here we are using Einstein summation convention), where:

• $a^{ij}\xi_i\xi_j \ge \lambda |\xi|^2$ whith $\lambda > 0$
• $\sum_{ij} \Vert a^{ij} \Vert_{\infty} + \sum_{i} \Vert b^{i} \Vert_{\infty} +\Vert c \Vert_{\infty} \le \Lambda$
• $a^{ij} \in C( \overline{\Omega} )$

The lemma says that there exists $R_0 > 0$ and a constant $C_0$ which depends only on $n$, $p$, $\frac{\Lambda}{\lambda}$ and the modulus of continuity of the $a^{ij}$ such that, whenever $0 < R \le R_0$, $B_R (x_0) \subset \Omega$, and $u \in W_{0}^{2,p}( B_R(x_0) )$ : $\Vert D^2 u \Vert_{L^p( B_R)} \le C_0 \{\frac{1}{\lambda}\Vert f \Vert_{L^p(B_R)} + R^{-2}\Vert u \Vert_{L^p(B_R)} \}$

The proof, initially goes like, assuming $\lambda = 1$: we freeze the coefficients of $L$ and get the constant coefficient equation ( assume the ball is centered at zero ):

$-a^{ij}(0) D_{ij}u = f + [a^{ij} - a^{ij}(0)]D_{ij}u - b^i D_i u - cu =: \overline{f}$

Using estimates for the costant coefficients equation we get:

$\Vert D^2 u \Vert_{L^p} \le C_1(n, p, \frac{\Lambda}{\lambda}) \Vert \overline{f} \Vert_{L^p}$

Now extimating the second term one gets:

$\Vert D^2 u \Vert_{L^p} \le C_1(n, p, \frac{\Lambda}{\lambda})\{ \Vert f \Vert_{L^p} + \omega(R)\Vert D^2 u \Vert_{L^p} + \Vert u \Vert_{W^{1,p}} \}$

Where $\omega(R)$ is the modulus of continuity of the $a^{ij}$.

Now one can obviously choose $R_1$ to be so small that $C_1 \omega(R) \le 1/2$ for every $R \le R_1$. taking to the left we obtain:

$\Vert D^2 u \Vert_{L^p} \le C\{ \Vert f \Vert_{L^p} + \Vert u \Vert_{W^{1,p}} \}$.

Now the authors say that an application of sobolev interpolation inequality yields the desired result. But I don't understand why we can't simply do the following reasoning ( obtaining a stronger result):

We can assume, for simplicity, that we choose as Sobolev norm $\Vert u \Vert_{W^{1,p}} = \Vert u \Vert_{L^p} + \Vert Du \Vert_{L^p}$. Now observe that, in general, if one has $v \in W_{0}^{1,p}(U)$, with $| U | < \infty$, then an application of the Poincaré inequality ( for $W_0$ !! ), yields:

$\Vert v \Vert_{L^p} \le c(n, p)|U|^{1/n} \Vert Dv \Vert_{L^p}$ ( see, for example, Giusti's book, Direct methods in the calculus of variations (corollary 3.1, page 95).

So we can estimate (remember that we are working on a ball):

$\Vert u \Vert_{W^{1,p}} = \Vert u \Vert_{L^p} + \Vert Du \Vert_{L^p} \le c(n,p)R(R+1) \Vert D^2 u \Vert_{L^p}$

Now we can take $R_0$ to be so small that $C c(n,p) R(R+1) < 1/2$. and thus we obtain:

$\Vert D^2 u \Vert_{L^p} \le C\frac{1}{\lambda}\Vert f \Vert_{L^p}$

Is the reasoning correct or am I doing something wrong?

Thank you very much.

• Should I ask on MathOverflow? – jJjjJ Sep 30 '17 at 13:47
• Why not? Assume $\phi_n \in C_{c}^{\infty}(\Omega)$ is a sequence of functions $\phi_n \to u$ in $W^{2,p}$. Thus we have $\phi_n \to u$ in $L^p$, $D_i \phi_n \to D_i u$ in $L^p$ and $D_{ij} \phi_n \to D_{ij} u$ in $L^p$. It is clear that $D_i \phi_n$ are compactly supported... EDIT: This was an answer to a user who said that $Du \not \in W_{0}^{1,p}$ – jJjjJ Sep 30 '17 at 15:08

Yes you are right. Since both $u$ and $Du$ have trace zero, you can use Poincaré inequality twice to bound $\Vert u \Vert_{W^{1,p}}$ by $\Vert D^2 u \Vert_{L^{p}}$. So the statement of the Lemma can be improved. However, in the proof of the main theorem, Theorem 4.2, the Lemma is applied to the function $v=\zeta u$ and with $f$ replaced by $\tilde f=\zeta f+hu+g Du$. So $\Vert \tilde f \Vert_{L^{p}}\le c \Vert f \Vert_{L^{p}}+c\Vert u \Vert_{W^{1,p}}$ and since now $u$ is no longer in $W^{2,p}_0$, you do have to use interpolation. The advantage in your observation is that you use interpolation only once and not twice.
• One more question. Do you know what Chen means, in global estimates, when he says let $u\in W^{2,p}( B_{R}^{+}$ be zero in a neighborhood of $\text( bdry)(B_{R}^{+}) \cap \{ x_n > 0\}$? – jJjjJ Oct 1 '17 at 9:00
• that $u$ is zero near the non-flat part of the boundary of $B^+_R$ but it may have nonzero trace on the flat part $B_R\cap\{x_n=0\}$. – Gio67 Oct 1 '17 at 11:00
• Yes that's what I thought too. But what does he mean? A ngb of the curved part in the precise term meaning, or maybe he means that there is an $r>0$ such that d(spt(u), curved part) > r ? – jJjjJ Oct 1 '17 at 12:04
• Usually it means a tubular neighborhood. Take $E=(\partial B^+_R)\setminus \{x_n=0\}$ and take $E_r=\{x\in B^+_r:\, \text{dist}(x,E)<r\}$. You want $u$ to be zero in $E_r$ for some $r>0$ small. – Gio67 Oct 1 '17 at 12:23