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Let $L = \sum_{i,j=1}^n \frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p*)$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Let $L = \sum_{i,j=1}^n -\frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p^*)$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Let $L = \sum_{i,j=1}^n \frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega) = (W_0^{1,p})^*$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Let $L = \sum_{i,j=1}^n \frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p*)$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Let $L = \sum_{i,j=1}^n \frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega) = (W_0^{1,p})^*$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newtown Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Let $L = \sum_{i,j=1}^n \frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega) = (W_0^{1,p})^*$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.