How to show that the weak derivative of Newtonian potential sequence converges

Question

I miss some problems in Calderon-Zygmund Inequality in 'Elliptic Partial Differential Equations of Second Order' written by Gilbarg.

Theorem 9.9

Let $f\in L^p(\Omega)$, $1 ＜ p ＜ \infty$, and let $w$ be a Newtonian potential of $f$. Then $w \in W^{2,p}(\Omega)$, $\Delta w=f$ a.e. and $$ (9.27)\qquad \|D^2w\|_p \leq C \|f \|_p$$ where $C$ depends only on $n$ and $p$. Furthermore, when $p=2$ we have $$(9.28) \qquad \|D^2w\|^2_2 = \|f \|^2_2$$

Proof

Let us deal first with the case $p=2$. If $f \in C^{\infty}_0(\mathbb R^n)$, we have $w \in C^{\infty}(\mathbb R^n)$ and by lemma 4.3, $\Delta w=f$. Consequently, for any ball $B_R$ containing the support of $f$ $$\int_{B_R}(\Delta w)^2=\int_{B_R} f^2$$ Applying Green's first identity (2.10) twice , we then obtain $$\int_{B_R}(D^2w)^2=\int_{B_R}\sum (D_{ij}w)^2=\int_{B_R} f^2+\int_{\partial B_R}Dw \frac{\partial}{\partial v}Dw$$ Using (2.14) we have $$Dw= \mathcal{O}(R^{1-n}), \quad D^2w= \mathcal{O}(R^{-n})$$ uniformly on $\partial B_R$ as $R\rightarrow \infty$, whence the identity $(9.28)$ follows. To extend $(9.28)$ to arbitrary $f \in L^2(\Omega)$, we observe first that, by Lemma 7.12, $N$ is bounded mapping from $L^p$ into itself for $1 \leq p ＜ \infty$. The full strength of theorem 9.9 in the case $p=2$ then follows by approximation. Indeed, if the sequence ${f_m}\subset C^{\infty}_0(\Omega)$ converges to $f$ in $L^2(\Omega),$ the sequence of Newtonian potentials ${Nf_m}$ converges to $w$ in $W^{2,2}(\Omega)$ where the Newtonian potential of $f$ is the function $w=Nf$ defined by convolution $$(9.26)\qquad w(x)=\int_{\Omega}\Gamma(x-y)f(y)dy$$ where $\Gamma (x-y) =\frac{1}{n(2-n) \omega _n} |x-y|^{2-n}$.$\cdots$

What puzzles me is that why $Nf_m$ converges to $w$ in $W^{2,2}(\Omega)$? By assumption, $f_m$ converges to $f$ in $L^2(\Omega)$. $N$ is a bounded mapping, then $Nf_m$ converges to $Nf=w$. I tried to prove this by definition and use lemma 4.1 and lemma 4.2. But when I want to show that $\int(D_{ij}w_m-D_{ij}w)^2$ is small enough, I found another question (lemma 4.1 and lemma 4.2 both need $f$ to be bounded).

Lemma 4.1

Let $f$ be bounded and integrable in $\Omega$, and let $w$ be the Newtonian potential of $f$. Then $w\in C^1(\mathbb R^n)$ and for any $x\in \Omega$,$$(4.8)\qquad D_iw(x)=\int_{\Omega}D_i\Gamma (x-y)f(y)dy$$ for $i=1,\dots, n$.

Lemma 4.2

Let $f$ be bounded and locally Holder continuous (with exponent $\alpha \leq 1$) in $\Omega$ and let $w$ be the Newtonian potential of $f$. Then $w \in C^2(\Omega)$, $\Delta w=f \in \Omega$ and for any $x \in \Omega$, $$D_{ij}w(x)=\int_{\Omega > _0}D_{ij}\Gamma(x-y)(f(y)-f(x))dy-f(x)\int_{\partial \Omega 0}D_i\Gamma(x-y)v_j(y)ds_y$$ for $i,j=1,\dots,n$. Here, $\Omega_0$ is any domain containing $\Omega$ for which the divergence theorem holds and $f$ is extended to vanish outside $\Omega$.

Answer 1

Let $B\subset\mathbb{R}^n$ such that $\Omega\subset\subset B$, we set $$(1)\,\,\,\,\, \overline{f}(x)=f(x)\,\,\mbox{in}\,\,\Omega,\,\, \overline{f}(x)=0\,\,\,\,\mbox{in}\,\,B\setminus\Omega\,\,\mbox{for all}\,\, f\in L^2(\Omega) $$.
Note that
$$(2)\,\,\,\,\,\,\,\overline{w}(x):=N\overline{f}\equiv w(x):=Nf$$ Let $f_m\in C_c^\infty(\Omega)$ such that $f_m\rightarrow f$ in $L^2(\Omega)$. Clearly we have $\overline{f}_m\rightarrow \overline{f}$ in $L^2(B)$and so $$ (3)\,\,\,\,\,\,\,\,\,\,\,\,N\overline{f}_m\rightarrow N\overline{f}\,\,\,\,\mbox{in}\,\,L^2(B) $$ by Theorem 8.8 of Gilbarg & Trudinger Book we have $$ \|\overline{w}_m-\overline{w}_k\|_{W^{2,2}(\Omega)}\leq C(\|\overline{w}_m-\overline{w}_k\|_{W^{1,2}(B)}+\|\overline{f}_m-\overline{f}_k\|_{L^2(B)}) $$ At the end of the proof of Theorem 8.8, the authors inform us that this estimate can be improved (see Problem 8.2) to $$ \|\overline{w}_m-\overline{w}_k\|_{W^{2,2}(\Omega)}\leq C(\|\overline{w}_m-\overline{w}_k\|_{L^2(B)}+\|\overline{f}_m-\overline{f}_k\|_{L^2(B)}) $$ from (1), (2) we conclude that $$ \|Nf_m-Nf_k\|_{W^{2,2}(\Omega)}\leq C(\|N\overline{f}_m-N\overline{f}_k\|_{L^2(B)}+\|f_m-f_k\|_{L^2(\Omega)}) $$ by (3) ${Nf_m}$ is a Cauchy sequence in $W^{2,2}(\Omega)$, therefore ${Nf_m}$ is convergent in $W^{2,2}(\Omega)$. Let $w\in W^{2,2}(\Omega)$ such that ${Nf_m}\rightarrow w$ in $W^{2,2}(\Omega)$, since $W^{2,2}(\Omega)\hookrightarrow L^2(\Omega)$ and $Nf_m\rightarrow Nf$ in $L^2(\Omega)$ then $Nf=w$, and this ends the proof.