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Lemma 7.13.2D57C89462C1B

Lemma 7.13.2.D57C89462C1B Let $D\subset \mathbb R^ d$ be open, and let $f\in L^2_{\mathrm{loc}}(D)$. If the distributional Laplacian $\Delta f$ belongs to $L^2_{\mathrm{loc}}(D)$, then $f\in W^{2,2}_{\mathrm{loc}}(D)$. In particular, this holds if $f$ is continuous and $\Delta f$ is a locally bounded measurable function.

Proof. Considering real and imaginary parts separately, we may assume that $f$ is real. Let $f_\varepsilon =f*\rho _\varepsilon $ be the mollifications of $f$. For every compact set $L\subset D$ they are smooth on a neighborhood of $L$ for small $\varepsilon $, and by proposition 2.5.3 and lemma 2.5.1, $f_\varepsilon \to f$ and $\Delta f_\varepsilon =(\Delta f)*\rho _\varepsilon \to \Delta f$ in $L^2(L)$ as $\varepsilon \to 0$.

We first show that the gradients converge. Let $\chi $ be a real test function, and let $h$ be a real smooth function on a neighborhood of $\operatorname{Supp}\chi $. Since $\int \partial _ a(\chi ^2h\, \partial _ ah)=0$ for each $a$, by Fubini's theorem and the fundamental theorem of calculus, we have

\[ \begin{aligned} \int \chi ^2|\nabla h|^2& =-\int \chi ^2h\Delta h-2\int \chi h\nabla \chi \cdot \nabla h\\ & \le \tfrac 12\int \chi ^2|\nabla h|^2 +\int \chi ^2|h\Delta h|+2\int h^2|\nabla \chi |^2, \end{aligned} \]

where we used $2xy\le x^2/2+2y^2$. Hence

\[ \int \chi ^2|\nabla h|^2\leq 2\| \chi \| _\infty ^2\| h\| _{L^2(S)}\| \Delta h\| _{L^2(S)}+4\| \nabla \chi \| _\infty ^2\| h\| _{L^2(S)}^2, \qquad S=\operatorname{Supp}\chi . \]

Let $L\subset D$ be compact, and choose $\chi \in C^\infty _ c(D)$ equal to one on $L$ (lemma 2.5.1). Apply the last inequality to $h=f_\varepsilon -f_\delta $. As $\varepsilon ,\delta \to 0$, the norms $\| f_\varepsilon -f_\delta \| _{L^2(\operatorname{Supp}\chi )}$ tend to zero, while the norms $\| \Delta f_\varepsilon -\Delta f_\delta \| _{L^2(\operatorname{Supp}\chi )}$ stay bounded. Hence the gradients $\nabla f_\varepsilon $ form a Cauchy family in $L^2(L)$. For $\psi \in C^\infty _ c(\operatorname{Int}L)$ we have $\int \partial _ af_\varepsilon \, \psi =-\int f_\varepsilon \, \partial _ a\psi \to -\int f\, \partial _ a\psi $, so the $L^2$ limits are the distributional first derivatives of $f$ on $\operatorname{Int}L$. Since $L$ was arbitrary, the first derivatives of $f$ belong to $L^2_{\mathrm{loc}}(D)$, and $\nabla f_\varepsilon \to \nabla f$ in $L^2(L)$ for every compact $L\subset D$.

For the second derivatives, let $L\subset D$ be compact, choose $\chi \in C^\infty _ c(D)$ equal to one on a neighborhood $N$ of $L$, and for small $\varepsilon $ put $v_\varepsilon =\chi f_\varepsilon $, extended by zero to $\mathbb R^ d$. These are smooth functions with compact support, and

\[ \Delta v_\varepsilon =\chi \Delta f_\varepsilon +2\nabla \chi \cdot \nabla f_\varepsilon +f_\varepsilon \Delta \chi \]

is a Cauchy family in $L^2(\mathbb R^ d)$, by the first part applied on $\operatorname{Supp}\chi $. For every real $v\in C^\infty _ c(\mathbb R^ d)$, integrating by parts twice gives

\[ \sum _{a,b=1}^ d\| \partial _ a\partial _ b v\| _2^2 =\sum _{a,b=1}^ d\int (\partial _ a^2v)(\partial _ b^2v) =\| \Delta v\| _2^2. \]

Applying this identity to $v_\varepsilon -v_\delta $, we see that all second derivatives of $v_\varepsilon $ form Cauchy families in $L^2(\mathbb R^ d)$. Testing against compactly supported smooth functions identifies their limits with the distributional second derivatives of $\chi f$, which agree with those of $f$ on $N$, where $\chi f=f$. Thus the second derivatives of $f$ are square-integrable on $L$. Since $L$ was arbitrary, $f\in W^{2,2}_{\mathrm{loc}}(D)$. The last assertion holds because continuous functions and locally bounded measurable functions are locally square-integrable. $\square$

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