4.1.28 Dirichlet energy and quasi-continuityEB60B98F8F53

Proof. We extend $u$ harmonically across $\partial B$, regularize the extension by convolution, and derive the identities from the divergence theorem for the regularizations. Since $u$ is bounded, hence locally integrable, the converse in part (2) of proposition 4.1.16 shows that $u=-G_ B\mu $; likewise $v=-G_ B\nu $. Choose $0<r<R$ with $\operatorname{Supp}\mu \subset B(0,r)$, and put $R'=R^2/r$. By part (2) of that proposition, $u$ extends to a function $\widetilde u$ on $B(0,R')$ that is harmonic on the annulus $r<|x|<R'$ and vanishes on $\partial B$. This extension is subharmonic on $B(0,R')$, because subharmonicity is a local property and the open sets $B$ and $\{ r<|x|<R'\} $ cover $B(0,R')$. Moreover, the distribution $\Delta \widetilde u$ on $B(0,R')$ equals $\mu $, extended by zero: the two agree on $B$, and both vanish on the annulus, so they agree on $B(0,R')$ by the last assertion of lemma 2.5.2.

Fix $r'$ with $r<r'<R$, put $Q=\overline{B(0,r')}$, and let $u_\varepsilon =\widetilde u*\rho _\varepsilon $ for $0<\varepsilon <\min \{ r'-r,R'-R\} $. Since $\widetilde u$ is harmonic on the annulus $r<|x|<R'$, the last assertion of theorem 3.1.7 gives $u_\varepsilon =\widetilde u$ on $r+\varepsilon <|x|<R'-\varepsilon $. Hence $u_\varepsilon $ is smooth on a neighborhood of $\overline B$, equals $u$ on $B\setminus Q$, vanishes on $\partial B$, and satisfies $\Delta u_\varepsilon =(\Delta \widetilde u)*\rho _\varepsilon =\mu *\rho _\varepsilon \geq 0$ on $B$ by proposition 2.5.3(1). In particular, $u_\varepsilon $ is a bounded subharmonic function on $B$ (a smooth function with nonnegative Laplacian is subharmonic, subsection 3.1.23), it tends to zero at $\partial B$, and its Riesz measure $(\mu *\rho _\varepsilon )\, \mathrm dx$ is supported in $\overline{B(0,r+\varepsilon )}\subset Q$. By (3.4.13), applied to $\widetilde u$, the functions $u_\varepsilon $ decrease to $u$ on $B$ as $\varepsilon \downarrow 0$.

For small $\varepsilon ,\delta >0$, the divergence identity (3.1.3) on $B$, applied to the field $u_\varepsilon \nabla u_\delta $, together with Fubini's theorem and the symmetry of $\rho _\delta $, gives

We let $\varepsilon $ and $\delta $ tend to zero. On $\operatorname{Supp}\mu $, the functions $\varphi _{\varepsilon ,\delta }=\widetilde u*\rho _\varepsilon *\rho _\delta $ are bounded by $\sup |\widetilde u|$ near $\overline B$, which is finite. By (3.4.13), $\widetilde u*\rho _\varepsilon $ decreases to $\widetilde u$ as $\varepsilon \downarrow 0$; convolving with $\rho _\delta \geq 0$ preserves order, so $\varphi _{\varepsilon ,\delta }$ decreases as either radius decreases, and $\varphi _{\varepsilon ,\delta }\geq \widetilde u*\rho _\delta \geq \widetilde u$. If $\varepsilon ,\delta \leq \eta \leq \eta '$, then $\varphi _{\varepsilon ,\delta }\leq \varphi _{\eta ,\eta }\leq \varphi _{\eta ',\eta }$, and $\varphi _{\eta ',\eta }\to \widetilde u*\rho _{\eta '}$ as $\eta \downarrow 0$ because $\widetilde u*\rho _{\eta '}$ is continuous. Letting $\eta \downarrow 0$ and then $\eta '\downarrow 0$, we see that $\varphi _{\varepsilon ,\delta }\to u$ pointwise on $\operatorname{Supp}\mu $ as $\max \{ \varepsilon ,\delta \} \to 0$.

By dominated convergence, $\int \nabla u_\varepsilon \mathbin {\cdot }\nabla u_\delta \, \mathrm d x\to -\int u\, \mathrm d\mu $. Expanding $\| \nabla u_\varepsilon -\nabla u_\delta \| _2^2$, we find that the gradients $\nabla u_\varepsilon $ form a Cauchy family in $L^2(B)$. Their limit is the distributional gradient of $u$, as one sees by integrating against test functions and using $u_\varepsilon \to u$ in $L^1_{\mathrm{loc}}(B)$. Hence $\nabla u\in L^2(B)$ and $\| \nabla u\| _2^2=\lim _\varepsilon \| \nabla u_\varepsilon \| _2^2=-\int u\, \mathrm d\mu $.

The function $u+v$ satisfies the same hypotheses, with Riesz measure $\mu +\nu $. Applying the first identity to $u$, $v$ and $u+v$ gives $2\int \nabla u\mathbin {\cdot }\nabla v\, \mathrm dx=-\int u\, \mathrm d\nu -\int v\, \mathrm d\mu $. Since $u=-G_ B\mu $ and $v=-G_ B\nu $, Tonelli's theorem and the symmetry of $G_ B$ (lemma 4.1.14) give

which proves the second line of the display. All these integrals are finite because $u$ and $v$ are bounded. The Cauchy–Schwarz inequality in $L^2(B)$ now gives the Cauchy–Schwarz inequality for the Green energy. Finally, $u_ K$ satisfies the hypotheses by proposition 4.1.16, and since $u_ K\geq -1$ and $\lambda _ K$ has total mass $C_ B(K)$, we get $\| \nabla u_ K\| _2^2=-\int u_ K\, \mathrm d\lambda _ K\leq C_ B(K)$. $\square$

Proof. The case of a bounded Green potential. Suppose first that $u$ is a bounded subharmonic function on $B$ that tends to zero at $\partial B$ and whose Riesz measure $\mu $ has compact support in $B$. Let $Q\subset B$ and $u_\varepsilon $ be the compact set and the regularizations given by the last assertion of proposition 4.1.29. These regularizations are continuous on $\overline B$, decrease to $u$, equal $u$ on $B\setminus Q$, and $\nabla u_\varepsilon \to \nabla u$ in $L^2(B)$. For $\delta >0$, the set $O_{\delta ,\varepsilon }=\{ u_\varepsilon -u>\delta \} $ is open, because $u_\varepsilon -u$ is lower semicontinuous, and it is contained in $Q$, so it has compact closure in $B$. Let $K\subset O_{\delta ,\varepsilon }$ be compact. The function $u_\varepsilon -u$ is a difference of two functions satisfying the hypotheses of proposition 4.1.29, and $\lambda _ K$ is supported on $K$, where $u_\varepsilon -u>\delta $. By the bilinear identity of that proposition, the Cauchy–Schwarz inequality and the bound $\| \nabla u_ K\| _2^2\leq C_ B(K)$,

Hence $C_ B(K)\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$, and taking the supremum over $K$ gives $C_ B(O_{\delta ,\varepsilon })\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$.

Choose $\delta _ j\downarrow 0$ and then radii $\varepsilon _ j\downarrow 0$ such that $C_ B(O_{\delta _ j,\varepsilon _ j})<\eta 2^{-j}$, and let $O=\bigcup _ jO_{\delta _ j,\varepsilon _ j}$, which is open with compact closure in $B$. By countable subadditivity (theorem 4.1.19), $C_ B(O)<\eta $. Outside $O$ we have $0\leq u_{\varepsilon _ j}-u\leq \delta _ j$ for all $j$, so the continuous functions $u_{\varepsilon _ j}$ converge uniformly to $u$ on $B\setminus O$. Hence $u$ is finite and continuous on $B\setminus O$, in particular on $L\setminus O$.

The case of a function bounded near $\overline B$. Suppose next that $u$ is bounded on a neighborhood of $\overline B$; subtracting a constant, we may assume that $u\leq -1$ there. Choose $r<R$ with $L\subset B(0,r)$, and $M>0$ with $-M<\inf _ Bu$. Let $q=Mu_{\overline{B(0,r)}}$, where the extremal function $u_{\overline{B(0,r)}}$ was computed in example 4.1.21. Then $q$ is subharmonic on $B$, equals $-M$ on $\overline{B(0,r)}$, is harmonic on $r<|x|<R$, and tends to zero at $\partial B$. The function $w=\max \{ u,q\} $ is subharmonic and bounded on $B$; it equals $u$ on $B(0,r)$, and it equals $q$ near $\partial B$, where $q>-1\geq u$. Hence $w$ tends to zero at $\partial B$, and its Riesz measure has compact support in $B$. The first case, applied to $w$, gives the required open set, since $w=u$ near $L$.

The general case. The function $u$ is bounded above near $\overline B$, so the functions $u_ N=\max \{ u,-N\} $, $N\geq 1$, are subharmonic and bounded near $\overline B$. Fix an open set $W$ with $L\subset W\Subset B$. By the second case, there are open sets $O_ N\subset W$ with $C_ B(O_ N)<\eta 2^{-N-1}$ such that $u_ N$ is continuous on $L\setminus O_ N$ (intersecting with $W$ does not affect this). On $L\setminus \bigcup _ NO_ N$, every $u_ N$ is continuous, and $\mathrm e^{u_ N}$ converges uniformly to $\mathrm e^ u$, because $0\leq \mathrm e^{u_ N}-\mathrm e^ u\leq \mathrm e^{-N}$. Hence $u$ is continuous there as a function with values in $[-\infty ,\infty )$. Let $\Omega _0$ be the connected component of $\Omega $ that contains the connected set $\overline B$. Since $u$ is not identically $-\infty $ on $\Omega _0$, the set $\{ u=-\infty \} \cap \Omega _0$ is polar by the definition of polarity. By corollary 4.1.24, $L\cap \{ u=-\infty \} $ is contained in an open set $O'\Subset B$ with $C_ B(O')<\eta /2$. The open set $O=O'\cup \bigcup _ NO_ N$ has compact closure in $B$ and capacity less than $\eta $ by countable subadditivity, and the restriction of $u$ to $L\setminus O$ is finite and continuous. $\square$