In this section, we will study regularity and some properties of the minimizer, in particular the set of non-1-Lebesgue points of \(D^2u\). We will expose a singular behavior of the Laplacian at all those points. Moreover we prove that minimizers are semiconvex, which can also be seen as a regularity property, having Aleksandrov’s theorem in mind.
For our arguments, we need some remarkable facts about the fundamental solution in two dimensions that were already discovered and applied to the biharmonic obstacle problem by Caffarelli and Friedman in [6, e.g., Equation (6.3)].
Lemma 3.1
(Fundamental solution of the biharmonic operator, cf. [23, Section 7.3]) Define \(F : {\mathbb {R}}^2 \times {\mathbb {R}}^2 \setminus \{(x,x): x \in {\mathbb {R}}^2 \} \rightarrow {\mathbb {R}}\) via
$$\begin{aligned} F(x,y) := \frac{1}{8\pi } |x-y|^2 \log |x-y| . \end{aligned}$$
Then F satisfies \(\Delta ^2 F(x,\cdot ) = \delta _x\) on \({\mathbb {R}}^2\), where \(\delta _x\) denotes the Dirac measure of \(\{x\}\). Then for each \(\beta \in (0,1]\) one has that \(F(\cdot ,y) \in W^{3,2-\beta }_{loc} ({\mathbb {R}}^2)\) for each \(y \in {\mathbb {R}}^2\). Moreover, for all \((x,y) \in {\mathbb {R}}^2\) such that \(x \ne y\) one has
$$\begin{aligned} \nabla _x F(x,y)&= - \nabla _y F(x,y) = \frac{1}{8\pi }(2 \log |x-y| + 1) (x-y), \end{aligned}$$
(3.1)
$$\begin{aligned} \partial _{x_ix_i}^2 F(x,y)&= \frac{1}{8\pi } \left( 1+ 2 \frac{(x_i-y_i)^2}{|x-y|^2} +2 \log |x-y| \right) \quad i = 1,2 , \end{aligned}$$
(3.2)
$$\begin{aligned} \partial ^2_{x_1x_2} F(x,y)&= \frac{1}{4\pi } \frac{(x_1-y_1) (x_2-y_2) }{|x-y|^2}. \end{aligned}$$
(3.3)
In particular,
$$\begin{aligned} \Delta _x F(x,y) = \frac{1}{2\pi } \left( \log |x-y| + 1 \right) , \end{aligned}$$
(3.4)
and \(\partial _{x_1x_1}^2F(x,\cdot ) - \partial _{x_2x_2}^2F(x,\cdot ), \partial _{x_1x_2}F(x, \cdot )\leqslant \frac{3}{8\pi }\) on \({\mathbb {R}}^2 \setminus \{x\}\) for each \(x \in {\mathbb {R}}^2\). Moreover, there is \(C > 0\) such that
$$\begin{aligned} |D_x^3F(x,y)| \leqslant \frac{C}{|x-y|} \quad \forall y \in {\mathbb {R}}^2 \setminus \{ x \} . \end{aligned}$$
(3.5)
Lemma 3.2
Let
\(x_0,y \in {\mathbb {R}}^2\)
and
Then
H
is decreasing on
\((0,\infty )\)
and its pointwise limit as
\(r \rightarrow 0\)
is given by
\(- \frac{1}{8\pi }\log |x_0 - y|\)
with the convention that
\(- \log 0 := \infty\)
Proof
The claim follows directly from [5, Proposition 4.4.11(6)] and [5, Proposition 4.4.15]. \(\square\)
The following result is very similar to crucial observations in [6].
Lemma 3.3
(Biharmonic measure representation, proof in Appendix 1) Let \(u \in {\mathcal {A}}(u_0)\) be a minimizer and \(\mu\) be as in (2.8). Further let \(\Omega _{\epsilon _0}\) be as in Corollary 2.10. Then there exists \(h \in C^\infty (\overline{\Omega _{\epsilon _0}^C})\) such that
$$\begin{aligned} u(x) = - \frac{1}{2}\int _\Omega F(x,y) \; \mathrm {d}\mu (y) + h(x) \quad \forall x \in \Omega _{\epsilon _0}^C, \end{aligned}$$
where F is the same as in Lemma 3.1.
The explicit representation of the minimizer will help to prove a first regularity result. The method used here is explained in the following lemma, whose proof is very straightforward by the definition of a weak derivative and Fubini’s theorem.
Lemma 3.4
(Kernel operators with measures) Let \(\Omega \subset {\mathbb {R}}^n\) be open and bounded and \(1 \leqslant p < \infty\). Let \(\alpha\) be a finite Borel measure on \(\Omega\) and let \(\lambda\) denote the n-dimensional Lebesgue measure on \(\Omega\). Let \(H :\Omega \times \Omega \rightarrow \overline{{\mathbb {R}}}\) be a Borel measurable function on \(\Omega \times \Omega\) such that
-
(1)
\((x,y) \mapsto H(x,y) \in L^p(\lambda \times \alpha )\)
-
(2)
For each \(y \in \Omega\), \(x \mapsto H(x,y)\) is weakly differentiable with \(\Omega \times \Omega\)-Borel measurable weak derivative \(\nabla _x H(x,y)\).
-
(3)
\((x,y) \mapsto \nabla _x H(x,y) \in L^p(\lambda \times \alpha )\).
Then \(A(x) := \int _\Omega H(x,y) \; \mathrm {d}\alpha (y)\) lies in \(W^{1,p}(\Omega )\) and its weak derivative satisfies
$$\begin{aligned} \nabla A(x) = \int _\Omega \nabla _x H(x,y) \; \mathrm {d}\alpha (y) . \end{aligned}$$
(3.6)
Using induction and the previous lemma, one easily obtains the following higher-order version.
Corollary 3.5
(Higher order derivatives) Let \(\Omega \subset {\mathbb {R}}^n\) be open and \(1 \leqslant p < \infty\). Let \(H : \Omega \times \Omega \rightarrow {\mathbb {R}}\) be Borel measurable on \(\Omega \times \Omega\) such that for each \(y \in \Omega\) the map \(x \mapsto H(x,y)\) lies in \(W^{k,p}(\Omega )\) and \(H, D_x H ,D_x^2 H ,\ldots D_x^k H \in L^p(\lambda \times \alpha )\) and all derivatives are all Borel measurable in \(\Omega \times \Omega\). Then \(A(x) := \int _\Omega H(x,y) d\alpha (y)\) lies in \(W^{k,p}(\Omega )\). Moreover one has
$$\begin{aligned} D^kA(x) = \int _\Omega D^kH(x,y) \; \mathrm {d}\alpha (y) \quad k = 1,\ldots ,n \quad \mathrm {a.e.} \; x \in \Omega . \end{aligned}$$
(3.7)
Corollary 3.6
(Sobolev regularity of minimizers) Let \(u\in {\mathcal {A}}(u_0)\) be a minimizer and \(\beta \in (0,1]\). Then \(u \in W^{3,2-\beta }(\Omega _{\epsilon _0}^C)\) for each \(\beta > 0\) and the set of non-1-Lebesgue points of \(D^2u\) in \(\Omega _{\epsilon _0}^C\) has Hausdorff dimension 0. Moreover, at every 1-Lebesgue point of \(D^2u\) which is not an atom of \(\mu\), one has
$$\begin{aligned} (D^2u)^*(x) =- \frac{1}{2} \int _\Omega D^2F(x,y)\; \mathrm {d}\mu (y) + D^2h(x), \end{aligned}$$
(3.8)
where F, \(\mu\) and h are given in Lemma 3.3.
Proof
For the \(W^{3,2-\beta }\)-regularity, we use the representation in Lemma 3.3 and Corollary 3.5. The requirements of Corollary 3.5 are satisfied if we can show that \(F,D_x F,D_x^2F\) and \(D_x^3F\) lie in \(L^{2-\beta }(\lambda \times \mu )\) (since the remaining requirements follow immediately from Lemma 3.1). We show this only for \(D_x^3F\), the other computations are very similar. Using (3.5), Tonelli’s Theorem and radial integration, we find
$$\begin{aligned}&\int _\Omega |D_x^3F(x,y) |^{2-\beta } \; \mathrm {d}( \lambda \times \mu )(x,y) = \int _\Omega \int _\Omega |D_x^3F(x,y)|^{2-\beta } \;\mathrm {d}x\; \mathrm {d}\mu (y) \\&\quad \leqslant \int _\Omega \int _\Omega \frac{C^{2-\beta }}{|x-y|^{2-\beta }} \;\mathrm {d}x\; \mathrm {d}\mu (y) \leqslant C^{2-\beta } \int _\Omega \int _{B_{\mathrm {diam}(\Omega )}(y)} \frac{1}{|x-y|^{2-\beta }} \;\mathrm {d}x\; \mathrm {d}\mu (y)\\&\quad \leqslant C^{2-\beta } \int _\Omega \int _0^{\mathrm {diam}(\Omega )} 2\pi \frac{r}{r^{2-\beta }} \;\mathrm {d}r\; \mathrm {d}\mu (y) \leqslant C^{2-\beta } \int _\Omega 2\pi \mathrm {diam}(\Omega )^\beta \; \mathrm {d}\mu (y) \\&\quad =2\pi C^{2-\beta } \mathrm {diam}(\Omega )^\beta \mu (\Omega ) < \infty . \end{aligned}$$
The \(W^{3,2-\beta }\)-regularity claim is shown. We conclude that \(D^2u \in W^{1,2-\beta }(\Omega _{\epsilon _0}^C)\) for each \(\beta > 0\). Since \(\Omega _{\epsilon _0}^C\) has Lipschitz boundary, \(D^2u\) extends to a function in \(W^{1,2-\beta }({\mathbb {R}}^n)\) (cf. [11, Thm.1, Sect.5.4]). From [12, Thm.1(i),(ii), Sect.4.8] follows that there is a Borel set \(E_\beta \subset \Omega\) of \(\beta\)-Capacity zero, such that the non-1-Lebesgue points are contained in \(E_\beta\). Now [12, Thm.4, Sect.4.7] implies that \({\mathcal {H}}^{2\beta }(E_\beta )= 0\) and hence the set of non-1-Lebesgue points is a \({\mathcal {H}}^{2\beta }\) null set. Equation (3.8) does not follow directly, since (3.7) only gives one representative of \(D^2u\). Let \(x_0\) be a 1-Lebesgue point of \(D^2u\). Then, according to Lemma 3.1
Since h is smooth, the last summand tends to \(\partial _{x_1x_1}^2h(x_0)\). We have already shown above that \(\partial _{x_1x_1}^2F = \frac{-1}{8\pi } \left( 1 + 2 \frac{(x_1 - y_1)^2}{|x-y|^2} + 2 \log |x-y| \right)\) lies in \(L^{2-\beta }(\lambda \times \mu )\). Therefore, we can interchange the order of the two integrations by Fubini’s Theorem. Hence
Now observe that
is decreasing in r because of Lemma 3.2 and hence the monotone convergence theorem yields
(Actually, the monotone convergence theorem is not exactly applicable since the integrand is not necessarily positive. This can however be fixed since \(\mu\) is finite and for each r the integrand is bounded from below by \(-\frac{1}{8\pi }\log \mathrm {diam}(\Omega )\). Adding and subtracting this quantity one obtains the claimed convergence). Therefore,
Observe that for \(y \ne x_0\) one has
Since \(\mu (\{x_0\})= 0\) the integrand converges \(\mu\)-almost everywhere to the right-hand side. This and fact that the expression is uniformly bounded in r by \(\frac{3}{8\pi }\) imply together with the dominated convergence theorem that
Plugging this into (3.10) we find
$$\begin{aligned} 2(\partial ^2_{x_1x_1} u)^*(x_0) = \int _\Omega \frac{-1}{8\pi } \left( 1 + 2 \frac{((x_0)_1 - y_1)^2}{|x_0-y|^2} + 2 \log |x_0-y| \right) \; \mathrm {d}\mu (y) + 2 \partial ^2_{x_1x_1} h(x_0) . \end{aligned}$$
The same techniques apply for \((\partial _{x_1x_2}^2u)^*\) and \((\partial _{x_2x_2}^2u)^*\). This proves (3.8). \(\square\)
Corollary 3.7
(Bounded mixed derivatives) Let \(u \in {\mathcal {A}}(u_0)\) be a minimizer. Then \(\partial ^2_{x_1x_2}u\) and \(\partial ^2_{x_1x_1} u - \partial ^2_{x_2x_2}u\) lie in \(L^\infty (\Omega _{\epsilon _0}^C)\). Moreover, each \(x_0 \in \Omega\) that is not an atom of \(\mu\) is a Lebesgue point of \(\partial ^2_{x_1x_2}u\) and \(\partial ^2_{x_1x_1} u -\partial ^2_{x_2x_2}u\).
Proof
For the fact that \(\partial ^2_{x_1x_1} u - \partial ^2_{x_2x_2} u \in L^\infty (\Omega _{\epsilon _0}^c)\) observe with the notation of (3.8) that almost everywhere one has
$$\begin{aligned} |\partial ^2_{x_1x_1} u - \partial ^2_{x_2x_2} u |&= \left| - \frac{1}{2} \int _\Omega (\partial ^2_{x_1x_1}F - \partial ^2_{x_2x_2} F) \; \mathrm {d}\mu (y) + \partial ^2_{x_1x_1} h - \partial ^2_{x_2x_2} h \right| \\&\leqslant \frac{3}{16 \pi } \mu (\Omega ) + 2|| D^2h||_\infty < \infty , \end{aligned}$$
where we used Lemma 3.1 in the last step. Similarly, one shows that \(\partial ^2_{x_1x_2} u \in L^\infty (\Omega _{\epsilon _0}^C)\). Now we show that each non-atom x of \(\mu\) is a 1-Lebesgue point of \(\partial ^2_{x_1x_2 }u\). By (3.8), it is sufficient to show that each non-atom of \(\mu\) is a 1-Lebesgue point of \(\int _\Omega \partial ^2_{x_1x_2}F(\cdot ,y) d\mu (y)\) as each point in \(\Omega _{\epsilon _0}^C\) is a Lebesgue point of \(D^2h\). We have already discussed in Corollary 3.6 that \(\partial ^2_{x_1x_2}F\) is \((\lambda \times \mu )\)-measurable. Moreover, it is product integrable as it is uniformly bounded.
By Fubini’s theorem
$$\begin{aligned}&\frac{1}{|B_r(x)|} \int _{B_r(x) } \left( \int _\Omega \partial _{x_1x_2}^2 F(z,y) \; \mathrm {d}\mu (y) \right) \; \mathrm {d}z \\&\quad \quad = \int _\Omega \left( \frac{1}{|B_r(x)|} \int _{B_r(x)} \partial ^2_{x_1x_2} F(z,y)\; \mathrm {d}z \right) \; \mathrm {d}\mu (y) . \end{aligned}$$
For each \(y \in \Omega \setminus \{x\}\) the expression in parentheses converges to \(\partial ^2_{x_1x_2} F(x,y)\) as \(r \rightarrow 0\) and since x is not an atom of \(\mu\) the expression converges to \(\partial ^2_{x_1x_2} F(x,y)\) \(\mu\)-almost everywhere. Moreover, Lemma 3.1 yields that the expression is uniformly bounded by \(\frac{3}{8\pi }\), and hence the dominated convergence theorem yields
$$\begin{aligned} \lim _{r\rightarrow 0 } \frac{1}{|B_r(x)|} \int _{B_r(x) } \left( \int _\Omega \partial _{x_1x_2}^2 F(z,y) \; \mathrm {d}\mu (y) \right) \; \mathrm {d}z = \int _\Omega \partial ^2_{x_1x_2} F(x,y) \; \mathrm {d}\mu (y). \end{aligned}$$
(3.11)
To show the Lebesgue point property, it remains to show that
$$\begin{aligned} \lim _{r \rightarrow 0} \frac{1}{|B_r(x)|} \int _{B_r(x)} \left| \int _\Omega \partial ^2_{x_1x_2} F(z,y) \; \mathrm {d}\mu (y) - \int _\Omega \partial ^2_{x_1x_2} F(x,y) \; \mathrm {d}\mu (y) \right| = 0 . \end{aligned}$$
This is immediate once one observes with the triangle inequality and Fubini’s theorem that
$$\begin{aligned}&\frac{1}{|B_r(x)|} \int _{B_r(x)} \left| \int _\Omega \partial ^2_{x_1x_2} F(z,y) \; \mathrm {d}\mu (y) - \int _\Omega \partial ^2_{x_1x_2} F(x,y) \; \mathrm {d}\mu (y) \right| \; \mathrm {d}z\\&\quad \quad \leqslant \int _\Omega \frac{1}{|B_r(x)|}\int _{B_r(x)} |\partial ^2_{x_1x_2} F(z,y) - \partial ^2_{x_1x_2} F(x,y)| \; \mathrm {d}z \; \mathrm {d}\mu (y) . \end{aligned}$$
The term on the right-hand side can be shown to tend to zero as \(r \rightarrow 0\) with the dominated convergence theorem using arguments similar to the discussion before (3.11). For \(\partial ^2_{x_1x_1} u - \partial ^2_{x_2x_2}u\) the analogous statement can be shown similarly. \(\square\)
Corollary 3.8
(The Laplacian of a minimizer) Let \(u \in {\mathcal {A}}(u_0)\) be a minimizer. Then for each \(x \in \Omega\) the quantity
exists in \([0, \infty ]\). Moreover, the map \(x \mapsto (\Delta u)^*(x)\) is superharmonic.
Proof
Recall that by \(\Delta u\) is weakly superharmonic by (2.5). By [30, Theorem 4.1] follows immediately that \((\Delta u)^*(x)\) exists in \({\mathbb {R}} \cup \{ \infty \}\) for all \(x \in \Omega\). By Corollary 2.13 it has to lie in \([0, \infty ]\), which shows the first part of the claim. From (3.8) and Lemma 3.1, we infer that
$$\begin{aligned} \Delta u (x) = - \frac{1}{4\pi } \int _\Omega (\log |x-y| + 1 ) \; \mathrm {d}\mu (y) + \Delta h (x) \quad a.e. . \end{aligned}$$
Similar to the discussion in (3.9) we can derive, using the special properties of the logarithm that
$$\begin{aligned} (\Delta u)^*(x) = - \frac{1}{4\pi } \int _\Omega (\log |x-y| + 1 ) \; \mathrm {d}\mu (y) + \Delta h (x) \quad \forall x \in \Omega _{\epsilon _0}^C. \end{aligned}$$
(3.12)
Note that \((\Delta u)^*\) is the so-called canonical representative of a weakly subharmonic function in the sense of [30, p.360]. To show that \((\Delta u)^*\) is subharmonic it suffices according to [30, Theorem 4.3] to show that \((\Delta u)^*\) is lower semicontinuous. For this let \((x_n)_{n = 1}^\infty \subset \Omega _{\epsilon _0}^C\) be such that \(x_n \rightarrow x \in \Omega _{\epsilon _0}^C\). Note that \(- \log |x_n - \cdot |\) is bounded from below independently of n by \(-\log \mathrm {diam}(\Omega )\). Thus Fatou’s lemma yields
$$\begin{aligned} \liminf _{n \rightarrow \infty } \int _\Omega - \log |x_n - y | \; \mathrm {d}\mu (y) \geqslant \int _\Omega \liminf _{n\rightarrow \infty } ( - \log |x_n - y| \; \mathrm {d}\mu (y) = -\int _\Omega \log |x-y| \; \mathrm {d}\mu (y) . \end{aligned}$$
(3.13)
Since \((\Delta u)^*(x_n)\) consists only of continuous terms and a positive multiple of the left-hand side in (3.13), one has \(\liminf _{n \rightarrow \infty } (\Delta u)^*(x_n) \geqslant (\Delta u)^*(x)\), that is \((\Delta u)^*\) is lower semicontinuous. As we already explained, this implies superharmonicity of \((\Delta u)^*\). \(\square\)
Remark 3.9
Note that the notation \((\Delta u)^*\) creates a slight ambiguity with (2.1), namely whenever the limit in the definition is infinite. It will always be clear from the context what convention is used, especially in view of the following consistency result.
Proposition 3.10
(Lebesgue points of superharmonic functions) Let \(f : \Omega \rightarrow {\mathbb {R}}\) be a nonnegative superharmonic function. Then each point where \(f< \infty\) is a 1-Lebesgue point of f.
Proof
By [3, Theorem 3.1.3] one has \(f(x) = \liminf _{y \rightarrow x} f(y)\) for each \(x \in \Omega\). In particular
$$\begin{aligned} f(x) = \lim _{r \rightarrow 0 } \inf _{B_r(x)} f . \end{aligned}$$
(3.14)
Now suppose that \(f(x)< \infty\). Then by the triangle inequality
As f is superharmonic we have
as \(r \rightarrow 0+\). Using this, \(f(x)< \infty\) and (3.14), we obtain that
Putting the previous results together, we obtain the following
Corollary 3.11
(Characterization of non-Lebesgue points) Each non-1-Lebesgue point x of \(D^2u\) is an atom of \(\mu\) or satisfies \((\Delta u)^*(x) = \infty\).
Proof
Suppose that x is neither an atom of \(\mu\) nor \((\Delta u)^*(x) = \infty\). By Corollary 3.8 and Proposition 3.10, we get that x is a 1-Lebesgue point of \(\Delta u\). By Corollary 3.7, we also know that x is a 1-Lebesgue point of \(\partial ^2_{x_1x_1} u - \partial ^2_{x_2x_2} u\) and \(\partial ^2_{x_1x_2} u\). Since all second derivatives of u are linear combinations of the mentioned quantities, x is a 1-Lebesgue point of \(D^2u\). The claim follows by contraposition. \(\square\)
We can refine the statement with the following observation that shows also singular behavior of \((\Delta u)^*\) at each atom of \(\mu\).
Lemma 3.12
Let \(u \in {\mathcal {A}}(u_0)\) be a minimizer. If \(x_0 \in \Omega\) is an atom of \(\mu\), then \((\Delta u)^*(x_0) = \infty\).
Proof
Suppose that \(x_0\) is an atom of \(\mu\) and set \({\widetilde{\mu }} := \mu - \mu (\{x_0\}) \delta _{x_0}\) which is also a finite measure. Using (3.12) we find with the notation from there that for each \(x \in \Omega _{\epsilon _0}^C\)
$$\begin{aligned} (\Delta u)^*(x)&= - \frac{1}{4\pi } \int _\Omega (\log |x-y| + 1 )\; \mathrm {d}\mu (y) + \Delta h(x) \\&= - \frac{1}{4\pi } (\log |x-x_0| + 1) \mu (\{x_0\}) - \frac{1}{4\pi }\int _\Omega (\log |x-y| + 1 ) \; \mathrm {d}{\widetilde{\mu }}(y) + \Delta h (x)\\&\geqslant - ||\Delta h ||_\infty - \frac{1}{4\pi }( \log \mathrm {diam}(\Omega ) + 1) {\widetilde{\mu }}(\Omega ) - \frac{\mu (\{ x_0\} )}{4\pi } ( 1 + \log |x-x_0| ) . \end{aligned}$$
Plugging in \(x= x_0\) we obtain finally that \((\Delta u)^*(x_0) = \infty\) as claimed. \(\square\)
Remark 3.13
The previous observations show that each non-1-Lebesgue point of \(D^2u\) satisfies \((\Delta u)^* = \infty\) and each atom of \(\mu\) is a non-1-Lebesgue point of \(D^2u\).
Lemma 3.14
(Semiconvexity) Let \(u \in {\mathcal {A}}(u_0)\) be a minimizer and set
$$\begin{aligned} A := \frac{\sqrt{5}}{2} \left( 2 ||D^2h||_\infty + \frac{3}{16\pi } \mu (\Omega ) \right) . \end{aligned}$$
Then at each \(x \in \Omega _{\epsilon _0}^C\) which is 1-Lebesgue point of \(D^2u\) the matrix \((D^2 u)^* + AI\) is positive semidefinite, where \(I = \mathrm {diag}(1,1)\) denotes the identity matrix. In particular, for each \(x_0 \in {\mathbb {R}}^2\), one has that \(x \mapsto u(x) + \frac{1}{2} A |x-x_0|^2\) is convex on \(\Omega _{\epsilon _0}^C\).
Proof
Let x be a Lebesgue point of \(D^2u\). By Remark 3.13, x is not an atom of \(\mu\). Note that if \(M = \begin{pmatrix} m_{11} &{} m_{12} \\ m_{12} &{} m_{22} \end{pmatrix} \in {\mathbb {R}}^{2\times 2}\) is a symmetric matrix then the eigenvalues of M are given by
$$\begin{aligned} \lambda _{1,2} = \frac{m_{11} + m_{22}}{2} \pm \sqrt{\frac{1}{4} (m_{11} - m_{22} )^2 + m_{12}^2 }. \end{aligned}$$
(3.15)
If \(M = (D^2u)^*(x) + AI\) then Corollary 2.13 implies that
$$\begin{aligned} \frac{m_{11} + m_{22}}{2} = (\Delta u)^* + 2A \geqslant 2 A . \end{aligned}$$
(3.16)
Using (3.8), the fact that x is not an atom of \(\mu\), and Lemma 3.1, we obtain
$$\begin{aligned} |m_{11}- m_{22}|&= \left| \frac{1}{2} \int _\Omega (\partial ^2_{x_1x_1} F(x,y) - \partial ^2_{x_2x_2}F(x,y) ) \; \mathrm {d}\mu (y) + \partial ^2_{x_1x_1}h - \partial _{x_2x_2}^2h \right| \\&\leqslant \frac{1}{2} \int _{\Omega \setminus \{x\} } |\partial ^2_{x_1x_1} F(x,y) - \partial ^2_{x_2x_2}F(x,y) | \; \mathrm {d}\mu (y) + 2 ||D^2h||_\infty \\&\leqslant \frac{3}{16\pi }\mu (\Omega ) + 2||D^2h||_\infty . \end{aligned}$$
Analogously one can show that
$$\begin{aligned} |m_{12}| \leqslant \frac{3}{16\pi } \mu (\Omega ) + 2 ||D^2h ||_\infty . \end{aligned}$$
Hence
$$\begin{aligned} \sqrt{\frac{1}{4} (m_{11} - m_{22} )^2 + m_{12}^2 } \leqslant \sqrt{ \left( 1 + \frac{1}{4}\right) \left( \frac{3}{16\pi } \mu (\Omega ) + 2 ||D^2h ||_\infty \right) ^2 } \leqslant A. \end{aligned}$$
Plugging this and (3.16) into (3.15) we find
$$\begin{aligned} \lambda _{1,2} \geqslant 2 A - A = A \geqslant 0. \end{aligned}$$
Thus we obtain that M is indeed positive semidefinite. For \(\epsilon >0\) let \(\rho _\epsilon\) be the standard mollifier. Set \(f_\epsilon (x) := \left( u(\cdot ) + \frac{1}{2}A |\cdot - x_0|^2\right) * \rho _\epsilon\). Observe that for \(\epsilon < \epsilon _0\), \(f_\epsilon \in C^2(\Omega _{\epsilon _0}^C)\) and \(D^2 f_\epsilon = (D^2 u + AI) * \rho _\epsilon\) on \(\Omega _{\epsilon _0}^C\). This matrix is positive semidefinite since for each \(z \in {\mathbb {R}}^2\)
$$\begin{aligned} z^TD^2f_\epsilon (x) z = (z^T(D^2u + AI) z * \rho _\epsilon )(x) \geqslant 0 , \end{aligned}$$
as \(\rho _\epsilon\) is nonnegative and \(D^2 u + AI\) is positive semidefinite almost everywhere. Hence \(f_\epsilon\) is convex. However \(f_\epsilon\) also converges to \(u + \frac{1}{2} A |\cdot - x_0|^2\) uniformly on \(\Omega _{\epsilon _0}^C\) as the latter function is continuous. It is easy to verify with the definition of convexity that uniform limits of convex functions are convex again. \(\square\)