For a <harmonic function> $u$, let
$$
M(r)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}u.
$$
The <divergence theorem> gives
$$
M'(r)=\frac1{|\partial B_r|}\int_{\partial B_r(x)}\partial_\nu u
=\frac1{|\partial B_r|}\int_{B_r(x)}\Delta u=0.
$$
Since $M(r)\to u(x)$ as $r\downarrow0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the <mean value property for harmonic functions>. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. <continuous function>[Continuity] makes $u$ constant on those spheres, and <connected space>[connectedness] propagates that value through the domain. Thus the <weak maximum principle for elliptic operators> gives
$$
\min_{\partial\Omega}u\leq u\leq\max_{\partial\Omega}u.
$$

For the derivative estimate, choose $r>0$ smaller than half the <distance to a set>[distance] from $\Omega'$ to $\partial\Omega$, and let $\rho_r$ be a smooth radial <mollifier> supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u*\rho_r=u$ on $\Omega'$. Hence, for every <multi-index> $\alpha$,
$$
D^\alpha u(x)=\int_\Omega D^\alpha\rho_r(x-y)u(y)\,dy,
$$
so <Holder inequality> gives
$$
\|D^\alpha u\|_{L^\infty(\Omega')}
\leq\|D^\alpha\rho_r\|_\infty\|u\|_{L^1(\Omega)}
\leq C(n,\alpha,\Omega,\Omega')\|u\|_{L^1(\Omega)}.
$$

Solved by gpt-5.6-sol high.

Solution

Paper 107 / 1 / a / Solution

For
$$
Lu=a^{ij}D_{ij}u+b^iD_iu+cu,
\qquad c\leq0,
$$
the <weak maximum principle for elliptic operators> states that $Lu\geq0$ implies
$$
\max_{\overline\Omega}u\leq\max\{0,\max_{\partial\Omega}u\}.
$$
In particular, a solution of $Lu=0$ cannot have a positive interior maximum exceeding its boundary maximum.

Because the coefficient matrix is positive definite and the closure of the smooth bounded domain is compact, strict ellipticity supplies a uniform lower bound after restricting to $\overline\Omega$. Rotate and translate coordinates so that $\Omega$ is bounded in the $x_1$ direction, and set $h=e^{\gamma x_1}$. For sufficiently large $\gamma$,
$$
Lh=e^{\gamma x_1}(\gamma^2a^{11}+\gamma b^1+c)>0.
$$
If $u+\varepsilon h$ had a positive interior maximum, its gradient would vanish and its <Hessian matrix> would be negative semidefinite there, giving $L(u+\varepsilon h)\leq0$. This contradicts $L(u+\varepsilon h)=\varepsilon Lh>0$. Comparing on the boundary and sending $\varepsilon\downarrow0$ proves the assertion.

Solved by gpt-5.6-sol high.

Paper 107 / 3 / a / Solution

Let $v$ range over a bounded subset of $C^{1,\beta}$. All jets $(x,v,Dv)$ then lie in one compact set, so the composed coefficients have uniform $C^{0,\alpha\beta}$ bounds and one ellipticity constant. The <weak maximum principle for elliptic operators> bounds $T(v)$ in $C^0$ by the boundary data and the bounded forcing term. The <Global Schauder estimate> consequently bounds $T(v)$ in $C^{2,\alpha\beta}$.

The <compact embedding of Hölder spaces>
$$
C^{2,\alpha\beta}(\overline\Omega)Subset C^{1,\beta}(\overline\Omega)
$$
then makes the image relatively compact. Hence $T$ is a compact operator on $C^{1,\beta}(\overline\Omega)$.

Solved by gpt-5.6-sol high.

Paper 107 / 3 / d / Solution

For a second-order elliptic operator $Lu=a^{ij}D_{ij}u+b^iD_iu+cu$ with $c\leq0$, the inequality $Lu\geq0$ on a bounded domain implies
$$
\max_{\overline\Omega}u\leq\max\{0,\max_{\partial\Omega}u\}.
$$
An exponential barrier reduces the non-strict inequality to a contradiction at a positive interior maximum.


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 Incoming links: Weak maximum principle for elliptic operators