Maximum Principles Harnack and Liouville in Rn

${\mathrm{\partial }}_{ii}\mathrm{\mid }x{\mathrm{\mid }}^2=2$ for each of the $n$ coordinates. Therefore $\mathrm{\Delta }{u}_{\epsilon }=\mathrm{\Delta }u+2n\epsilon \ge 2n\epsilon >0$.

At an interior local maximum $q$, each restriction $t\mapsto u_{\epsilon }(q+t{e}_i)$ would have second derivative at zero at most zero. Summing would give $\mathrm{\Delta }{u}_{\epsilon }(q)\le 0$, incompatible with the strict positive value already obtained. Hence no such maximum exists.

Choose $R>0$ with $\bar{\mathrm{\Omega }}\subseteq {\bar{B}}_R(0)$. The closure is nonempty compact. For each $\epsilon >0$, continuity makes $u_{\epsilon }=u+\epsilon \mathrm{\mid }x{\mathrm{\mid }}^2$ attain its maximum on that closure.

The maximizer cannot lie in $\mathrm{\Omega }$, so it lies in $\mathrm{\partial }\mathrm{\Omega }$. In particular the boundary is nonempty; it is closed and bounded, hence compact, and $m={\max }_{\mathrm{\partial }\mathrm{\Omega }}u$ exists.

For every $x\in \bar{\mathrm{\Omega }}$, $u(x)\le u_{\epsilon }(x)\le m+\epsilon R^2$. Letting $\epsilon \downarrow 0$ yields $u(x)\le m$. Since a boundary maximizer belongs to the closure, equality of the maxima follows.

Put $v=-u$. It has the same regularity and $\mathrm{\Delta }v=-\mathrm{\Delta }u\ge 0$.

Apply the weak maximum principle to $v$ and multiply its equality by $-1$. Maxima of $-u$ are negatives of minima of $u$, which proves the claimed equality.

The function $w=u-v$ is continuous on the closure, belongs to $C^2(\mathrm{\Omega })$, and satisfies $\mathrm{\Delta }w\ge 0$ and $w\le 0$ on the boundary.

The weak maximum principle gives ${\max }_{\bar{\mathrm{\Omega }}}w={\max }_{\mathrm{\partial }\mathrm{\Omega }}w\le 0$, exactly the required comparison.

Write $m(t)$ for the sphere average at $a$. The radial identity gives $m\mathrm{\prime }(t)=\frac{t}{n}\frac{1}{\mathrm{\mid }{B}_t\mathrm{\mid }}{\int }_{B_t(a)}\mathrm{\Delta }u\ge 0$. Continuity gives $m(t)\to u(a)$ as $t\downarrow 0$, so $m(r)\ge u(a)$.

For the converse, two integrations of the one-variable fundamental theorem along $a+t\theta$ give $u(a+r\theta )=u(a)+r\mathrm{\nabla }u(a)\cdot \theta +r^2{\int }_0^1(1-s){\theta }^T{D}^{2}u(a+sr\theta )\theta ds$. The continuity of $D^{2}u$ makes its remainder after replacing the Hessian by $D^{2}u(a)$ uniformly $o(r^2)$ for unit $\theta$. Reflection and permutation symmetry give sphere averages of ${\theta }_i$ and ${\theta }_i{\theta }_j$ equal to zero for $i\mathrm{\ne }j$, and of ${\theta }_i^2$ equal to $1\mathrm{/}n$. Thus $m(r)=u(a)+r^2\mathrm{\Delta }u(a)\mathrm{/}(2n)+o(r^2)$.

Polar coordinates give the ball average $\frac{n}{r^n}{\int }_0^r{t}^{n-1}m(t)dt\ge u(a)$. For signed $u$, apply the nonnegative polar formula to its positive and negative parts on the bounded ball; both are bounded and integrable. Replacing $u$ by $-u$ proves both reversed inequalities.

Integrating the expansion with the radial weights gives ball average $u(a)+r^2\mathrm{\Delta }u(a)\mathrm{/}(2(n+2))+o(r^2)$. Either assumed mean inequality, divided by $r^2>0$ and followed by $r\downarrow 0$, forces $\mathrm{\Delta }u(a)\ge 0$. Negation gives the superharmonic converse.

Put $M=u(a)$ and $E=\{x\in \mathrm{\Omega }:u(x)=M\}$. This is nonempty and relatively closed by continuity.

For $x\in E$ choose $r>0$ with ${\bar{B}}_r(x)\subset \mathrm{\Omega }$. The ball mean inequality gives ${\int }_{B_r(x)}(M-u)\le 0$, while the integrand is nonnegative. If it were positive at a point, continuity would give a positive lower bound on a smaller ball of positive volume, contradicting that integral inequality. Thus $u=M$ throughout $B_r(x)$, and $E$ is open.

If $\mathrm{\Omega }\setminus E$ were nonempty, it and $E$ would separate $\mathrm{\Omega }$ into disjoint nonempty relatively open sets. Connectedness therefore gives $E=\mathrm{\Omega }$.

For an attained global maximum $M$, the set $E=\{u=M\}$ is nonempty and relatively closed. For $x\in E$ take ${\bar{B}}_r(x)\subset \mathrm{\Omega }$. The ball mean property makes ${\int }_{B_r(x)}(M-u)=0$.

The integrand is nonnegative and continuous. A positive value would make its integral positive on a small ball; hence it vanishes everywhere on this ball. Thus $E$ is open, and connectedness gives $E=\mathrm{\Omega }$.

If the attained extremum is a minimum, apply the preceding argument to the harmonic function $-u$. Constancy of $-u$ is constancy of $u$.

If $u(a)=0$ at some $a\in \mathrm{\Omega }$, it attains its global minimum there. The strong harmonic maximum/minimum principle makes it constant, hence identically zero.

If there is no such point, nonnegativity forces $u(x)>0$ at every point. These alternatives exhaust the possibilities.

Comparison applied to $(u,v)$ and then $(v,u)$ gives $u\le v$ and $v\le u$, respectively, proving uniqueness.

For possibly different boundary values put $b={\max }_{\mathrm{\partial }\mathrm{\Omega }}\mathrm{\mid }u-v\mathrm{\mid }$. It is finite because the boundary is nonempty compact and the difference is continuous. Compare $u$ with $v+b$ and $v$ with $u+b$; their Laplacians agree. Hence $\mathrm{\mid }u-v\mathrm{\mid }\le b$ on the closure.

For a sequence with common Laplacian, the bound just proved applies to every pair of terms. Uniformly convergent boundary traces are uniformly Cauchy, so the solutions are uniformly Cauchy on the closure. Completeness of the real numbers supplies the pointwise limit and the same Cauchy bound makes convergence uniform. The limit is continuous there, as follows by combining a uniform error bound with continuity of a fixed term.

Put $b={\max }_{\mathrm{\partial }\mathrm{\Omega }}\mathrm{\mid }u\mathrm{\mid }$ and $v(x)=b+\frac{L}{2}(d^2-x_1^2)$. Since $0\le x_1\le d$ on the closure, $v\ge b$ there, and $\mathrm{\Delta }v=-L\le \mathrm{\Delta }u,\mathrm{\Delta }(-u)$.

Both $u$ and $-u$ are at most $v$ on the boundary, so comparison gives $\mathrm{\mid }u\mathrm{\mid }\le v\le b+L{d}^2\mathrm{/}2$ on the closure.

The parameter $\alpha$ is positive, so the denominator defining $c$ is positive. Substitution at radii $R\mathrm{/}2$ and $R$ gives the two boundary values.

Writing $z=x-a$, Cartesian differentiation gives ${\mathrm{\partial }}_{ii}v=c(4{\alpha }^2{z}_i^2-2\alpha )e^{-\alpha \mathrm{\mid }z{\mathrm{\mid }}^2}$ and hence $\mathrm{\Delta }v=2c\alpha (2\alpha \mathrm{\mid }z{\mathrm{\mid }}^2-n)e^{-\alpha \mathrm{\mid }z{\mathrm{\mid }}^2}\ge 0$ on the closed annulus. This is subharmonicity.

At $p$ on the outer sphere, the supplied direction is $\nu =(p-a)\mathrm{/}R$. The radial derivative gives ${\mathrm{\partial }}_{\nu }v(p)=-2c\alpha R{e}^{-\alpha R^2}<0$, also equal to the one-sided quotient limit.

On the compact inner sphere, continuity and strict inequality give $\epsilon =M-{\max }_{\mathrm{\mid }x-a\mathrm{\mid }=R\mathrm{/}2}u(x)>0$. Take the exponential barrier $v$, equal to one on that sphere and zero on the outer sphere.

Here is the additional Hopf route to the strong subharmonic principle. Suppose $U$ is a domain and $f\in C^2(U)$ satisfies $\mathrm{\Delta }f\ge 0$ and $f\le M$ in $U$, and $E=\{f=M\}$ is nonempty but not all of $U$. It is relatively closed. Its complement $G$ is nonempty open, and some $y\in E$ is a relative boundary point of $G$; otherwise $E$ and $G$ would separate $U$. Choose $d>0$ with ${\bar{B}}_d(y)\subset U$ and $x\in G$ with $\mathrm{\mid }x-y\mathrm{\mid }<d\mathrm{/}4$.

Continuity from inside the tangent ball gives $u\le M$ on its entire outer sphere. On the annulus, $w=u+\epsilon v-M$ is subharmonic and continuous on the closure, and its values are at most zero on both boundary spheres. The weak maximum principle gives $w\le 0$ throughout the annulus.

Set $r={\inf }_{z\in E}\mathrm{\mid }x-z\mathrm{\mid }$. Openness of $G$ gives $r>0$, and $r\le \mathrm{\mid }x-y\mathrm{\mid }<d\mathrm{/}4$. A closest point $q\in E$ exists: minimize distance on the nonempty compact set $E\cap {\bar{B}}_{d\mathrm{/}2}(y)$; points of $E$ outside this set have distance from $x$ greater than $d\mathrm{/}4$, so this also minimizes over all of $E$. The ball $B_r(x)$ lies in $G$, its closure lies in $U$, and $q$ is on its sphere.

For $0<t<R\mathrm{/}2$, $u(p)-u(p-t\nu )\ge \epsilon v(p-t\nu )$. Divide by $t>0$ and pass to the assumed finite limit: ${\mathrm{\partial }}_{\nu }u(p)\ge -\epsilon {\mathrm{\partial }}_{\nu }v(p)>0$.

Apply the boundary conclusion already proved in step 3.1 to $f$ on $G$, with this tangent ball and boundary point $q$. It gives a positive outward derivative. But $q\in U$ is an interior maximum of the differentiable function $f$, so all its first derivatives are zero and this directional derivative is zero. Therefore $E=U$ and $f$ is constant. This alternative uses no mean inequality.

Fix $d=\mathrm{\mid }x-a\mathrm{\mid }$ and $d<s<R$. The closed ball of radius $s-d$ centered at $x$ lies in $B_R(a)$, and its open ball lies in $B_s(a)$. Nonnegativity and the ball mean property yield $u(x)\le \mathrm{\mid }{B}_{s-d}{\mathrm{\mid }}^{-1}{\int }_{B_s(a)}u=(s\mathrm{/}(s-d){)}^{n}u(a)$.

Let $s\uparrow R$ to obtain the first inequality. It is also valid for $x=a$, with equality. No integral at radius $R$ is needed.

For $\mathrm{\mid }x-a\mathrm{\mid }\le r<R$, divide the segment from $a$ to $x$ into $m$ equal steps. Each has length at most $(R-r)\mathrm{/}4$. Every segment point centers a ball of radius $\rho =(R-r)\mathrm{/}2$ inside $B_R(a)$. Apply the first inequality from each endpoint of a step to the other, using these radius-$\rho$ balls: either value is at most $2^n$ times the other. Multiplying along the $m$ steps proves both comparisons with $C$. This uses no division by a value of $u$, so also covers $u(a)=0$.

An additional consequence is Gantumur’s growth estimate. If an entire harmonic $u$ satisfies $u(x)\ge -p(\mathrm{\mid }x\mathrm{\mid })$ for a nonnegative nondecreasing function $p$, fix $r=\mathrm{\mid }x\mathrm{\mid }>0$ and apply the first inequality to $u+p(2r)\ge 0$ on $B_{2r}(0)$. It gives $u(x)\le 2^n(u(0)+p(2r))-p(2r)\le 2^n(u(0)+p(2r))$. At $x=0$ the sharper expression after subtracting $p(0)$ gives the same conclusion, since $u(0)+p(0)\ge 0$.

Call a ball admissible if its radius is positive and its fourfold closed ball lies in $\mathrm{\Omega }$. Every point centers such a ball by openness. Fix one admissible base ball, using nonemptiness of $\mathrm{\Omega }$. Let $E$ be the union of all admissible balls joined to it by a finite sequence of overlapping admissible balls.

The set $E$ is nonempty and open. If $x\in \mathrm{\Omega }$ is a relative closure point of $E$, an admissible ball centered at $x$ meets $E$, hence meets a ball in one of the finite sequences. Appending the new ball proves it is in the reachable family and $x\in E$. Thus $E$ is relatively closed, and connectedness implies $E=\mathrm{\Omega }$.

Compactness gives a finite subcover of $K$ by admissible balls. Each selected ball meets a reachable ball by the preceding conclusion, and can be appended to a finite sequence from the base. Take the union of these finitely many finite sequences together with the base ball. The resulting finite family covers $K$ and its overlap graph is connected. If $K$ is empty, take just the base ball. All its fourfold closed balls are compact by Heine–Borel and remain in $\mathrm{\Omega }$.

If $u$ has an interior zero, it vanishes throughout $\mathrm{\Omega }$, and the inequality is immediate. Otherwise it is strictly positive. Fix the finite admissible-ball family of the chain lemma, of size $N$, independently of $u$.

If $x,y\in B_r(a)$ from this family, then $B_{3r}(y)\subset B_{4r}(a)\subset \mathrm{\Omega }$ and $\mathrm{\mid }x-y\mathrm{\mid }<2r$. Ball Harnack gives $u(x)\le 3^{n}u(y)$. Thus any two points of one of the small balls are comparable by the same constant.

For $x,y\in K$, connect balls containing them by a simple path in the finite connected graph, with at most $N$ vertices. Choose one point in each of its finitely many overlaps and apply the local comparison in successive balls. It gives $u(x)\le 3^{nN}u(y)$. Taking the supremum in $x$ and the infimum in $y$ proves the claim with $C=3^{nN}$.

If no point has a bounded-above scalar sequence $u_j(x)$, each scalar sequence is increasing and unbounded above, hence tends to $+\mathrm{\infty }$. Otherwise fix $a\in \mathrm{\Omega }$ at which it is bounded above; there $u_j(a)$ converges to a finite real number.

For any nonempty compact $K\subset \mathrm{\Omega }$, apply compact Harnack on $K\cup \{a\}$ to the harmonic nonnegative difference $u_j-u_i$, $j\ge i$. It yields ${\sup }_K\mathrm{\mid }{u}_j-u_i\mathrm{\mid }\le C(u_j(a)-u_i(a))$. The right side tends to zero as $i,j\to \mathrm{\infty }$, uniformly in those indices.

The real completeness property gives a pointwise limit, and the uniform Cauchy bound proves convergence uniformly on $K$. Empty compact sets impose no condition. Since $K$ was arbitrary, convergence is locally uniform and the harmonic-limit theorem makes the limit harmonic.

Translate $a$ to zero. Put $q=\mathrm{\mid }x-y\mathrm{\mid }$, $A=R^2-\mathrm{\mid }x{\mathrm{\mid }}^2$, and $P(x,y)=A{q}^{-n}\mathrm{/}(R{\omega }_{n-1})$, where $\mathrm{\mid }y\mathrm{\mid }=R$. The sphere has finite positive measure ${\omega }_{n-1}{R}^{n-1}$. On compact interior sets $q$ is bounded away from zero; all $x$-derivatives of the kernel times bounded $g$ have a constant integrable majorant. The mean value theorem bounds their difference quotients likewise. Dominated convergence therefore permits differentiation of every order under the sphere integral and proves continuity of those derivatives.

Cartesian differentiation gives $\mathrm{\nabla }{q}^{-n}=-n(x-y)q^{-n-2}$, $\mathrm{\Delta }{q}^{-n}=2n{q}^{-n-2}$ and $\mathrm{\Delta }A=-2n$. The product rule gives $\mathrm{\Delta }(A{q}^{-n})=2n{q}^{-n-2}(A-q^2+2x\cdot (x-y))=2n{q}^{-n-2}(R^2-\mathrm{\mid }y{\mathrm{\mid }}^2)=0$. Thus both $h$ and $I(x)=\int P(x,y)dS(y)$ are smooth harmonic functions.

Rotation invariance of surface measure and the kernel makes $I$ constant on each sphere centered at zero. Its spherical mean equals $I(0)$ by harmonic mean values; since it is already constant on that sphere, $I(x)=I(0)$. At zero, $P(0,y)=1\mathrm{/}({\omega }_{n-1}{R}^{n-1})$, so $I(0)=1$. Also $P>0$ in the ball.

Fix $p\in \mathrm{\partial }{B}_R$. The mass identity gives $h(x)-g(p)=\int P(x,y)(g(y)-g(p))dS(y)$. For a given $\epsilon >0$, take $\delta >0$ so that $\mathrm{\mid }g(y)-g(p)\mathrm{\mid }<\epsilon$ when $\mathrm{\mid }y-p\mathrm{\mid }<\delta$. That part of the integral has absolute value at most $\epsilon$. On the remaining sphere, if $\mathrm{\mid }x-p\mathrm{\mid }<\delta \mathrm{/}2$, then $\mathrm{\mid }x-y\mathrm{\mid }\ge \delta \mathrm{/}2$, and $P(x,y)\le (R^2-\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{/}(R{\omega }_{n-1}(\delta \mathrm{/}2{)}^n)$ uniformly tends to zero. The remaining integral is bounded by this supremum times $2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\partial }{B}_R\mathrm{\mid }$. Thus $h(x)\to g(p)$, proving continuity on the closed ball.

Any two such harmonic replacements have identical boundary data on a bounded ball and are continuous on its closure, so classical Dirichlet uniqueness makes them equal. Translation restores the stated formula at $a$.

The ball mean property and the continuous mean-value theorem make $u$ smooth. Its partial derivatives are therefore harmonic by the smooth derivative lemma, whose smoothness hypothesis is now satisfied.

On every sphere of radius $0<s<R$, the trace of $u$ is smooth. Harmonic replacement and uniqueness represent $u$ inside the smaller ball by this trace. Differentiate the kernel at its center and write $y=a+s\theta$. It gives $\mathrm{\nabla }u(a)=\frac{n}{s}\frac{1}{\mathrm{\mid }\mathrm{\partial }{B}_s\mathrm{\mid }}{\int }_{\mathrm{\partial }{B}_s(a)}u(y)\theta dS(y)$, while the undifferentiated center formula gives $u(a)=\mathrm{\mid }\mathrm{\partial }{B}_s{\mathrm{\mid }}^{-1}{\int }_{\mathrm{\partial }{B}_s(a)}u$. Derivative passage is justified by the same compact-sphere bounds as in replacement.

Taking vector norms and using $\mathrm{\mid }\theta \mathrm{\mid }=1$ bounds the gradient by $(n\mathrm{/}s)$ times the average of $\mathrm{\mid }u\mathrm{\mid }$. This is at most $nS\mathrm{/}s$ in the bounded signed case, and exactly $nu(a)\mathrm{/}s$ as an upper bound in the nonnegative case. Let $s\uparrow R$ to obtain both estimates.

For a nonnegative entire harmonic function, the positive estimate holds at each fixed $a$ for every $R>0$. Let $R\to \mathrm{\infty }$ to obtain $\mathrm{\nabla }u(a)=0$. This is also a second route to one-sided Liouville after shifting and, if necessary, negating the function.

If $u\ge b$, set $v=u-b$; if $u\le b$, set $v=b-u$. In either situation $v$ is nonnegative and entire harmonic.

Fix $x,y\in {\mathbb{R}}^n$. For every $R>\mathrm{\mid }x-y\mathrm{\mid }$, ball Harnack at center $y$ gives $v(x)\le (R\mathrm{/}(R-\mathrm{\mid }x-y\mathrm{\mid }){)}^{n}v(y)$. Let $R\to \mathrm{\infty }$ to obtain $v(x)\le v(y)$. Interchanging $x,y$ gives equality. Therefore $v$, and hence $u$, is constant.

Nonnegativity is a global lower bound by the finite real number zero.

The one-sided Liouville theorem therefore applies and yields constancy. The constant may be zero.

The ball mean property and the continuous mean-value theorem give smoothness of $u$. Each ${\mathrm{\partial }}_{i}u$ is therefore an entire harmonic function; boundedness of the gradient bounds its absolute value. Liouville makes it a constant $c_i$.

For fixed $x$, the fundamental theorem along the segment $t\mapsto tx$ gives $u(x)-u(0)={\int }_0^1\mathrm{\nabla }u(tx)\cdot xdt=c\cdot x$. Set $b=u(0)$.

Translate $p=0$ and choose $0<R<1$ so that ${\bar{B}}_R\subset \mathrm{\Omega }$. On the punctured domain, ball mean values and the smoothness theorem make $u$ smooth; its trace on $\mathrm{\partial }{B}_R$ is smooth. Let $h$ be its harmonic replacement on $B_R$, and $w=u-h$. Then $w$ is harmonic off zero and vanishes on the outer sphere. The replacement is bounded on the closed ball.