The Fredholm alternative for an elliptic Dirichlet problem and part (b) give an inverse $S:Y\to X$ of norm at most $K\ge 1$. ChooseOn the closed radius-$\epsilon$ ball $\mathcal{B}\subset X$, set $T(v)=S(\mathcal{Q}(v)+f)$. The stated inequalities giveandThus $T$ is a contraction mapping of the complete metric space $\mathcal{B}$ into itself. The Banach fixed-point theorem gives $u=T(u)$, solving the required nonlinear elliptic boundary value problem. This small-data existence for a nonlinear elliptic Dirichlet problem also gives uniqueness within this small ball; it does not assert global uniqueness.

Failure of the bound would give normalized $v_j$ with $\Vert v_j{\Vert }_{\infty }=1$, zero Dirichlet boundary condition, and $\Vert L{v}_j{\Vert }_{C^{0,\alpha }}\to 0$. The global Schauder estimate bounds them in $C^{2,\alpha }$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$ to $v$ with $Lv=0$, zero boundary data and $\Vert v{\Vert }_{\infty }=1$. The uniform Hölder seminorm bounds pass to the limit, so $v\in C^{2,\alpha }$. This contradicts the trivial null space, provingThe global Schauder estimate then gives $\Vert u{\Vert }_{C^{2,\alpha }}\le K\Vert f{\Vert }_{C^{0,\alpha }}$, the bounded-inverse estimate used next.

Interpret strict ellipticity as uniform ellipticity up to the boundary, as required for the global elliptic boundary value problem. Mere pointwise positive definiteness in the interior is insufficient: $L=x^2{D}^2$ on $(0,1)$ has trivial homogeneous Dirichlet kernel but $Lu=1$ has no $C^2$ solution up to $x=0$.

Set $X=\{u\in C^{2,\alpha }(\overline{\Omega }):u{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The Fredholm alternative for an elliptic Dirichlet problem says $L:X\to Y$ is a Fredholm operator of index zero. Its null space $N$ is finite dimensional, its range is closed, and the range codimension equals $\dim N$. Hence $\boxed{N=\{0\}⟺L:X\to Y\text{ is invertible}}$; in this case every forcing has a unique solution.

Otherwise there are nonzero homogeneous solutions and finitely many compatibility conditions: $Lu=f$ is solvable exactly when $\ell (f)=0$ for every continuous linear functional on $Y$ vanishing on $L(X)$. When solvable, its solutions are $u_0+N$. This Fredholm solvability condition can be expressed with formal-adjoint null solutions for smoother coefficients. The functional formulation avoids differentiating the merely $C^{0,\alpha }$ coefficients.

Take ${\Omega }^{\prime }$ nonempty. If the bound failed, normalization would give nonnegative $v_j$ with supremum one, zero boundary values, ${\inf }_{{\Omega }^{\prime }}{v}_j\to 0$, and $\Vert L{v}_j{\Vert }_{C^{0,\alpha }}\to 0$. The global Schauder estimate bounds them in the Hölder space $C^{2,\alpha }$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$, $0<\beta <\alpha$, to $v\ge 0$ with $Lv=0$, zero boundary data and supremum one.

Points approaching the infima have a subsequence converging in $\overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence gives an interior zero of $v$, and part (c), the strong minimum principle for elliptic operators after replacing $c$ by $\min (c,0)$, gives $v=0$. This contradicts its supremum one, proving

Nonnegativity lets us dispense with a sign restriction on $c$. Set $c_0=\min (c,0)$ and $L_0=a^{ij}{D}_{ij}+b^i{D}_i+c_0$. ThenThe strong minimum principle for elliptic operators forces $u$ to vanish identically if it has an interior zero. Otherwise it is strictly positive:Connectedness is part of the usual domain hypothesis; on a disconnected open set this conclusion applies separately to each component.

For zero Dirichlet boundary condition, the global Schauder estimate isThe constant depends on $n,\alpha$, domain regularity, the ellipticity lower bound, and coefficient $C^{0,\alpha }$ norms. No sign restriction on $c$ is needed. Nonzero boundary data contribute the $C^{2,\alpha }$ norm of an extension.

Assume $c\le 0$, bounded coefficients and uniform ellipticity near the boundary point, an interior tangent ball there, and enough regularity for the boundary normal derivative. The Hopf boundary point lemma says that a nonnegative maximum at $x_0$ with strictly smaller values inside the tangent ball satisfiesfor the outward unit normal. If $c=0$, the maximum can have either sign. Without existence of a classical derivative, one states a strict one-sided derivative bound.

The PDF also asks to deduce the strong maximum principle for elliptic operators; that request is missing from the TeX transcription. Suppose $v$ attains a nonnegative global maximum $M$ inside a connected domain but is not constant. A ball in $\{v<M\}$ can be chosen relatively compact in the domain and tangent to $\{v=M\}$ at an interior point $z$: use the distance from a nearby point to the closed maximum set. The Hopf boundary point lemma on that ball gives a nonzero outward derivative at $z$, contradicting $Dv(z)=0$ at an interior maximum. Therefore $v$ is constant. When $c=0$, the same proof works for a maximum of either sign.

The cone condition implies $u(\lambda x)=\lambda u(x)$, so $Du(\lambda x)=Du(x)$. The gradient is homogeneous of degree zero, and $v=|Du{|}^2$ attains its global maximum on the compact unit sphere. For $n\ge 2$ the punctured space is connected. Part (b) gives $\mathcal{L}v\ge 0$ for a locally uniformly elliptic operator with no zeroth-order term. The strong maximum principle for elliptic operators makes $v$ constant. HencePositive definiteness gives $D^{2}u=0$. Thus $u=a\cdot x+b$, and degree-one homogeneity forces $b=0$: $\boxed{u(x)=a\cdot x}$. This proves that a minimal graphical cone is a plane. Strictly, the graph restricted to $x\ne 0$ is the plane with its origin omitted; its closure is the full plane through the origin.

Integration by parts gives the minimal surface equation for a graph in nondivergence form:Its eigenvalues are between $(1+|p{|}^2{)}^{-1}$ and one. On each relatively compact subdomain, $Du$ is bounded, so it is a uniformly elliptic operator. Since $u\in C^2$, its coefficient functions are locally Lipschitz. The interior Schauder estimate gives $C_{\mathrm{loc}}^{2,\beta }$ regularity for every $0<\beta <1$; differentiating and repeatedly applying regularity estimates gives $\boxed{u\in C^{\infty }(\Omega )}$.

Differentiate in direction $k$. Each $u_k$ satisfies $\mathcal{L}{u}_k=0$, withFor $v=|Du{|}^2$ this yieldsIf $u\in C^2(\overline{\Omega })$, the principal coefficients are uniformly elliptic globally and the drift is bounded. The weak maximum principle for elliptic operators bounds $v$ by its boundary maximum. Boundary values are limits of interior values by continuity, proving the gradient maximum principle for a minimal graph:

Assume uniform ellipticity $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ with $\lambda >0$, bounded coefficients, and $c\le 0$. The weak maximum principle for elliptic operators isWhen $c=0$, zero can be omitted. The sign restriction is necessary, as shown by the failure of the weak maximum principle with a positive zeroth-order coefficient.

Let $M=\max (0,{\sup }_{\partial \Omega }u)$ and $w=u-M$, so $Lw=Lu-cM\ge 0$ and $w\le 0$ on the boundary. For sufficiently large $k$, $q=e^{k{x}_1}$ satisfiesIf $w$ is positive somewhere, $w+\epsilon q$ retains a positive interior maximum above all its boundary values for small $\epsilon >0$. At that point its gradient is zero and its Hessian matrix is negative semidefinite; ellipticity and $c\le 0$ give $L(w+\epsilon q)\le 0$, contradicting its strict positivity. This exponential perturbation proof of the weak maximum principle with drift also works with $M={\sup }_{\partial \Omega }u$ of either sign when $c=0$.

The conclusion holds in dimension two. Shift and change sign to obtain a nonnegative harmonic function $v$. Its circular mean satisfiesHere the angular second derivative in the polar-coordinate Laplacian integrates to zero. Thus $m(r)=a\log r+b$; nonnegativity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions and nonnegativity giveThe removable singularity for a bounded harmonic function fills in the origin, and the harmonic Liouville theorem makes the extension constant.

The conclusion fails for $n\ge 3$: $\boxed{u(x)=|x{|}^{2-n}}$ is positive and nonconstant, and its radial Laplacian $u^{\prime \prime }+(n-1)u^{\prime }/r$ vanishes away from zero. For $n=1$, $\boxed{u(x)=|x|}$ is a nonconstant function bounded below whose second derivative vanishes on both components of the punctured line.

Shift $u$ by a constant and, if necessary, change sign to obtain an entire nonnegative harmonic function $v$. The Harnack inequality for harmonic functions on arbitrarily large balls centred at $x_0$ gives $v(x)\le 3^{n}v(x_0)$ for every $x$. Thus $v$ is globally bounded. The interior derivative estimate for a harmonic function now gives $|Dv(x)|\le C\Vert v{\Vert }_{\infty }/R$ on balls of arbitrarily large radius $R$. Letting $R\to \infty$ proves $Dv=0$, hence $u$ is constant. This proves the one-sided harmonic Liouville theorem in every dimension.

For $x,z\in B_{\rho }(y)$, the open balls satisfy $B_{\rho }(x)\subset B_{3\rho }(z)\subset B_{4\rho }(y)$. Nonnegativity and the mean value property for harmonic functions giveTaking the supremum over $x$ and the infimum over $z$ proves the Harnack inequality for harmonic functions:This includes a zero infimum, which forces $u=0$ on the smaller ball.

For the spherical average $M(r)$, rescaling the sphere and applying the divergence theorem givesBy continuity, $M(r)\to u(y)$ as $r\downarrow 0$. Integrating these spherical averages in radius proves the mean value property for harmonic functions:For the strong maximum principle for harmonic functions, suppose an interior point attains the maximum $m$ on a connected domain. The continuous nonnegative function $m-u$ has zero average on each sufficiently small ball around that point, so it vanishes there. The maximum set is therefore open as well as closed, and connectedness makes it the entire domain. Apply the same argument to $-u$ for a minimum.

The Beurling–Gelfand spectral radius formula states that for every element $a$ of a complex Banach algebra,For a Hermitian element of a C-star algebra $a=a^{\ast }$, the C-star identity gives $\Vert a^2\Vert =\Vert a{\Vert }^2$. Iterating,The subsequence $n=2^m$ in the spectral-radius limit therefore proves $\boxed{r(a)=\Vert a\Vert }$.

Let $\theta :A\to B$ be a unital C-star homomorphism. If $a-\lambda 1$ is invertible in $A$, its inverse is sent to an inverse of $\theta (a)-\lambda 1$ in $B$. Thus ${\sigma }_B(\theta (a))\subseteq {\sigma }_A(a)$ without assuming continuity of $\theta$. For Hermitian $a$, $\theta (a)$ is also Hermitian, soFor arbitrary $x\in A$, apply this inequality to $x^{\ast }x$ and the C-star identity in both algebras:Hence $\boxed{\Vert \theta (x)\Vert \le \Vert x\Vert }$ for every $x$.

Now assume $\theta$ is injective. For a fixed $x$, it is enough to prove norm preservation on $C=C^{\ast }(1,x^{\ast }x)$, a commutative unital C-star algebra: the preceding square-norm identity will then recover the norm of $x$. By the Commutative Gelfand--Naimark theorem, identify $C$ with $C(K)$. Let $D=\overline{\theta (C)}\subseteq B$. It is also a commutative unital C-star algebra, and its character space ${\Phi }_D$ is compact. Each $\chi \in {\Phi }_D$ restricts along $\theta$ to a character of $C(K)$, hence to evaluation at a point $q(\chi )\in K$. The map $q:{\Phi }_D\to K$ is continuous, so $L=q({\Phi }_D)$ is closed.

If $L\ne K$, the Urysohn lemma provides a nonzero $f\in C(K)$ vanishing on $L$. Then $\chi (\theta (f))=f(q(\chi ))=0$ for every $\chi \in {\Phi }_D$. Faithfulness of the Gelfand transform in $D$ gives $\theta (f)=0$, contradicting injectivity. Thus $q$ is onto. The isometric Gelfand transform of $D$ now givesFinally,This proves that an injective C-star homomorphism is isometric without assuming its image is closed before norm preservation has been established.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.

By the Riesz-Markov-Kakutani representation theorem, the continuous dual space of $C(K)$ is isometrically the space of finite regular Borel measures, signed for real scalars and complex for complex scalars:Here $|\mu |$ is the variation measure, whose total mass is the total variation norm of a measure.

If $f_n\rightharpoonup 0$ in $C(K)$, evaluation at each $t\in K$ gives $f_n(t)\to 0$. The Uniform boundedness principle also gives $M={\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$. For any $\mu \in M(K)$, the dominated convergence theorem with respect to the finite positive measure $|\mu |$ yieldssince $|f_n(t){|}^2\to 0$ pointwise and $|f_n(t){|}^2\le M^2$. ThusThe same proof works for complex squares, because the absolute-value domination remains valid.

The Hahn-Banach separation theorem for two convex sets states that disjoint nonempty convex sets $A,B$ in a real locally convex space, with $A$ open, admit a nonzero continuous linear functional $\ell$ and $\alpha \in \mathbb{R}$ such thatThere need not be a positive uniform gap between the two sets.

For a proof, set $C=A-B$. It is open and convex, and $0\notin C$. Choose $c_0\in C$, let $V=C-c_0$, and let $p$ be the Minkowski functional of $V$. Because $V$ is open, convex and contains zero, $p$ is a finite sublinear functional, and $V=\{x:p(x)<1\}$. Put $z=-c_0$. Since $z\notin V$, $p(z)\ge 1$. On $\mathbb{R}z$ define ${\ell }_0(tz)=tp(z)$. For $t\ge 0$ this equals $p(tz)$; for $t<0$ it is nonpositive and hence at most $p(tz)$. The Hahn-Banach theorem extends it to a linear $\ell$ with $\ell \le p$. Both $p(x)$ and $p(-x)$ are bounded near zero, so this domination makes $\ell$ continuous; it is nonzero because $\ell (z)=p(z)\ge 1$. For $c\in C$,Thus $\ell (a)<\ell (b)$ for every $a\in A,b\in B$. Define $\alpha ={\sup }_{a\in A}\ell (a)$, which is finite because any fixed $b\in B$ bounds it above. Openness of $A$ and nonzeroness of $\ell$ ensure $\ell (a)<\alpha$ for every $a\in A$, and $\alpha \le \ell (b)$ for every $b\in B$. This proves the stated separation.

For the sequence criterion, suppose $f\in S_{X^{\ast }}$ and $f(x_i)\ge \varrho$. If each $t_i\ge 0$, thenConversely, the assumed inequality implies that every convex combination of the $x_i$ has norm at least $\varrho$. Hence $D=\{x:\Vert x\Vert <\varrho \}$ is disjoint from $C=\mathrm{conv}\{x_i:i\ge 1\}$. Apply the separation theorem with the open set $D$ to obtain $\ell \ne 0$ andThen $f=\ell /\Vert \ell \Vert$ has norm one and satisfies $f(x_i)\ge \varrho$ for every $i$. ThereforeThe PDF confirms the “if and only if” that is garbled in the TeX transcription.