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Choose continuous representatives using Morrey's inequality. On the closed unit ball they are uniformly bounded by $CM$ and satisfyThe Arzela-Ascoli theorem therefore gives a uniformly convergent subsequence with a continuous limit. For $p=\infty$, use the uniform Lipschitz estimate instead; this part remains valid at that endpoint.

For weak lower semicontinuity of the Hilbert norm, weak convergence gives $⟨u_n-\varphi ,\varphi ⟩\to 0$, and henceTaking the lower limit proves $\boxed{\Vert \varphi \Vert \le \underset{n}{\mathrm{liminf}}\Vert u_n\Vert }$.

For the nonlinear integral, use local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Select a subsequence attaining the lower limit of the integrals. The Rellich-Kondrachov compactness theorem on successively larger balls gives a diagonal subsequence converging strongly in local L2 space to $\varphi$. After a further subsequence it converges almost everywhere. Since $1-\cos s\ge 0$, the Fatou lemma givesEach integral is finite because $0\le 1-\cos s\le s^2/2$. The argument uses local compactness rather than convexity: $1-\cos s$ is not a convex function on the whole real line.

On a bounded Lipschitz domain $\Omega \subset {\mathbb{R}}^d$, the Rellich-Kondrachov compactness theorem states that $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le p<\infty$ andwith $1/q=0$ for $q=\infty$. Thus $q<p^{\ast }$ when $p<d$, every finite $q$ is allowed when $p=d$, and the embedding into continuous functions with their uniform norm is compact when $p>d$.

For the main proof, a Sobolev extension operator puts a bounded sequence into a common compactly supported region of ${\mathbb{R}}^d$. The Sobolev fundamental theorem of calculus on lines gives the uniform translation estimate $\Vert u(\cdot +h)-u{\Vert }_p\le |h|\Vert \nabla u{\Vert }_p$. Convolution with a mollifier therefore approximates the sequence uniformly in Lp space. At any fixed smoothing scale, its derivatives and supremum are uniformly bounded, so the Arzela-Ascoli theorem gives a convergent subsequence. A diagonal choice and the uniform approximation give strong Lp space convergence. The finite measure of $\Omega$ handles $q<p$, while Lp interpolation inequality with the bounded Sobolev embeddings upgrades this to the larger subcritical finite Lp spaces; for $p>d$, Morrey's inequality gives uniform equicontinuity directly and the Arzela-Ascoli theorem gives uniform convergence. Boundedness of the domain is essential.

The Arzela-Ascoli theorem says that a uniformly bounded, equicontinuous family of continuous functions on a compact metric space is relatively compact in the uniform norm. Thus every sequence in that family has a uniformly convergent subsequence, and its limit is continuous.

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzela-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.

Use the paper's classical norm convention: $|u{|}_{2;U}$ is the derivative supremum norm through order two, not an $L^2$ norm. Write $H(u;U)=[D^{2}u{]}_{\alpha ;U}$ and $Q(u,f)=|u{|}_{2;B_1}+|f{|}_{0,\alpha ;B_1}$, where the latter is the full Hölder norm of the forcing.

Fix $0<\delta <1$. If the asserted estimate failed, rescaling the functions by their global Hessian matrix Hölder seminorm would produce $u_j,f_j$ withIn particular $Q(u_j,f_j)\to 0$. Choose $x_j,y_j\in B_{1/2}$ for whichThe supremum norm of $D^2{u}_j$ tends to zero, so $r_j=|y_j-x_j|\to 0$. Let $P_j$ be the quadratic Taylor polynomial of $u_j$ at $x_j$, and defineThe domains contain $B_{1/(2r_j)}$, and $U_j,D{U}_j,D^2{U}_j$ vanish at zero. MoreoverThe right side is bounded by $[f_j{]}_{\alpha }|z{|}^{\alpha }$ and tends to zero locally uniformly. The normalized Hessian matrix bound controls $|D^2{U}_j(z)|\le |z{|}^{\alpha }$; integration along line segments then bounds $D{U}_j,U_j$ on every compact Euclidean ball. Arzela-Ascoli theorem and a diagonal subsequence argument give convergence in $C^2$ on compact subsets to an entire harmonic function $U$, with $[D^{2}U{]}_{\alpha ;{\mathbb{R}}^n}\le 1$ and $D^{2}U(0)=0$.

After a further subsequence $z_j=(y_j-x_j)/r_j\to z_{\ast }$ with $|z_{\ast }|=1$. The normalization ensures $|D^{2}U(z_{\ast })|\ge \delta /2$. But each second derivative of $U$ is harmonic and has finite global Hölder seminorm with exponent below one. The supplied Liouville theorem makes each derivative constant; it is the globally Hölder case of the polynomial-growth Liouville theorem for harmonic functions. Its value at zero makes it zero, a contradiction. This blow-up compactness proof of an interior Schauder estimate establishes

To absorb the term on the larger Euclidean ball, rescale this estimate to Euclidean balls contained in $B_s$. For $1/2\le r<s<1$, pairs separated by less than $(s-r)/2$ are controlled in a Euclidean ball centred at their first point; farther pairs are controlled directly by $\Vert D^{2}u{\Vert }_{\infty }$. ThusThis is the covering step in the Simon absorption lemma. Take $r_j=1-2^{-j-1}$ and fix $\delta <2^{-2-\alpha }$. Iterating makes the remainder ${\delta }^{j}H(u;B_{r_j})$ tend to zero, since the global seminorm is finite; the error terms form a convergent geometric series. Consequently

For the requested failure of the weaker estimate, take $u_m=m^{-2}\sin (m{x}_1)$ and $f_m=-\sin (m{x}_1)$. The derivative supremum norm $|u_m{|}_2$ and $\Vert f_m{\Vert }_{\infty }$ remain bounded, but comparison at $0$ and $\pi e_1/(2m)$, both in $B_{1/2}$ for large $m$, givesThis proves the necessity of Hölder forcing for Schauder estimates. Every member of the sequence is smooth; failure comes from the increasing frequency, not from any lack of individual regularity.

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzela-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.

The Arzela-Ascoli theorem says that a subset $\mathcal{F}\subseteq C(K)$ is relatively compact in the uniform norm exactly when it is equicontinuous and pointwise bounded.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subseteq C(K,\mathbb{R})$ is uniformly dense if it contains the constants and separates points of $K$. In the complex version, a complex subalgebra $A\subseteq C(K,\mathbb{C})$ must additionally be closed under complex conjugation.

For an example, take $K=[0,1]$ andThis family is bounded in the uniform norm, but it is not equicontinuous at $1$. Equivalently, every subsequence converges pointwise to the discontinuous function that is zero on $[0,1)$ and one at $1$, so no subsequence converges uniformly. Thus $\mathcal{F}$ is bounded but not relatively compact.

Let $K=\overline{T(B_X)}$, which is compact. Every $y^{\ast }$ in the unit ball of the continuous dual space $Y^{\ast }$ restricts to a function on $K$. These functions are uniformly bounded and equicontinuous, sinceBy the Arzela-Ascoli theorem, every sequence $(y_n^{\ast })$ in the unit ball has a subsequence that converges uniformly on $K$. Along this subsequence,The continuous dual space $X^{\ast }$ is complete, so $(T^{\ast }{y}_n^{\ast })$ converges in norm. Every sequence in the image of the unit ball therefore has a convergent subsequence, proving the forward direction of the Schauder theorem for compact operators:

For $F_1$, the functions satisfy $\Vert \sin (\pi nx)/n{\Vert }_{\infty }\le 1/n$, so the sequence converges uniformly to zero. Its closure is the sequence together with its limit and is compact. The same argument applies to $F_2$, since $\Vert \sin (\pi nx)/\sqrt{n}{\Vert }_{\infty }\le n^{-1/2}$. Thus $F_1$ and $F_2$ are relatively compact.

The family $F_3$ is not equicontinuous. At $x=0$ its members vanish, but for $x_n=1/(2n)\to 0$,By the Arzela-Ascoli theorem, a relatively compact subset of $C[0,1]$ must be equicontinuous. Hence $F_3$ is not relatively compact.

A family $\mathcal{S}\subseteq C(K)$ is equicontinuous if for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such thatfor every $y\in U$ and every $f\in \mathcal{S}$. The Arzela-Ascoli theorem says that, for a compact Hausdorff space $K$, a subset of $C(K)$ has compact closure in the uniform norm if and only if it is equicontinuous and pointwise bounded. On compact $K$, equicontinuity and pointwise boundedness together imply uniform boundedness.

For necessity, a compact closure is totally bounded. Given $\epsilon >0$, choose a finite $\epsilon /3$-net $f_1,\dots ,f_m$. The finitely many continuous functions $f_j$ are simultaneously continuous near each point, and comparison with a nearby $f_j$ proves equicontinuity of the whole family. Evaluation at any fixed point is a continuous map $C(K)\to \mathbb{R}$ or $\mathbb{C}$, so compactness also gives pointwise boundedness.

Conversely, equicontinuity gives, around each $x\in K$, a neighborhood on which every $f\in \mathcal{S}$ oscillates by less than $\epsilon /3$. Compactness supplies a finite such cover with selected points $x_1,\dots ,x_m$. Pointwise boundedness makes the set of vectorsa bounded subset of a finite-dimensional normed vector space, so it has a finite $\epsilon /3$-net. Choose one function of $\mathcal{S}$ for every nonempty cell of this net. If two functions have evaluation vectors in the same cell, then comparison at a selected $x_j$ and the two oscillation bounds show that their uniform distance is less than $\epsilon$. Thus $\mathcal{S}$ is totally bounded. Since $C(K)$ is a Banach space, its closure is complete; a complete totally bounded metric space is compact, proving the theorem.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzela-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzela-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

Apply Arzela-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

Fix $B_r⋐B_R⋐B$. Local uniform convergence bounds $|u_k{|}_{0;B_R}$ uniformly. Interior Schauder estimates therefore bound $u_k$ uniformly in $C^{2,\alpha }(B_r)$. The Arzela-Ascoli theorem gives a subsequence converging in $C^2(B_r)$. Its limit must be the locally uniform limit $v$, and passing to the limit in $L{u}_k=0$ gives $Lv=0$. A diagonal argument over $r\uparrow 1$ proves

The sequence $(f_n)$ is equicontinuous on $[0,1]$ when, for every $\epsilon >0$, there is a $\delta >0$ such thatfor every $n$. The Arzela-Ascoli theorem says that a uniformly bounded equicontinuous sequence of continuous real-valued functions on a compact interval has a uniformly convergent subsequence.

The Arzela-Ascoli theorem makes the family of iterates relatively compact in $C([0,T_1])$. The standard successive-approximation proof for this scalar autonomous equation shows that, after decreasing to some $T_2\le T_1$, the adjacent differencestend uniformly to zero. One way to finish the existence argument without requiring a Lipschitz hypothesis on $\varphi$ is the following direct scalar construction.

If $\varphi (0)=0$, the function $f(t)=0$ is already a solution. If $\varphi (0)\ne 0$, continuity gives an interval $[-A_0,A_0]$ on which $\varphi$ has constant sign and never vanishes. DefineThen $F$ is continuously differentiable with $F^{\prime }(0)\ne 0$. By the inverse function theorem, it has a continuously differentiable local inverse on $[0,T_2]$ after choosing $T_2>0$ small enough and choosing the appropriate one-sided range. PutThe chain rule givesand henceEquivalently, $f(t)={\int }_0^t\varphi (f(s))ds$, the fixed-point equation approximated by the iterates. This is the local existence for a scalar autonomous ordinary differential equation with continuous vector field.

On a relatively compact ball, mollification commutes with the Laplacian, so $u_{\epsilon }=u\ast {\eta }_{\epsilon }$ is a smooth harmonic function. Repeated interior derivative estimates, together with the Caccioppoli inequality, bound every derivative of $u_{\epsilon }$ on a smaller ball by the local $L^2$ norm of $u$, uniformly as $\epsilon \downarrow 0$. The Arzela-Ascoli theorem and a diagonal argument give a smooth local limit, while mollification gives $u_{\epsilon }\to u$ in $L_{\mathrm{loc}}^2$. Thus the limit equals $u$ almost everywhere. After choosing this smooth representative, $u$ is harmonic pointwise. This is the Weyl lemma for an $H^1$ weak solution.

Fix $B⋐\Omega$ and $x_0\in B$. Choose $u_j$ in the Perron family with $u_j(x_0)\uparrow \overline{u}(x_0)$, replace successive terms by finite maxima using part (i), and take their harmonic lifts on $B$. The lifts remain in the family, are increasing, and are uniformly bounded. Interior estimates and the Arzela-Ascoli theorem give a harmonic limit $h$ on $B$ with $h\le \overline{u}$ and $h(x_0)=\overline{u}(x_0)$.

If $h(y)<\overline{u}(y)$ somewhere in $B$, take another family member larger than $h(y)$ and repeat the maximum-and-lift construction. Its harmonic limit $H$ satisfies $H\ge h$ and $H(x_0)=h(x_0)$. The strong minimum principle for elliptic operators forces $H=h$, contradicting the strict inequality at $y$. Thus $h=\overline{u}$ on $B$. Since $B$ was arbitrary, $\overline{u}$ is smooth and harmonic in $\Omega$. This is the Perron method for the Dirichlet problem.

The interior Schauder estimate iswhere $C$ depends only on $n$, $\alpha$, the ellipticity constant, and the stated coefficient bounds.

It is enough first to prove the estimate for the Hölder seminorm of $D^{2}u$. The Interpolation inequality in Holder spaces controls $Du$ and $D^{2}u$ by a small multiple of $[D^{2}u{]}_{\alpha }$ plus $\Vert u{\Vert }_0$; the small term is later absorbed. Freezing $a^{ij}$ at the centre of each ball and moving coefficient differences and lower-order terms to the right reduces the local estimate toOnce this is known, the Simon absorption lemma gives the desired estimate on $B_{1/2}$.

For completeness, prove the frozen-coefficient estimate by contradiction. If it failed, choose solutions $u_k$, points $x_k$, and scales ${\rho }_k\downarrow 0$ at which the scale-invariant Hölder quotient is almost maximal. Let $q_k$ be the quadratic Taylor polynomial of $u_k$ at $x_k$ and defineThen $v_k(0)=D{v}_k(0)=D^2{v}_k(0)=0$, the Hessians have uniformly bounded local $C^{0,\alpha }$ seminorms, and the normalization makes their oscillation nonzero on a fixed ball. The localized equation iswhere the coefficient matrices converge locally uniformly to one constant positive-definite matrix and the given normalized error $g_k$ tends locally uniformly to zero.

The Arzela-Ascoli theorem produces a locally $C^2$ convergent subsequence with limit $v$ satisfying a constant-coefficient elliptic equation on ${\mathbb{R}}^n$. A linear rotation and scaling turn it into a harmonic function. The normalization gives $D^{2}v(0)=0$ but a nonconstant Hessian, while maximality of the scaled quotient gives growth at most $C(1+|y{|}^{2+\alpha })$. Applying the Liouville theorem to derivatives shows that every second derivative is constant because $0<\alpha <1$. This contradicts the normalized Hessian oscillation. The frozen estimate follows, and interpolation plus absorption completes the proof.