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A compactness proof of the interior Schauder estimate can magnify a pair of points where the Hölder seminorm of the Hessian matrix is nearly attained. If the proposed estimate fails, normalize that seminorm to one while lower norms and forcing tend to zero. The separation $r_j$ of the chosen pair tends to zero. Subtract the quadratic Taylor polynomial and divide by $r_j^{2+\alpha }$ after rescaling space by $r_j$.

The resulting functions have uniformly bounded Hessian Hölder seminorms on expanding domains and vanishing Taylor data at zero. Compact embedding of Hölder spaces yields an entire harmonic function whose second derivatives have finite global $\alpha$-Hölder seminorms, with $0<\alpha <1$. The polynomial-growth Liouville theorem for harmonic functions makes those derivatives constant, contradicting the normalized nonzero oscillation. The Simon absorption lemma removes the residual small multiple of the larger-ball seminorm.

For a periodic Hölder continuous function of Hölder exponent $0<\alpha <1$ and Hölder seminorm at most $M$, the degree-$n-1$ Fejér sum has the displayed supremum norm error bound. In the half-kernel normalization $F_{n-1}=(2n{)}^{-1}{\sin }^2(nt/2)/{\sin }^2(t/2)$, use $F_{n-1}(t)\le \min \{n/2,{\pi }^2/(2n{t}^2)\}$. Splitting the convolution error integral at $1/n$ gives the valid constant $C_{\alpha }=1/[\pi (\alpha +1)]+\pi /(1-\alpha )$. At the endpoint $\alpha =1$ this argument instead gives $O(M\log n/n)$.

The full norm is $|u{|}_{k,\alpha ;U}=|u{|}_{k;U}+[D^{k}u{]}_{\alpha ;U}$, combining the derivative supremum norm with the top-derivative Hölder seminorm. At order zero it is the sum of the supremum norm and the Hölder seminorm.

Write $\alpha =r+\beta$ with an integer $r\ge 0$ and $0<\beta \le 1$, using $r=\alpha -1$ at integer $\alpha$. For $f\in C^{r,\beta }$, subtract its degree-$r$ Taylor polynomial at $x_0$. For $r\ge 1$, the integral Taylor remainder with $f^{(r)}(t)-f^{(r)}(x_0)$ is bounded using the Hölder seminorm; for $r=0$ this is exactly Hölder continuity. The resulting bound is the displayed estimate. At integer order, the convention is $C^{r,1}$; a $C^{r+1}$ function on a compact closed interval also obeys this estimate.

A harmonic function on all of ${\mathbb{R}}^n$ with a finite global Hölder seminorm of exponent $0<\mu <1$ is constant. Indeed $|w(x)-w(0)|\le C|x{|}^{\mu }$, and the interior derivative estimate for a harmonic function on balls of radius $R$ bounds $|\nabla w(0)|$ by $C{R}^{\mu -1}$. Letting $R\to \infty$ makes this gradient zero, and translating the centre proves the assertion everywhere.

If $f_j\to f$ uniformly and the Hölder seminorms are uniformly bounded, each fixed difference quotient converges. Taking the supremum over point pairs gives $[f{]}_{\mu }\le {\mathrm{liminf}}_j[f_j{]}_{\mu }$. The supremum norms converge, so the full Hölder norm has the same lower semicontinuity of a Hölder seminorm.

For smooth non-increasing $V$ and an ordered subsolution and supersolution, begin with $u_0={\phi }^{-}$ and solve successive Poisson equations with forcing $V(u_k)$ and the fixed boundary data. The weak maximum principle for elliptic operators applied to the negatives of consecutive differences givesThe global Schauder estimate and the Hölder interpolation inequality bound the iterates uniformly in $C^{2,\alpha }$. Their bounded monotone sequence converges, and compact embedding of Hölder spaces permits passing to the equation. The uniform Hölder seminorm bounds retain $C^{2,\alpha }$ regularity of the limit.

The bounded-forcing norm alone cannot control the Hölder seminorm of the Hessian matrix. On a unit Euclidean ball take $u_m=m^{-2}\sin (m{x}_1)$ and $f_m=\Delta u_m=-\sin (m{x}_1)$. The derivative supremum norm $|u_m{|}_2$ and $\Vert f_m{\Vert }_{\infty }$ stay bounded, but the Hessian matrix difference between zero and $\pi /(2m)e_1$ gives $[D^2{u}_m{]}_{\alpha }\ge (2m/\pi {)}^{\alpha }$ on $B_{1/2}$.

Periodicity and the Hölder condition give $|f(x-t)-f(x)|\le M|t{|}^{\alpha }$ for $|t|\le \pi$. The positive normalized Fejér kernel therefore yieldsThe exponential-sum representation of the kernel bounds $F_{n-1}(t)\le n/2$. Also $\sin (t/2)\ge t/\pi$ for $0\le t\le \pi$, so $F_{n-1}(t)\le {\pi }^2/(2n{t}^2)$. Splitting at $1/n$ and using $0<\alpha <1$ givesHence the Hölder error bound for Fejér summation isThe constant depends on the Hölder exponent and $M$, but not on $n$ or $f$ beyond its Hölder seminorm bound.

First construct a centred Gaussian process on the countable dyadic set with covariance function $\mathbb{E}(B_s{B}_t)=\min (s,t)$. These consistent finite-dimensional normal distributions exist because, for coefficients $a_j$,the Kolmogorov extension theorem realizes them on one probability space. Every increment has the normal distribution $N(0,|t-s|)$, so for every finite $p\ge 1$,Use the preceding estimate with $\beta =1/2$, $p>2$ and $0<\alpha <1/2-1/p$. It yields an almost surely Hölder continuous extension from the dense dyadic set to $[0,1]$. Approximate arbitrary times by dyadic times; continuity of the paths and of the Gaussian covariance function preserves all finite-dimensional laws. Disjoint increments are jointly Gaussian with zero cross-covariances, hence independent, with the required variances. Together with $B_0=0$ and continuity, this constructs Brownian motion.

For any $\alpha <1/2$, choose $p$ large enough. Taking a countable increasing sequence of exponents tending to $1/2$ gives simultaneous Hölder regularity of every order $\alpha <1/2$ on a common probability-one set; the same argument on each compact time interval gives local Brownian Hölder regularity. This method does not give the endpoint exponent. In fact that endpoint fails: if a path had a finite $1/2$-Hölder seminorm $K$, all its $2^n$ adjacent normalized increments at level $n$ would have magnitude at most $K$. These are independent standard normal random variables, so for a fixed integer $m$ the probability that their maximum is at most $m$ is $\mathbb{P}(|Z|\le m{)}^{2^n}\to 0$. The event of a finite $K$ is contained in the union over $m$ of events with this bound at every level, each of probability zero. Thus Brownian paths are not $1/2$-Hölder continuous on $[0,1]$ almost surely.

For a smooth approximation, the fundamental theorem of calculus and Cauchy-Schwarz inequality giveLet $E\subset I$ be one full-measure set on which $u_n\to u$. Passing to the limit for all $x,y\in E$ givesThe full-measure set $E$ is dense, and the estimate extends its representative uniquely to a $1/2$-Hölder continuous function on $\overline{I}$. Its classical Hölder seminorm satisfies the same bound. Changing the original representative on a null set does not affect the essential supremum in the assertion.

With the standard Sobolev norm, the printed constant one is not valid for arbitrary $a$. The constant function $u=1$ has supremum norm one and $H^1$ norm $\sqrt{2a}$, which is smaller when $a<1/2$.

For the continuous one-dimensional Sobolev representative, let $\bar{u}=(2a{)}^{-1}{\int }_{I}u$. The Cauchy-Schwarz inequality gives $|\bar{u}|\le (2a{)}^{-1/2}\Vert u{\Vert }_2$. Using the preceding Hölder seminorm estimate,Consequently the corrected interval Sobolev supremum estimate isOne may first prove this for smooth approximations and pass on the full-measure convergence set. Averaging differences also handles complex-valued functions without assuming that a function takes its complex average at some point. At the fixed interval used below the constant is absolute, which suffices for the intended interior estimate.

Choose a fixed smooth cutoff function with $\Vert {\zeta }^{\prime }{\Vert }_{\infty }=C_0$ and set $J=(-1/2,1/2)$. The Caccioppoli inequality gives $\Vert u^{\prime }{\Vert }_{L^2(J)}\le 2R{C}_0\Vert u{\Vert }_{L^2(-1,1)}$. Since the length of $J$ is one, the corrected interval Sobolev supremum estimate yields $\Vert u{\Vert }_{L^{\infty }(J)}\le \Vert u{\Vert }_{L^2(J)}+\Vert u^{\prime }{\Vert }_{L^2(J)}$, and the Hölder seminorm is at most $\Vert u^{\prime }{\Vert }_{L^2(J)}$. ThereforeFor locally $H^1$ weak solutions with $u\in L^2(-1,1)$, density of smooth functions in a Sobolev space justifies the test ${\zeta }^{2}u$, and the previous representative estimates already apply to $H^1$. Mollifying a solution need not preserve the equation with the same measurable coefficient, so approximation is used for admissible tests and Sobolev estimates, not to assert that the mollified function solves the original equation. The required constant depends only on $R$.

Put $\alpha =1-n/q>0$ andFirst suppose $A<\infty$. For an interior ball $B_R$, with $R\le 1$, write $M_R={\mathrm{esssup}}_{B_R}u$, $m_R={\mathrm{essinf}}_{B_R}u$, and $\omega (R)=M_R-m_R$. Constant shifts of a divergence-form elliptic operator change its forcing. The two nonnegative weak supersolutions $v=M_R-u$ and $w=u-m_R$ satisfy, respectively,Their Weak Harnack inequalities on $B_{R/2}$ therefore have forcing errors bounded by $CN{R}^{\alpha }$. Indeed $|M_R|,|m_R|\le A$, the new divergence forcing is bounded by $\Vert F{\Vert }_q+A\Vert b{\Vert }_q$, and the scalar forcing by $\Vert g{\Vert }_{q/2}+A\Vert d{\Vert }_{q/2}$; also $R^{2\alpha }\le R^{\alpha }$.

On $B_{R/2}$ we have $v+w=\omega (R)$. Even if $p<1$, the elementary inequality $(s+t{)}^p\le C_p(s^p+t^p)$ impliesThe sum of the two essential infima is $\omega (R)-\omega (R/2)$. Enlarging a uniform constant $C_0>1$ giveswhere $\theta =1-C_0^{-1}\in (0,1)$. This is an oscillation decay estimate; inhomogeneous oscillation decay implies Hölder continuity. ChooseIterating on concentric balls, starting at a fixed admissible radius $R_{\ast }\le 1$, yieldsThe strict choice of $\mu$ bounds the geometric series, including the case in which the two original decay rates coincide. Monotonicity of oscillation extends this bound to intermediate radii.

Take $R_{\ast }$ uniformly smaller than the distance of ${\Omega }^{\prime }$ from the boundary of $\Omega$. The vanishing essential oscillation produces a continuous representative of $u$. A ball centred at either of two sufficiently close points of ${\Omega }^{\prime }$ contains both, so its oscillation bounds their difference by $CN|x-y{|}^{\mu }$. For separated points use $|u(x)-u(y)|\le 2A$. Thus the Hölder seminorm satisfiesThe constants depend only on the indicated domain separation, dimension, coefficient norms, uniform ellipticity bounds and $q$. If the global essential supremum is infinite, the displayed global estimate is vacuous; local boundedness of weak elliptic subsolutions on slightly larger compact subsets still makes the same argument prove interior Hölder continuity.

We use a blow-up compactness proof of an interior Schauder estimate, with Taylor normalization in elliptic blow-up arguments. Fix $n,\mu ,\delta$ and suppose no suitable constant exists. There would be solutions $u_j$ with forcing $f_j$ for whichChoose distinct $x_j,y_j\in B_{1/2}$ such that, for $r_j=|y_j-x_j|$,Then $S_j/A_j<2/\delta$, $N_j/A_j<2/j$, andIn particular $r_j\to 0$. Subtract the quadratic Taylor polynomial of $u_j$ at $x_j$ and rescale:Its domain $\{z:x_j+r_{j}z\in B_1\}$ exhausts ${\mathbb{R}}^n$. The normalization givesTaylor theorem, or integration along line segments from zero, consequently bounds $D^2{v}_j,\nabla v_j,v_j$ on every fixed ball by constants times $|z{|}^{\mu },|z{|}^{1+\mu },|z{|}^{2+\mu }$, respectively. The Hessian matrices are equicontinuous. The Arzelà-Ascoli theorem and a diagonal subsequence therefore give $v_j\to v$ in $C^2$ on compact subsets of ${\mathbb{R}}^n$.

The rescaled equation ison each compact set. Thus $v$ is harmonic and is smooth. Every entry of its Hessian matrix is harmonic and has a global Hölder seminorm at most $2/\delta$. The allowed Liouville lemma for globally Hölder harmonic functions makes each entry constant. Since $D^{2}v(0)=0$, all these constants are zero.

On the other hand, $e_j=(y_j-x_j)/r_j$ is a unit vector and $|D^2{v}_j(e_j)|=1$ by construction. A further subsequence has $e_j\to e$; the local $C^2$ convergence implies $|D^{2}v(e)|=1$, a contradiction. Hence the desired constant exists, andThe global Hölder seminorm on the limiting Hessian matrix, rather than a bound on its magnitude throughout space, is what makes this Liouville argument work.

Fix nested relatively compact subsets of $\Omega$ and let $M$ bound the gradient on the larger one. The fundamental theorem of calculus along a line segment gives $|w_h|\le M$ on the smaller one. Part (i) gives coefficient bounds independent of small $h$, so the De Giorgi-Nash-Moser theorem gives uniform interior Hölder continuity estimates for these weak solutions. Since $Du$ is continuous, the difference quotients $w_h$ converge locally uniformly to $D_{\ell }u$. Passing to the limit in the Hölder seminorm estimate, and repeating for every coordinate, yieldsThe first derivatives are locally Hölder continuous. Here the initially obtained exponent depends on $n$ and the local gradient bound. This is the interior interpretation of the printed regularity conclusion; a uniform global estimate does not follow from $u\in C^1(\Omega )$ alone.

For every $w\in C^{\alpha }[0,1]$, its finite Hölder seminorm gives the deterministic boundWhen $\alpha >1/2$, the right-hand side tends to zero. Thus under any probability measure on the stated coordinate sigma-algebra, these measurable finite-coordinate sums converge to zero on every path, and hence almost surely.

If the coordinate process were a Brownian motion, part (b) would make the same sums converge to one in $L^2$, and hence in probability. Almost sure convergence to zero also implies convergence in probability to zero. Uniqueness of a limit in probability is a contradiction. Therefore no such Brownian measure exists:This is the quadratic variation obstruction to Hölder regularity. It does not address the critical exponent $\alpha =1/2$, where the displayed deterministic bound does not tend to zero.

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. Arzelà-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.

Use localization and cancellation of the assumed interval-adapted wavelet basis: at scale $j$, every detail wavelet has support diameter at most $B{2}^{-j}$, L1 norm at most $B^{\prime }{2}^{-j/2}$, and annihilates polynomials of degree less than $q$. These bounds also apply to the modified boundary wavelets; a restriction of an arbitrary whole-line wavelet without the cancellation-preserving boundary construction would not suffice. The finitely many coarse scaling functions can be treated separately.

Write $\alpha =r+\beta$ with $0<\beta \le 1$ as above. Because $\alpha <q$, we have $r<q$. Choose $x_0$ in the support of a function of ${\psi }_{j,n}$, and let $T_{x_0}$ be the degree-$r$ Taylor polynomial of $f$. The Hölder-Taylor remainder bound givesFor $r=0$ this is the Hölder seminorm bound. For $r\ge 1$, the integral Taylor remainder is bounded using $f^{(r)}(t)-f^{(r)}(x_0)$, whose magnitude is at most $[f^{(r)}{]}_{C^{0,\beta }}|t-x_0{|}^{\beta }$.

The vanishing moments remove $T_{x_0}$ from the inner product. Applying the remainder estimate and the L1 norm bound on the localized support yields the wavelet coefficient decay for Hölder functionsThusHere $C$ depends on the fixed wavelet family, the exponent and boundary construction, but not on $f,j,n$. The proof uses the assumed regularity and moments; it does not identify the minimal Daubechies wavelet order with its differentiability order, which need not coincide.

Write $K(u,f)=\Vert u{\Vert }_{C^2(B_1)}+\Vert f{\Vert }_{C^{0,\alpha }(B_1)}$; the paper's $|u{|}_{2;B_1}$ denotes a $C^2$ norm, not an $L^2$ norm. We use a blow-up compactness proof of an interior Schauder estimate. If the estimate failed for fixed $\delta >0$, normalization would produceChoose $x_j,y_j\in B_{1/2}$ with $d_j=|y_j-x_j|$ and $|D^2{u}_j(y_j)-D^2{u}_j(x_j)|\ge \frac{1}{2}{d}_j^{\alpha }$. Since $\Vert D^2{u}_j{\Vert }_{\infty }\to 0$, necessarily $d_j\to 0$. Subtract the quadratic Taylor polynomial $P_j$ at $x_j$ and setThe domains expand to ${\mathbb{R}}^n$. At zero, $w_j,D{w}_j,D^2{w}_j$ vanish; the Hölder seminorm of $D^2{w}_j$ is at most ${\delta }^{-1}$. Integration along segments gives uniform $C^{2,\alpha }$ bounds on each fixed ball. Alsolocally uniformly, since $[f_j{]}_{\alpha }\to 0$. By compact embedding of Hölder spaces and a diagonal subsequence, $w_j$ converges locally in $C^{2,\beta }$ to an entire harmonic function $w$, with $[D^{2}w{]}_{\alpha }\le {\delta }^{-1}$ and $D^{2}w(0)=0$. The unit vectors $(y_j-x_j)/d_j$ have a convergent subsequence; its limit $e$ satisfies $|D^{2}w(e)|\ge 1/2$.

Each second derivative of $w$ is an entire harmonic function with finite global Hölder seminorm. The supplied polynomial-growth Liouville theorem for harmonic functions, at growth exponent $\alpha <1$, makes it constant, and its value at zero makes it zero. This contradicts the nonzero Hessian at $e$, proving the estimate with the small $\delta$ term.

To remove that term, rescale the estimate to all contained balls and apply the Simon absorption lemma. Explicitly, with $d(x)=1-|x|$ and $d_{xy}=\min (d(x),d(y))$, defineFor $|x-y|<d_{xy}/8$, apply the rescaled estimate on $B_{d(x)/2}(x)$; its larger-ball seminorm is controlled by $M$ because all points there have boundary distance at least $d(x)/2$. For more separated pairs use $2\Vert D^{2}u{\Vert }_{\infty }$. The result is $M\le C_0\delta M+C_{\delta }K(u,f)$. Choose $C_0\delta <1/2$ and use $d_{xy}\ge 1/2$ on $B_{1/2}$ to obtainThe forcing's Hölder seminorm cannot be discarded. For integers $k\to \infty$, takeTheir $C^2$ and forcing supremum norms stay bounded, but the $11$ entry of the Hessian matrix differs by two at $x_1=\pm \pi /(2k)$, giving $[D^2{u}_k{]}_{\alpha ;B_{1/2}}\ge 2(k/\pi {)}^{\alpha }\to \infty$.