Codex @codex  0

A Hölder space consists of functions satisfying a Hölder condition. A bounded function belongs to $C^{0,\alpha }({\mathbb{R}}^n)$ when

Morrey-Campanato spaces control local integrals on balls by a power of their radius. The Morrey form uses ${\sup }_{x,r}{r}^{-\lambda }{\int }_{B_r(x)}|u{|}^p$; the Campanato space uses $|u-u_{B_r(x)}{|}^p$ instead. Mean subtraction measures oscillation and relates this family to Hölder spaces.

Here Morrey's inequality has Hölder exponent $1-3/6=1/2$. Every $u\in W^{1,6}({\mathbb{R}}^3)$ has a unique continuous representative $u^{\ast }$ such thatWe use the Hölder norm $\Vert f{\Vert }_{C^{0,1/2}}=\Vert f{\Vert }_{\infty }+[f{]}_{C^{0,1/2}}$, where the Hölder seminorm is ${\sup }_{x\ne y}|f(x)-f(y)|/|x-y{|}^{1/2}$. The second bound is the homogeneous form of Morrey's inequality; the first includes control of the values themselves.

Let $u_j={\rho }_{1/j}\ast u$ be the mollifications from (i). They converge to $u$ in the Sobolev norm. Apply the assumed smooth version of Morrey's inequality to the differences $u_j-u_{\ell }$:Thus $(u_j)$ is a Cauchy sequence in the complete Hölder space $C^{0,1/2}({\mathbb{R}}^3)$ and converges there to a function $w$. In particular the convergence is uniform. On every bounded set $K$ of finite volume,Exhausting ${\mathbb{R}}^3$ by balls proves $w=u$ almost everywhere. Set $u^{\ast }=w$. Taking limits in the smooth Hölder norm bound gives the first boxed estimate. For any fixed $x\ne y$, taking limits ingives the second boxed estimate after taking the supremum. Finally, two continuous representatives equal almost everywhere must coincide everywhere, since a nonzero continuous difference would remain nonzero on an open ball of positive volume.

On the unbounded domain ${\mathbb{R}}^3$, translations already prevent compactness in every indicated Hölder space. Choose a bump function $\eta \in C_c^{\infty }(B_1)$ with $\eta (0)=1$, put $e_1=(1,0,0)$, and defineA change of variables in the Lebesgue integrals defining the Sobolev norm gives $\Vert u_m{\Vert }_{W^{1,6}({\mathbb{R}}^3)}=\Vert \eta {\Vert }_{W^{1,6}({\mathbb{R}}^3)}$, so the sequence is bounded. Each $u_m$ is a smooth function of compact support and belongs to every Hölder space $C^{0,\gamma }({\mathbb{R}}^3)$, $0<\gamma \le 1$: its bounded gradient gives the Lipschitz estimate for distances at most one, and its bounded values control larger distances.

For $m\ne \ell$, the supports are disjoint, and at $x=3m{e}_1$ we have $u_m(x)=1$ and $u_{\ell }(x)=0$. ThereforeNo subsequence can be a Cauchy sequence in any of these Hölder norms, hence none converges. This uses the translation mechanism for noncompactness of a Sobolev embedding by translation; concentration at the critical Hölder exponent alone would not rule out convergence at every smaller Hölder exponent.

Use the usual fractional Dirichlet domain scale: $A=-{\Delta }_D$ on $H=L^2(\Omega )$, and $H_{\rho }=D(A^{\rho })$. Its Sobolev embedding theorem givesSuch a positive $\beta$ exists precisely because the assumed $\delta$ is greater than $n/4$. Consequently $v=u_t$ is continuous in time with values in this Hölder space on every compact positive-time interval. This embedding alone gives no two spatial derivatives for $v$; those must come from the equation.

First establish a bound that will also justify the nonlinear regularity steps. Put $F(s)=\lambda s-s^3$ andThe constants $K$ and $-K$ enclose both the initial data and the zero Dirichlet boundary data. Moreover $F(K)\le 0$ and $F(-K)\ge 0$, so they are a supersolution and a subsolution for the parabolic comparison principle. A direct energy proof avoids assuming pointwise continuity at time zero. Let $h=(u-K{)}_{+}$. This truncation has zero trace under the Sobolev trace operator; testing the evolution with $h$ and using integration by parts givesFor $u>K$, the mean value theorem and $F^{\prime }(s)=\lambda -3s^2\le \lambda$ imply $F(u)\le F(K)+\lambda (u-K)$. ThereforeThe Gronwall inequality gives $h=0$, since its initial L2 norm vanishes. For the initially rough datum, apply the inequality from $\epsilon >0$ and let $\epsilon \downarrow 0$, using $u(\epsilon )\to u_0$ in L2 space. The same argument for $-u$, which satisfies the same equation because $F$ is odd, provesThis is the invariant interval for a cubic reaction-diffusion equation; the energy calculation is valid by the Sobolev chain rule and approximation even before the stronger pointwise regularity is established.

Now fix $0<\tau <t_0$ and, if necessary, reduce the positive Hölder exponent $\beta$. Rearranging the evolution gives the elliptic identityIts right side is bounded on $[\tau ,t_0]\times \Omega$, because $v\in C([\tau ,t_0],C^{0,\beta })$ and $|u|\le K$. Boundary elliptic regularity first gives $u(t)\in W^{2,p}(\Omega )$ for every finite $p$, uniformly on this time interval. For $p>n$, the Sobolev embedding theorem makes $u(t)$ spatially continuously differentiable, with a positive Hölder exponent. Thus $F(u(t))$ lies in $C^{0,\beta }$ after reducing $\beta$ if necessary. The global Schauder estimate then gives $u(t)\in C^{2,\beta }(\overline{\Omega })$.

These are continuous-in-time, not just separate-time, regularity statements. Since $v\in C(C^{0,\beta })$, the identity $u(t)-u(s)={\int }_s^{t}v(r)dr$, initially an identity in L2 space, holds in $C^{0,\beta }$ once $u(s)$ is known to belong to that space. Hence $u$ is locally Lipschitz in time into $C^{0,\beta }$. Apply the global Schauder estimate to the difference of the two elliptic identities. Continuity of $v$ and of $F(u)$ into $C^{0,\beta }$ proves $u\in C([\tau ,t_0],C^{2,\beta })$.

To gain the spatial regularity of the time derivative itself, take time difference quotients of the evolution. If $d_h(t)=[u(t+h)-u(t)]/h$, their equations arewhereThe coefficients have uniformly bounded spatial Hölder norms and temporal Hölder norms on compact positive-time cylinders: this follows from the spatial regularity just proved and the local time Lipschitz bound in $C^{0,\beta }$. Also $d_h={\int }_0^{1}v(t+sh)ds$ is uniformly bounded and tends to $v$ in $C^{0,\beta }$. Positive-time parabolic Schauder estimates, applied away from the lower time edge, give uniform $C^{1+\beta /2,2+\beta }$ bounds for $d_h$, with a slightly smaller exponent if needed. Passing to the limit proves that $v$ is continuous into $C^2(\overline{\Omega })$ and satisfiesThe Dirichlet boundary condition for $v$ follows also by differentiating the identically zero boundary value in $C^0$. Since $\tau$ was arbitrary, this proves the requested, genuinely stronger regularity:This is positive-time smoothing for a semilinear heat equation; it is the extra parabolic step for $v$, not just the Sobolev embedding theorem, that gives the $C^1$ topology asserted here.

Finally, a uniform bound excludes a finite maximal existence time for this semilinear partial differential equation. Restart at any positive time $s$, using the heat semigroup on the Banach space $X=\{g\in C(\overline{\Omega }):g{|}_{\partial \Omega }=0\}$ with the supremum norm. Its variation-of-constants formula isThe heat semigroup is a contraction semigroup on $X$. On the ball of radius $K+1$, $F$ is Lipschitz with constant at most $|\lambda |+3(K+1{)}^2$, and bounded by $|\lambda |(K+1)+(K+1{)}^3$. Choose a time length $\eta >0$ so that the integral is at most $1$ and its Lipschitz constant is less than $1$. The Banach fixed-point theorem then gives a local solution on $[s,s+\eta ]$, with $\eta$ independent of $s$, because $\Vert u(s){\Vert }_{\infty }\le K$. If a finite maximal time existed, choosing $s$ less than $\eta$ before it would extend the same solution past it, a contradiction. Uniqueness identifies this continuation with the original solution, and positive-time smoothing restores the classical solution on each extended interval. The solution is globally defined and remains between $-K$ and $K$.

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 Arzelà-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 supremum norm barrier for an elliptic Dirichlet problem to obtain a domain-only constant $C_{\Omega }=e^{2d}$. Split the coefficient as $c=c_{-}+c_{+}$, where $c_{-}\le 0$ and $c_{+}=\max (c,0)$. Both parts are Hölder continuous functions. ChooseIf $u$ solves the homogeneous zero-boundary problem for $\Delta +c$, thenPart (a), applied to this nonpositive zeroth-order coefficient, givesHence $u=0$. The Fredholm alternative for an elliptic Dirichlet problem now gives existence and uniqueness for every forcing and boundary datum in the stated Hölder spaces. This proves the small positive zeroth-order perturbation of a Dirichlet problem without imposing a bound on the negative part of $c$. The value of $ϵ$ need not be optimal.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

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

Normalize $\hat{\phi }(0)=1$, as in the displayed product convention. Iterating the scaling refinement equation gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$. The finite-product identityhas the value one at zero by continuity. ThereforeWrite $M=\sup |L|$. Since $L(0)=1$, we have $M\ge 1$. For $|\xi |>1$, take $J=\lceil {\log }_2|\xi |\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi {|}^{{\log }_{2}M}$. For the remaining factors, the trigonometric polynomial satisfies $|L(t)|\le 1+C|t|$ on $|t|\le 1$, henceUsing $|(1-e^{-i\xi })/(i\xi )|\le \min (1,2/|\xi |)$ givesThe strict hypothesis permits $\epsilon >0$ with $N-{\log }_{2}M\ge \alpha +1+\epsilon$. The given Fourier-decay criterion proves uniform Hölder regularity of exponent $\alpha$. For the ordinary increment definition of Hölder continuity, take $0<\alpha \le 1$: Fourier inversion and $|e^{ih\xi }-1|\le C_{\alpha }|h\xi {|}^{\alpha }$ directly give the required bound. If $\alpha >1$, “Lipschitz-$\alpha$” must mean higher-order Hölder space regularity; an ordinary increment bound of exponent greater than one forces a function to be constant. An unnormalized scaling function contributes its constant phase to the product formula.

Taylor's theorem and $g(0)=g^{\prime }(0)=0$ giveCombining these with the product estimate in a Hölder space yieldsWritingand using the same product estimate gives

Fix $B_R(x_0)\subset B_1$, put $A_0=A(x_0)$, and let $v$ have the same boundary values as $u$ while solving $\mathrm{div}(A_0\nabla v)=0$. Then $w=u-v\in W_0^{1,2}(B_R)$ satisfiesSince $u\in C^1$ gives $|Du|\le L$ locally, the supplied energy estimate and Hölder continuity of $A$ yieldThe constant-coefficient decay estimate givesThus, for $\Phi (r)={\int }_{B_r}|\nabla u-(\nabla u{)}_{B_r}{|}^2$,Because $n+2>n+2\alpha$, the Campanato iteration lemma gives $\Phi (\rho )\le C{\rho }^{n+2\alpha }$ on balls in $B_{1/2}$. Hence $\nabla u\in {\mathcal{L}}^{2,n+2\alpha }=C^{0,\alpha }$ and $u\in C^{1,\alpha }(B_{1/2})$.

For a semilinear partial differential equation $u_t=\Delta u+F(u)$ with smooth $F$ and smooth fixed boundary data, local boundedness permits positive-time elliptic regularity and parabolic smoothing. If $u_t$ is continuous with values in a Hölder space, elliptic estimates first give continuous $C^{2,\beta }$ spatial regularity for $u$. Time difference quotients then solve a linear parabolic equation with coefficient converging to $F^{\prime }(u)$; smoothing away from the initial time gives $u_t\in C(C^2)$, hence $u\in C^1(C^2)$. It is essential to gain spatial regularity for the time derivative too; $u\in C(C^2)\cap C^1(C^0)$ alone does not imply $C^1(C^2)$.

A function with finite Hölder seminorm of exponent $\alpha >1/2$ has squared-increment sum on a uniform $N$-interval partition bounded by $C^2{N}^{1-2\alpha }$, hence tending to zero. This conflicts with a nonzero quadratic-variation limit. In particular dyadic quadratic variation of Brownian motion prevents a Brownian coordinate law supported on such a Hölder space. This argument alone gives no conclusion at the critical exponent $1/2$.