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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.

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 continuous representative of $\bar{u}$ furnished by the one-dimensional Sobolev embedding theorem, and define $u^{\ast }(x)=\bar{u}(|x|)$ on $A$. The one-dimensional Sobolev representative is absolutely continuous; Holder inequality applied to its fundamental theorem of calculus givesAlso $\Vert \bar{u}{\Vert }_{\infty }\le \Vert \bar{u}{\Vert }_6+\Vert {\bar{u}}^{\prime }{\Vert }_6$: select a point where $|\bar{u}|$ is no larger than its $L^6$ norm on the unit-length interval and apply the same fundamental theorem of calculus. Thus the one-dimensional Hölder norm is controlled by $\Vert \bar{u}{\Vert }_{W^{1,6}(1,2)}$.

The two Hölder seminorms must be compared using the correct distances. By the reverse triangle inequality,Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two Hölder seminorms. Their supremum norms are equal as well. Part (i) now givesThe representative agrees with $u$ almost everywhere, and the assignment is linear and injective on the radial Sobolev space. This proves the requested continuous embedding. The inner radius being positive makes the radial profile Sobolev norm comparable to its weighted radial version.

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 $K$.

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.

Let $M$ bound the piecewise Hölder norms, and let $B=\Vert f{\Vert }_{\infty }$. At scale $j$, classify a wavelet as bad when its localized support of a function meets a discontinuity, and as good otherwise. Compact support and bounded overlap imply at most $C_{0}K$ bad wavelets per scale and at most $C_1{2}^j$ total wavelets. A good support lies in a single smooth piece, so the preceding wavelet coefficient decay for Hölder functions gives $|c_{j,n}|\le CM{2}^{-j(\alpha +1/2)}$. The permitted bounded-function estimate gives $|c_{j,n}|\le CB{2}^{-j/2}$ for bad coefficients.

Consider a specific N-term approximation: retain all coarse scaling functions and all wavelets through scale $J$, and also retain the bad coefficients at scales $J<j\le L$, where $L=\lceil 2\alpha J\rceil$. Since $\alpha >1/2$, $L\ge J$ for large $J$. Its term count is bounded byBy the Parseval identity, its squared L2 norm error is the sum of omitted squared coefficients. The good tail satisfiesand the remaining bad tail satisfiesChoose $J=\lfloor {\log }_2(N/(2C_2^{\prime }))\rfloor$ with a fixed sufficiently large $C_2^{\prime }$; then the $O(KJ)$ extra terms fit the remaining budget for all sufficiently large $N$, while $2^J$ is comparable to $N$. Padding to $N$ terms does not increase the error. The best N-term approximation is at least as accurate as this constructed selection, soThe constant may depend on the finite number of jumps, the piecewise Hölder norms, $B$ and the fixed wavelet family. Values exactly at the jumps are immaterial in $L^2$. The improvement comes from retaining only a bounded number of jump-crossing coefficients at each finer scale, as in best N-term wavelet approximation of piecewise Hölder functions.