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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 $O(M\log n/n)$.

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.

Here is the one-parameter Kolmogorov continuity theorem. Suppose a real-valued stochastic process on $[0,1]$ satisfies, for some $p>0$, $q>0$ and finite $C$,Then it has a continuous modification of a stochastic process whose paths are Hölder continuous of every positive exponent $\gamma <q/p$. A modification means equality almost surely at each fixed time, not an a priori simultaneous equality at all times. Other compact time intervals follow by rescaling; the vector-valued version uses a norm in place of absolute value.

For the proof fix $0<\gamma <q/p$ and use the dyadic grids $D_n=\{j{2}^{-n}:0\le j\le 2^n\}$. Let $D_n^{\ast }$ be the largest adjacent increment on this grid. The union bound and Markov inequality giveThe right-hand side is summable. The Borel-Cantelli first lemma shows that, outside a null set, an almost-surely finite random constant $K$ can absorb the finitely many early grids so that

We now use dyadic increment chaining. For two distinct dyadic points $s,t$, choose $n$ with $2^{-(n+1)}<|t-s|\le 2^{-n}$. Approximate them from below by $s_j,t_j\in D_j$, for $j\ge n$. Each refinement moves by zero or one grid step, while $s_n,t_n$ are at most two grid steps apart. The sequences eventually equal the original dyadic points. Telescoping giveswhere a deterministic finite $C_{\gamma }$ absorbs the geometric series and $2^{-n}<2|t-s|$. Thus the sample function on the dense union of dyadic grids has a unique Hölder-continuous extension to $[0,1]$. Define this extension to be $\tilde{X}$ on the probability-one event, and set it to zero on the exceptional null set.

For a fixed $t$, take dyadic $t_j\to t$. The moment condition gives $X_{t_j}\to X_t$ in probability, while the constructed extension gives $X_{t_j}\to {\tilde{X}}_t$ almost surely. Uniqueness of a probability limit yields ${\tilde{X}}_t=X_t$ almost surely. This proves the required modification property. Performing the argument for a countable increasing sequence of exponents tending to $q/p$ gives one version with all smaller positive Hölder exponents, since its extensions agree on the dyadic points.

For Brownian motion, Gaussian increments give, for any integer $r\ge 2$,Use $p=2r$ and $q=r-1$. The theorem gives every exponent $\alpha <(r-1)/(2r)=1/2-1/(2r)$. For any $0<\alpha <1/2$, choose $r$ large enough. Intersect probability-one events over integer $r$ and compact intervals $[0,N]$ to obtainThe usual continuous Brownian version has the same property: its continuous paths agree with the constructed version on a countable dense set and hence everywhere, outside one null set. The Hölder constant depends on the compact interval. This is the precise local meaning of Brownian Hölder regularity; it is not a uniform Hölder bound on the entire unbounded time axis.

A sufficient set of assumptions is that $L$ is a densely defined closed operator on the Hilbert space $H$, generating an analytic semigroup $S(t)$, and that $f\in C^{0,\theta }([0,T],H)$ for some $0<\theta \le 1$. For an operator criterion, one may assume that $A=\omega I-L$ is a sectorial operator of angle less than $\pi /2$, with dense domain. The analytic semigroup then satisfies, on a fixed finite time interval,The Hölder continuity assumption means $\Vert f(t)-f(s){\Vert }_H\le K|t-s{|}^{\theta }$. It supplies time regularity of the forcing, not spatial membership of $f(t)$ in $D(L)$.

With initial datum $u_0\in H$, a classical solution of an abstract Cauchy problem meanswith the evolution identity holding in $H$ at positive times and $u(t)\to u_0$ in $H$. Equivalently, the positive-time domain continuity is in the graph norm. If the definition requires differentiability also at zero, assume additionally $u_0\in D(L)$; the proof below then gives $u\in C^1([0,T],H)\cap C([0,T],D(L))$.

The variation-of-constants formula gives the candidateThe integral is a Bochner integral. Boundedness of $S$ and continuity of $f$ make the expression continuous into $H$, and its integral term tends to zero with $t$. Thus it is a mild solution of an abstract Cauchy problem with the correct initial value. The issue is showing that its integral term lies in the domain of the unbounded linear operator $L$.

Write $v(t)={\int }_0^{t}S(r)f(t-r)dr$. For $\epsilon >0$, the truncated integral $v_{\epsilon }(t)={\int }_{\epsilon }^{t}S(r)f(t-r)dr$ lies in $D(L)$. Subtract the current forcing before applying $L$:Here the last term follows by integrating $LS(r)f(t)=\frac{d}{dr}S(r)f(t)$ away from zero; it does not assume that $f(t)\in D(L)$. The first integrand satisfiesSince $v_{\epsilon }(t)\to v(t)$ in $H$ and the displayed images also converge, the closed operator property givesThis is the reason for the Hölder continuity assumption: the unsplit estimate would have the nonintegrable factor $r^{-1}$. Merely knowing that $f$ is continuous does not provide the integrable bound needed here. More generally, a modulus of continuity $\eta$ satisfying ${\int }_0^1\eta (r)dr/r<\infty$ suffices for this argument; a positive Hölder exponent is a convenient sufficient hypothesis.

The formula also proves continuity of $Lv$ at positive times. On compact positive-time intervals, split the integral at a small $r=\epsilon$: its short part is uniformly bounded by $C{\epsilon }^{\theta }$, while the remaining part is continuous by the strongly continuous semigroup property and ordinary dominated convergence. Together with analytic smoothing of $S(t)u_0$, this makes $Lu$ continuous at positive times. The semigroup property givesBecause $u(t)\in D(L)$, division of this identity by $h$ after subtracting $u(t)$ gives the continuous right derivative $Lu(t)+f(t)$. The left derivative agrees by the same identity and continuity of $Lu+f$. Thus $u^{\prime }=Lu+f$, proving existence of the classical solution of an abstract Cauchy problem on the entire given finite interval, and in particular local existence.

For uniqueness, the difference $h$ of two such classical solutions of an abstract Cauchy problem satisfies $h^{\prime }=Lh$ and $h(0)=0$. For $0<\epsilon <t$, differentiation of $S(t-s)h(s)$ on $[\epsilon ,t]$ gives zero, so $h(t)=S(t-\epsilon )h(\epsilon )$. Letting $\epsilon \downarrow 0$ yields $h(t)=0$. Finally, if $u_0\in D(L)$, then $LS(t)u_0=S(t)L{u}_0\to L{u}_0$, and the expression for $Lv(t)$ tends to zero. Hence $u^{\prime }(t)\to L{u}_0+f(0)$ and the stronger time-zero regularity follows. The solution is unique; the forcing regularity controls the singularity in the differentiated integral.

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 $K$.