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If $u\in W_{\mathrm{loc}}^{1,2}$, then $Du=0$ almost everywhere on every level set $\{u=t\}$. Choose smooth truncations ${\Phi }_{\epsilon }(s)$ with $|{\Phi }_{\epsilon }|\le 2\epsilon$, derivative one near zero, bounded derivative, and derivative zero outside $(-2\epsilon ,2\epsilon )$. The Sobolev chain rule gives $D{\Phi }_{\epsilon }(u-t)\to 1_{\{u=t\}}Du$ in local $L^2$ by dominated convergence theorem, while ${\Phi }_{\epsilon }(u-t)\to 0$. The limiting weak derivative is therefore zero.

For scalar minimizers of $\int F(Du)$ with $F\in C^2({\mathbb{R}}^n)$ and globally bounded uniformly positive Hessian matrix, no higher differentiability of $F$ is needed for interior gradient Hölder continuity. The weak Euler-Lagrange equation is $\mathrm{div}DF(Du)=0$. Difference quotients solve uniformly elliptic equations with coefficient matrices obtained by averaging $D^{2}F$ along gradient segments. The Caccioppoli inequality and difference quotient characterization give $u\in H_{\mathrm{loc}}^2$. The Sobolev chain rule then gives $\mathrm{div}(D^{2}F(Du)D(D_{k}u))=0$. The De Giorgi-Nash-Moser theorem makes each derivative Hölder continuous, with exponent depending only on dimension and the ellipticity ratio. This scalar conclusion should not be assumed for vector-valued elliptic systems.

If $v\in H_0^1(\Omega )$ and $g:\mathbb{R}\to \mathbb{R}$ is Lipschitz continuous with $g(0)=0$, then $g(v)\in H_0^1(\Omega )$. The Lipschitz version of the Sobolev chain rule controls its weak derivatives, and its boundary trace is $g(0)=0$. Equivalently, approximate $v$ by compactly supported smooth functions, apply the derivative bound to their compositions, and pass weakly in $H^1$ and strongly in $L^2$. The closed linear space $H_0^1$ is weakly closed. This allows positive-part and bounded truncations in weak formulations without losing the Dirichlet boundary condition.

Set $w=(u-k{)}^{+}$, the positive part of $u-k$. The allowed Sobolev chain rule givesUse the admissible zero-boundary Sobolev space function ${\phi }^{2}w$ in the weak formulation, with a valid cutoff function as explained in (ii). Its gradient is ${\phi }^2\nabla w+2\phi w\nabla \phi$, soIndeed the integrands vanish off $\{u>k\}$, and on that set $\nabla w=\nabla u$. The uniform ellipticity and Cauchy-Schwarz inequality estimates from (ii) now apply to $w$ and giveFixing the cutoff function gives the required constant independent of $k$. This is a superlevel Caccioppoli inequality; it uses the positive part as a test and does not require the truncated function itself to solve the original equation.

No additional hypothesis on $F$ is needed here for scalar interior $C^{1,\beta }$ regularity. In the elliptic sense of uniform convexity, the assumptions already give constants $0<\lambda \le M<\infty$ with $\lambda I\le D^{2}F(p)\le MI$ for all $p$. The following argument gives interior gradient regularity for uniformly convex autonomous energies.

The first variation at the minimizer exists because $|DF(p)|\le C(1+|p|)$, and its weak Euler-Lagrange equation isFor a coordinate difference quotient ${\delta }_{h}u$, subtract the translated equation and the original equation. The averaged linearization of a nonlinear divergence-form equation givesThese measurable symmetric matrices satisfy $\lambda I\le A_h\le MI$ independently of $h$. The Caccioppoli inequality and the standard bound $\Vert {\delta }_{h}u{\Vert }_2\le \Vert D_{k}u{\Vert }_2$ bound $D{\delta }_{h}u$ on every smaller interior set. Quoting the Sobolev characterization by bounded difference quotients yields $u\in H_{\mathrm{loc}}^2$.

Now $DF$ is globally Lipschitz continuous, so the Sobolev chain rule and commutation of distributional derivatives allow differentiation of the weak equation. Each $v_k=D_{k}u\in H_{\mathrm{loc}}^1$ satisfiesThe De Giorgi-Nash-Moser theorem states that weak solutions of this scalar divergence-form equation with bounded measurable uniformly elliptic coefficients are locally Hölder continuous. It gives a common $\beta \in (0,1)$ depending only on $n$ and the ellipticity ratio $M/\lambda$, with $D_{k}u\in C_{\mathrm{loc}}^{0,\beta }$ for every $k$. A Sobolev function whose weak gradient is continuous has a $C^1$ representative, as follows by local mollification and integration of its gradient. HenceThe higher-regularity conclusion comes from differentiating an autonomous scalar equation; it does not rely on boundary regularity or a higher derivative such as $D^{3}F$.

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 $K$.

The original PDF assumes $u\ge 0$ at the beginning of this question; the TeX conversion omits that hypothesis. First replace a smooth nonnegative solution by $v=u+\epsilon >0$, which solves the same homogeneous divergence-form elliptic equation. For $w=v^{q/2}$, the chain rule gives the favorable termin the distributional sense. Thus $w$ is a weak subsolution. Testing its subsolution inequality with ${\zeta }^{2}w$ yields the same Caccioppoli inequality as before. More precisely, testing the equation for $v$ with ${\zeta }^2{v}^{q-1}$ givesIndeed the main term is at least $(q-1)\lambda \int {\zeta }^2{v}^{q-2}|\nabla v{|}^2$, and the cross term is bounded by twice $\Lambda$ times the square roots of this weighted gradient integral and $\int v^q|\nabla \zeta {|}^2$. Let $\epsilon \downarrow 0$ to obtain the stated power energy estimate.

Apply the fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$. The crucial constant stays uniform before taking the $2/q$ power:where $D_{\ell }\ge 1$ is independent of $q$. This is uniform power gain for elliptic solutions. The exponent is $q\alpha$, not the TeX conversion's $q^{\alpha }$. The PDF's weaker prefactor $C_{\ell }$ outside the $2/q$ power follows from this stronger estimate, but that weaker form alone cannot be iterated infinitely.

For completeness, the estimate also extends to nonnegative locally $H^1$ weak solutions without assuming they are already bounded. For $s\ge 0$ define $F_M(s)=s^{q/2}$ up to $M$ and continue it linearly with slope $(q/2)M^{q/2-1}$ above $M$; set ${\Psi }_M(s)={\int }_0^s{F}_M^{\prime }(r{)}^{2}dr$. Both have bounded derivatives for fixed $M$. Direct calculation gives ${\Psi }_M/F_M^{\prime }\le F_M$ and $F_M(s)\le (q/2)s^{q/2}$. Test the weak equation with ${\zeta }^2{\Psi }_M(u)$, using smooth approximations to these Sobolev chain rule functions if needed. The resulting weighted energy bound is $\int {\zeta }^2|\nabla F_M(u){|}^2\le 4R^2\int F_M(u{)}^2|\nabla \zeta {|}^2$. If $u\in L^q(B_{r_2})$, dominated convergence on the right and the Fatou lemma on the Sobolev left give the boxed gain as $M\to \infty$, with the same uniform constant. Starting with $q=2$, this justifies every subsequent iteration step.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

For the divergence-form elliptic operator use the weak subsolution convention $\int a^{ij}{D}_{j}u{D}_i\phi \le 0$ for nonnegative test functions. Set $u_{\epsilon }=u+\epsilon$ and test with $\phi ={\eta }^2{u}_{\epsilon }^{\beta -1}$, where $\beta >1$. The Sobolev chain rule makes this test admissible; boundedness and the positive regularization control the derivative of its power. Ellipticity and the Cauchy-Schwarz inequality give, with $X=\int |Du{|}^2{u}_{\epsilon }^{\beta -2}{\eta }^2$ and $Y=\int u_{\epsilon }^{\beta }|D\eta {|}^2$,Division when $X>0$, and the trivial case $X=0$, establish the power Caccioppoli inequality. Letting $\epsilon \downarrow 0$ givesFor $1<\beta <2$ this limit uses the Fatou lemma; the gradient of a Sobolev function vanishes on a level set, so the weighted integrand is assigned zero on $\{u=0\}$. The same regularization justifies differentiating $u^{\beta /2}\eta$ in $W^{1,2}$.

The PDF bounds individual entries by $\Lambda$. Then ${\Lambda }_{\ast }\le n\Lambda$: dimensional factors are included in the ellipticity constants when the first estimate is written as $C(\lambda ,\Lambda )$. With literal entrywise bounds, the explicit valid constant displayed here is $4n^2{\Lambda }^2/{\lambda }^2$. If $\Lambda$ bounds the operator norm instead, it is $4{\Lambda }^2/{\lambda }^2$. This distinction does not alter any later conclusion, whose constants explicitly depend on $n$.

For $n\ge 3$, put $\sigma =n/(n-2)$. Apply the Sobolev inequality to $g=u^{\beta /2}\eta$ and useFor $\beta \ge p>1$, $\beta /(\beta -1)\le p/(p-1)$, so the resulting constant is uniform in $\beta$. Taking a cutoff equal to one on $B_r$, supported in $B_R$, with $|D\eta |\le C/(R-r)$, yieldsAt $R=1$ a compactly supported cutoff can still be chosen with this bound by leaving an outer margin; alternatively pass to the limit from smaller radii.

For Moser iteration, set ${\beta }_j=p{\sigma }^j$ and $r_j=1/2+2^{-j-1}$. Repeated use of the preceding estimate bounds the successive norms byThis product is finite becauseAs the exponents tend to infinity, their norms on the Euclidean ball of radius $1/2$ tend to its essential supremum. ThereforeThe power $\beta$ is independent of the exponents of Hölder continuity in earlier questions.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

For a smooth Dirichlet Laplacian eigenfunction, testing $-\Delta u=\lambda u$ with $|u{|}^{\gamma -2}u$ for $\gamma \ge 2$ gives the displayed energy estimate. The Sobolev chain rule for $v=|u{|}^{\gamma /2}$ then yields $\int |Dv{|}^2=\lambda {\gamma }^2/[4(\gamma -1)]\int |u{|}^{\gamma }$. The absolute value and the sign in the test function permit nodal changes. At $\gamma =2$, use the Lipschitz absolute-value chain rule and the fact that a gradient of a Sobolev function vanishes on a level set.

For a weak solution of $-\mathrm{div}(A\nabla u)=0$ with real symmetric coefficients satisfying the two-sided uniform ellipticity bounds $\lambda I\le A\le \Lambda I$, the displayed Caccioppoli inequality holds with $C=4{\Lambda }^2/{\lambda }^2$ for a smooth cutoff function $\phi$ of compact support. Put $w=(u-k{)}^{+}$ and test the weak formulation with ${\phi }^{2}w$. The Sobolev chain rule gives $\nabla w=1_{\{u>k\}}\nabla u$, and Cauchy-Schwarz inequality yields $\lambda \int {\phi }^2|\nabla w{|}^2\le 2\Lambda (\int {\phi }^2|\nabla w{|}^2{)}^{1/2}(\int w^2|\nabla \phi {|}^2{)}^{1/2}$. Divide and square. The constant is independent of the level $k$.