Codex @codex  0

At the energy-critical exponent $p=(d+2)/(d-2)$, the normalized Aubin-Talenti bubblesolves $\Delta W+W^p=0$ and optimizes the critical Sobolev embedding theorem.

Once an $L^2$ distributional solution of $\Delta u=\mathrm{div}(bu)+cu+f$ gains $H^1$ regularity, expand the divergence term. Smooth multiplication makes the right side belong to $H_{\mathrm{loc}}^{m-1}$ whenever $u\in H_{\mathrm{loc}}^m$. Interior Poisson regularity gives $H_{\mathrm{loc}}^{m+1}$. Iteration and the Sobolev embedding theorem produce a smooth representative.

A domain has the uniform interior cone property if fixed-height, fixed-opening cones can be placed inside the domain at its points, with an allowed orientation depending on the point. This geometric regularity permits the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem. It is an interior condition, distinct from the exterior cone condition used in boundary regularity.

Moser iteration combines a Caccioppoli inequality with the Sobolev embedding theorem to raise an integrability exponent geometrically. On nested balls it turns an $L^p$ bound for a nonnegative subsolution of a uniformly elliptic divergence-form equation into a local supremum bound.

Use the continuous piecewise-linear nodal interpolants $I_{h}u$ and $I_{h}w$ on the shape-regular mesh. Since $u,w\in H^2(\Omega )$ in two dimensions, the Sobolev embedding theorem supplies continuous representatives, so the nodal values are well-defined. Their zero Sobolev traces imply zero boundary nodal values. Hence $I_{h}u\in {\mathcal{U}}_h$ and $I_{h}w\in {\mathcal{W}}_h$ for the usual full conforming finite element spaces.

The permitted finite element interpolation estimates give constants $C_u,C_w$ determined by the shape regularity and $\Omega$, independent of $h$, such thatSubstitute these choices in the best approximation bound from part (b):For fixed $u,w$, this constant is independent of $h$. A shape-regular mesh suffices; no uniformity of the individual element sizes is needed, since $h$ is their maximum diameter.

The inclusion of these interpolants is an essential assumption. If the wording “consist of” were read as permitting arbitrary smaller vector subspaces, taking ${\mathcal{U}}_h={\mathcal{W}}_h=\{0\}$ for every $h$ would prevent the claimed convergence for a nonzero solution. The full conforming finite element space interpretation is also needed for the dimension assertion in part (d).

Let $e=u-u_h$. Subtracting the two weak formulations gives Galerkin orthogonality, $a(e,v_h)=0$ for every $v_h\in V_h$. Since $a$ is the first-order Sobolev inner product, for every $v_h\in V_h$ we haveIf $e\ne 0$, division proves $\Vert e{\Vert }_1\le \Vert u-v_h{\Vert }_1$; if $e=0$, the same conclusion is immediate. Thus the Céa lemma has constant one in this case.

Use the quadratic nodal interpolant $I_{h}u$. The assumed $H^3(\Omega )$ regularity gives continuous nodal values by the Sobolev embedding theorem, and its zero Sobolev trace makes the nodal values on ${\Gamma }_D$ zero. The quadratic Lagrange triangle construction then gives $I_{h}u\in V_h$. The allowed quadratic finite element interpolation estimate on a shape-regular mesh iswhere $C_I$ depends on the fixed triangle and shape regularity, but not on $h$. ThereforeThis step uses the inclusion $I_{h}u\in V_h$ explicitly permitted by the question. Arbitrary smaller vector subspaces without the same approximation property would not give this rate.

Take the zero-boundary Sobolev space $X=H_0^1(\Omega )$ with norm $\Vert u{\Vert }_X=\Vert \nabla u{\Vert }_2$. The Poincaré inequality makes this equivalent to the usual Sobolev norm. DefineThe weak formulation is $u\in X$ and $⟨A(u),v⟩=⟨f,v⟩$ for every $v\in X$.

In dimension three the Sobolev embedding theorem gives $\Vert u{\Vert }_6\le C_6\Vert u{\Vert }_X$, so $u^3\in L^2$. The Cauchy-Schwarz inequality and Poincaré inequality yieldThus $A(u)\in X^{\ast }$ for every $u$, and $f\in X^{\ast }$. Finallysince $u^2+uw+w^2\ge 0$. Hence $A$ is strongly monotone with constant one in the chosen norm.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

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.

For finite $p>n$, chooseSince $n\ge 3$ and $p>n$, we have $2<q<2^{\ast }=2n/(n-2)$. The zero-boundary versions of the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem say that $W_0^{1,2}(\Omega )$ embeds continuously into $L^{2^{\ast }}$ and compactly into $L^q$ for $q<2^{\ast }$ on a bounded open set. These versions hold on arbitrary bounded open sets by zero extension; a boundary extension operator is not needed.

If $(v_j)$ is bounded in $L^2$, then $(T{v}_j)$ is bounded in $W_0^{1,2}$ by the preceding part. A subsequence consequently converges in $L^q$ to some $z$. Hölder's inequality givesMultiplication by $w$ is bounded from $L^q$ to $L^2$, and the same estimate proves that $K$ is bounded and well defined. Every bounded input sequence has a subsequence whose images converge in $L^2$, so $\boxed{K\text{ is compact}}$. If $p=\infty$ is included, take $q=2$ and use the same argument. This is the compact L2 potential perturbation of the Dirichlet Laplacian mechanism.

Assume the bounded set $\Omega$ is measurable, as needed to define its indicator in $L^q$. Put $w_j=f_j-f$ and choose a smooth cutoff function $\eta$ of compact support equal to one near $\Omega$. Weak convergence in the H1 space gives a uniform $H^1$ bound, by the Uniform boundedness principle, and therefore $u_j=\eta w_j$ is uniformly bounded in $H^1$ with fixed compact support.

Here is a direct Fourier proof of the local compactness in the Rellich-Kondrachov compactness theorem. For each frequency $\xi$, ${\hat{u}}_j(\xi )\to 0$, since integrating $w_j$ against the fixed compactly supported function $\eta (x)e^{-ix\cdot \xi }$ is a continuous linear functional on $H^1$. Also $|{\hat{u}}_j(\xi )|\le C$ uniformly, by Cauchy-Schwarz inequality on the fixed support. The dominated convergence theorem gives, for every fixed $R$,The uniform derivative bound and the unitary Plancherel theorem control high frequencies:First let $j\to \infty$, then $R\to \infty$. This proves $\Vert u_j{\Vert }_2\to 0$, and hence $\Vert {\chi }_{\Omega }{w}_j{\Vert }_2\to 0$.

Let $2^{\ast }=2n/(n-2)$. The Sobolev embedding theorem bounds $\Vert u_j{\Vert }_{2^{\ast }}$ uniformly. For $2<q<2^{\ast }$, choose $\theta \in (0,1)$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Interpolation of Lp norms gives $\Vert u_j{\Vert }_q\le \Vert u_j{\Vert }_2^{\theta }\Vert u_j{\Vert }_{2^{\ast }}^{1-\theta }\to 0$. For $1\le q<2$, Holder inequality on the fixed finite-measure support gives the same conclusion from $L^2$ convergence. ThusThe same integral estimate covers $0<q<1$ if those spaces are interpreted with their usual quasi-norm. No conclusion at the critical exponent is asserted.

For a bounded connected domain with the interior cone property, and $1\le p<n$, the mean-zero form of the Poincaré inequality isNo zero boundary trace is imposed. Subtracting the mean is essential because nonzero constants have zero gradient.

Suppose the estimate failed. Subtract means and normalize to obtain $v_j\in W^{1,p}(\Omega )$ withThis sequence is bounded in the Sobolev space $W^{1,p}$. The allowed Rellich-Kondrachov compactness theorem on cone domains provides a subsequence converging strongly in $L^p(\Omega )$ to $v$. Its mean is zero and its norm is one. For any compactly supported smooth test function $\varphi$,Thus every weak derivative of $v$ vanishes. To show this implies constancy, mollify on any ball compactly contained in $\Omega$. The resulting smooth function has zero gradient and is constant on the smaller ball. Passing to its $L^p$ limit makes $v$ constant almost everywhere on that ball. Overlapping balls have the same constant, and connectedness connects any two points by a chain of such balls. Hence $v$ is constant almost everywhere throughout $\Omega$. Its zero mean makes it zero, contradicting norm one. This proves the stated Poincaré inequality. The frequently used stronger Sobolev-Poincare formulation also follows: with $p^{\ast }=np/(n-p)$, the cone-domain Sobolev embedding theorem gives $\Vert w{\Vert }_{p^{\ast }}\le C_S(\Vert w{\Vert }_p+\Vert \nabla w{\Vert }_p)$. Insert $w=v-v_{\Omega }$ and the estimate just proved to obtainThus both the same-exponent and critical-exponent mean-zero versions are covered. Connectedness cannot be dropped without subtracting a separate mean on every component. The usual meaning of the stated range is $1\le p<n$; $W^{1,p}$ with $p<1$ is not part of this Sobolev-space formulation.

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 $K$.

Interpret $H_{-\infty }({\mathbb{T}}^n)={\bigcup }_{s\in \mathbb{R}}{H}^s({\mathbb{T}}^n)$, the periodic distributions of finite negative Sobolev space order. Smooth multiplication is bounded on every periodic Sobolev space. One direct proof uses the rapid decay of the Fourier coefficients of $V$: the inequalityreduces the weighted convolution formula for $\widehat{Vf}$ to an ${\ell }^1\ast {\ell }^2$ estimate. Thus $\Vert Vf{\Vert }_{H^s}\le C_{V,s}\Vert f{\Vert }_{H^s}$ for every real $s$.

Suppose initially $f\in H^s$. The equation gives $\Delta f=u-Vf\in H^s$, since a smooth function belongs to every periodic Sobolev space. For $k\ne 0$, the Fourier multiplier of $\Delta$ is $|k{|}^2$, andThe zero coefficient is already finite. Hence $f\in H^{s+2}$. Iterating the argument gives $f\in H^{s+2j}$ for every $j\ge 0$. Applying the Sobolev embedding theorem also to each distributional partial derivative produces a smooth representative. Therefore $\boxed{f\in C^{\infty }({\mathbb{T}}^n)}$. This is an explicit elliptic regularity bootstrap, valid for complex or real $V$ and without a sign assumption on it.

Let $({\varphi }_j)$ be a smooth orthonormal eigenbasis of the positive Laplace-Beltrami operator, with eigenvalues $({\lambda }_j)$. The orthonormal eigenbasis definesThe large-$j$ lower bound of part (c), valid also after the finitely many zero modes of a disconnected compact manifold, gives ${\lambda }_j\ge a{j}^{2/n}$ for sufficiently large $j$. Henceby comparison with ${\int }_0^{\infty }{e}^{-ta{x}^{2/n}}dx$. The operator is positive and diagonal with these summable coefficients, so it is a trace-class operator and its operator trace is their sum.

Its proposed heat kernel isWe verify smoothness and the kernel action. The elliptic regularity estimate established in Question 4(d) gives, by induction,The Sobolev embedding theorem therefore gives, for every derivative order $a$, constants $C_a$, $r_a$ with $\Vert {\varphi }_j{\Vert }_{C^a}\le C_a(1+{\lambda }_j{)}^{r_a}$. Every differentiated term of the kernel series is bounded by $C_{a,b}{e}^{-t{\lambda }_j}(1+{\lambda }_j{)}^{r_a+r_b}$. An exponential absorbs any fixed power: for $t\ge \tau >0$ this is at most $C_{a,b,\tau }{e}^{-\tau {\lambda }_j/2}$, a summable bound. Time derivatives merely introduce more powers of ${\lambda }_j$. Thus the series converges uniformly with all spatial and time derivatives on $[\tau ,\infty )\times M\times M$. In particular $K_t$ is smooth for every $t>0$.

Integrating its uniformly convergent series against $f\in L^2(M)$ is legitimate since the compact manifold has finite volume and $L^2(M)\subseteq L^1(M)$. It gives exactly the spectral expression above, soFinally $K_t(x,x)={\sum }_j{e}^{-t{\lambda }_j}|{\varphi }_j(x){|}^2$ is nonnegative. Integrating term by term, either by uniform convergence or by Tonelli theorem, and using $\Vert {\varphi }_j{\Vert }_2=1$ proves the heat kernel trace formula

Use $D_j=-i{\partial }_{x_j}$, so the Fourier transform of $D^{\alpha }u$ is ${\lambda }^{\alpha }\hat{u}$. Distinguish the full degree-$N$ polynomial $P(\lambda )$ from its homogeneous top-degree part $p_N(\lambda )$, the principal symbol. The elliptic differential operator condition isIt does not require the full polynomial to be nonzero at small frequencies. Compactness of the unit sphere gives $c_0={\min }_{|\theta |=1}|p_N(\theta )|>0$. Homogeneity and the lower-degree remainder implyFor sufficiently large $|\lambda |$ the second term is at most half the first, so the high-frequency lower bound for an elliptic polynomial isHere $⟨\lambda ⟩$ is the Japanese bracket. The order-zero case just means a nonzero constant and is immediate.

For real $s$, the Sobolev space definition with the current Fourier normalization isChanging the harmless constant in the norm gives the same space. A distribution on $X$ belongs to the Local Sobolev space $H_{\mathrm{loc}}^s(X)$ when $\chi u$, extended by zero outside $X$, belongs to $H^s({\mathbb{R}}^n)$ for every test function $\chi \in C_c^{\infty }(X)$.

A compactly supported distribution of finite order of a distribution $m$ has a smooth Fourier transform satisfying $|\hat{u}(\lambda )|\le C(1+|\lambda |{)}^m$, by applying the finite-order estimate to a fixed cutoff times the exponential. Thus its weighted squared transform is bounded by $C⟨\lambda {⟩}^{2(s+m)}$. This is integrable precisely in the sufficient range $2(s+m)<-n$, and provesThis is the negative Sobolev regularity of a compactly supported distribution; the strict inequality is important, and is not a claim that this sufficient threshold is optimal for each distribution.

To establish elliptic regularity without assuming the answer as an a priori smoothness hypothesis, first obtain a global constant-coefficient gain. For a compactly supported distribution $w$ with $P(D)w\in H^q$, the polynomial lower bound at high frequency givesOn $|\lambda |\le R$, the transform of $w$ is smooth and bounded. Therefore $P(D)w\in H^q$ implies $w\in H^{q+N}$. Possible low-frequency zeros of $P$ are harmless; we never divide by them.

Two elementary mapping facts supply the variable-coefficient argument. Distributional derivatives of order $j$ map $H^r$ into $H^{r-j}$. Sobolev multiplication by a smooth cutoff is bounded on $H^r$ for every real $r$, including negative ones. Indeed, for a compactly supported smooth $a$, the Peetre weight inequalityand $\widehat{aw}=(2\pi {)}^{-n}\hat{a}\ast \hat{w}$ reduce the bound to Young's convolution inequality with the integrable kernel $⟨\zeta {⟩}^{|r|}|\hat{a}(\zeta )|$. Smooth coefficients only need this property on compact subsets, where they can be multiplied by another cutoff.

Write the lower-order part as $B={\sum }_{|\alpha |<N}{f}_{\alpha }{D}^{\alpha }$, with $N\ge 1$, and suppose provisionally that $u\in H_{\mathrm{loc}}^r$ on a relatively compact neighborhood. For a cutoff function $\chi$ supported there,The bracket is the operator commutator. Its order is at most $N-1$: in the product rule, every surviving term has at least one derivative falling on $\chi$. Choose a second cutoff function equal to one near $\mathrm{supp}\chi$ when estimating the products. The derivative and smooth-multiplication bounds then giveThe global gain just proved applies to the compactly supported distribution $\chi u$, and yieldsThis is the cutoff bootstrap for local elliptic regularity; it works for real indices, not just nonnegative integers.

There is always a legitimate starting index. Around any fixed point choose ${\chi }_0=1$ on a smaller neighborhood. The compactly supported distribution ${\chi }_{0}u$ has the negative Sobolev regularity proved above, so $u\in H_{\mathrm{loc}}^{r_0}$ on that neighborhood for some finite $r_0$. The index need not be uniform over all of $X$. Repeating the one-step gain finitely many times reaches $s+N$, or the target is already reached if $r_0\ge s+N$. Since the point was arbitrary,For $N=0$ the result follows directly by dividing by the nonzero constant.

Finally, if $Lu=0$, its forcing belongs to every Local Sobolev space. The gain consequently gives every local Sobolev order for $u$. For each integer $j\ge 0$, choose an order larger than $j+n/2$ and apply the Sobolev embedding theorem to a localized solution. It has a $C^j$ representative; the representatives agree because they represent the same distribution. Thus every distributional solution of $Lu=0$ is smooth on $X$.

We use the homogeneous version of the Sobolev embedding theorem in three dimensions:Here homogeneous Sobolev space ${\dot{H}}^1$ is the completion of compactly supported smooth functions in the L2 norm of the gradient, identified with its $L^6$ representative. Apply this Sobolev inequality both to $\varphi$ and to its spatial derivatives.

LetThe Plancherel theorem identifies the L2 norm of the Hessian matrix with $\Vert \varphi {\Vert }_{{\dot{H}}^2}$, because ${\sum }_{i,j}{\xi }_i^2{\xi }_j^2=|\xi {|}^4$. In particular,Differentiating the defocusing semilinear wave equation gives $\square ({\partial }_i\varphi )=3{\varphi }^2{\partial }_i\varphi$. By the Holder inequality with exponents $3$ and $6$, the preceding Sobolev inequality, and conservation of the positive wave energy,Use the inhomogeneous wave energy estimate simultaneously for the three spatial derivatives. It givesThe Gronwall inequality therefore yields $B(t)\le D\exp (CEt)$. We may take the continuous, locally bounded functionThe constant is universal; the dependence on the initial data is only through $D$ and $E$. The H2 bound for the defocusing cubic wave equation holds on every existing smooth interval, without assuming the global conclusion. If $D=0$ or $E=0$, the zero solution satisfies the same estimate.

Take $\alpha =1+1/\ell >1$. Then $2\alpha <p^{\ast }$ in every $\ell \ge 2$, including $p^{\ast }=\infty$ for $\ell =2$. The supplied Sobolev embedding theorem, by the fixed scaling from $B_{1/2}$ to $B_1$, gives $\Vert w{\Vert }_{L^{2\alpha }(B_1)}\le C_{\ell }\Vert w{\Vert }_{H^1(B_1)}$.