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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.

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 $D_j=-i{\partial }_{x_j}$, 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 $\Vert w{\Vert }_{L^{2\alpha }(B_1)}\le C_{\ell }\Vert w{\Vert }_{H^1(B_1)}$.

Choose $0\le \zeta \le 1$, equal to one on $B_{r_1}$, supported in $B_{r_2}$, with $|\nabla \zeta |\le C/(r_2-r_1)$. Extend $w=\zeta u$ by zero to $B_1$. The Caccioppoli inequality before discarding the cutoff weight givesAlso $\Vert w{\Vert }_2\le \Vert u{\Vert }_{L^2(B_{r_2})}$. Since $R>1$ and $r_2-r_1\le 1$, both terms are controlled by $CR(r_2-r_1{)}^{-1}\Vert u{\Vert }_2$. Applying the Sobolev embedding theorem to $w$ givesUsing a fixed ambient ball prevents the Sobolev constant from acquiring an unintended dependence on either radius.

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.

Choose nested interior open sets $U\subset \subset V\subset \subset W\subset \subset \Omega$. For small $\sigma$, part (b) gives a smooth equation on $V$ of the formThe divergence-forcing interior H1 estimate isOne can obtain this estimate directly by testing the smooth equation against ${\eta }^2{u}_{\sigma }$ and applying Young inequality, with a cutoff function equal to one on $U$ and supported in $V$. The approximate identity and the $L^2$ convolution bound give, uniformly in $\sigma$,with analogous bounds for $(cu{)}_{\sigma }$ and $f_{\sigma }$. The coefficients need only be bounded on $W$. Therefore $u_{\sigma }$ is bounded in $H^1(U)$. It converges to $u$ in $L^2(U)$, and weak sequential compactness in a Hilbert space in this Sobolev space gives $u\in H^1(U)$. Since $U$ was arbitrary, $u\in H_{\mathrm{loc}}^1(\Omega )$.

Now expand the distributional derivative:Its right side belongs to $L_{\mathrm{loc}}^2$, so the interior $H^2$ elliptic regularity estimate gives $u\in H_{\mathrm{loc}}^2$. More generally, if $u\in H_{\mathrm{loc}}^m$ for an integer $m\ge 1$, multiplication by the smooth coefficients puts the right side in $H_{\mathrm{loc}}^{m-1}$. Interior elliptic regularity then gives $u\in H_{\mathrm{loc}}^{m+1}$. This elliptic regularity bootstrap proves $u\in H_{\mathrm{loc}}^k$ for every integer $k$.

For any nonnegative integer $\ell$, choose $k>\ell +n/2$. The Sobolev embedding theorem on compact interior subsets gives $u\in C_{\mathrm{loc}}^{\ell }$. Taking all $\ell$ proves $u\in C^{\infty }(\Omega )$, for its smooth representative. The first gain from $L^2$ to $H^1$ is the step supplied by the mollified equation; assuming $H^1$ in the initial definition would miss that step.

The Sobolev embedding theorem $H^2(\Omega )\hookrightarrow L^{\infty }(\Omega )$ in dimension three, combined with the preceding estimate, givesOne direct justification of this embedding is a bounded Sobolev extension operator into $H^2({\mathbb{R}}^3)$, followed by Fourier inversion and the Cauchy-Schwarz inequality: ${\int }_{{\mathbb{R}}^3}(1+|\xi {|}^2{)}^{-2}d\xi <\infty$ bounds the inverse Fourier integral by the $H^2$ norm.

For radial forcing, the preceding part makes $u=U(r)$ and $f=F(r)$. Set $a=1/2$, $b=2$ and $H(r)={\int }_a^r{s}^{2}F(s)ds$. The radial equation gives $r^2{U}^{\prime }(r)=C_0+H(r)$. Both zero boundary values implyMoreover $\Vert H{\Vert }_{\infty }\le ({\int }_a^b{s}^{2}ds{)}^{1/2}({\int }_a^b{s}^2|F{|}^{2}ds{)}^{1/2}\le C\Vert f{\Vert }_2$. Since $r\ge a$, this proves the radial Poisson gradient estimate:

For a nonradial counterexample choose $x_0$ in the shell and a nonconstant $\phi \in C_c^{\infty }(B(0,1))$. For sufficiently small $\epsilon$, set $u_{\epsilon }(x)={\epsilon }^{1/2}\phi ((x-x_0)/\epsilon )$ and $f_{\epsilon }=\Delta u_{\epsilon }$. These are smooth with support away from the boundary. Scaling givesThus the failure of L2-to-Linfinity Poisson gradient bounds in three dimensions is explicit, despite the valid $H^2$ and value-supremum estimates.

Take $\psi \in L^2(U)$, the natural initial-data assumption needed for the requested $L^2$ convergence but not explicitly stated in this subpart. Use the genuine Dirichlet eigensystem from part (b)(iii) and put ${\psi }_m=(\psi ,w_m)$. The spectral construction of a parabolic solution isEach finite sum has zero Dirichlet boundary condition and satisfies $u_t+Lu=0$.

To justify all derivatives, fix $\epsilon >0$. For any integers $k,j\ge 0$, the squared $X_k$ norm of a tail of the $j$th time derivative isuniformly for $t\ge \epsilon$. Use the corrected Sobolev domains of powers of an elliptic Dirichlet operator, not the inaccurate unqualified assertion in the PDF. Their norm equivalence gives convergence in every ordinary $H^k$. The Sobolev embedding theorem then gives convergence of every desired spatial derivative by choosing $k$ large enough. Thus termwise differentiation is justified, $u\in C^{\infty }(U_T)$, and $u(t,\cdot )\in C^{\infty }(\overline{U})$ for every $t>0$. The trace remains zero, and $u_t+Lu=0$ holds pointwise.

Finally Parseval identity givesby the dominated convergence theorem, since each factor tends to zero and is bounded by one. No boundary compatibility of the initial $L^2$ data is required for positive-time smoothness.

Let $\alpha$ be a multi-index with $|\alpha |\le k$. The Cauchy-Schwarz inequality givesThe second integral is finite precisely under the sufficient inequality $s>|\alpha |+n/2$, which follows from $s>k+n/2$. Hence $(i\xi {)}^{\alpha }\hat{u}$ is Lebesgue integrable for every $|\alpha |\le k$.

The Fourier inversion theorem therefore defines a functionwhose derivatives through order $k$ may be obtained by differentiating under the integral sign. Those derivatives are continuous, so $u^{\ast }\in C^k({\mathbb{R}}^n)$. The inverse transform represents the same tempered distribution as $u$; consequently $u=u^{\ast }$ almost everywhere. This is the Sobolev embedding theorem $H^s({\mathbb{R}}^n)\hookrightarrow C^k({\mathbb{R}}^n)$ in this range.