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For $n\ge 3$ and a bounded smooth domain, a Dirichlet Laplacian eigenfunction with eigenvalue $\lambda$ satisfies the displayed estimate. A signed power test for a Laplacian eigenfunction and the zero-boundary Sobolev inequality give a Moser iteration with exponents $p_j=2[n/(n-2){]}^j$. The geometric series ${\sum }_{j}1/p_j=n/4$ gives the exact power of $\lambda$, while ${\sum }_j(\log p_j)/p_j<\infty$ controls the remaining constant. The constant depends only on dimension, because the zero-boundary Sobolev inequality follows by extension by zero to the whole space.

On a bounded three-dimensional domain, interpolation between $L^2$ and the Sobolev inequality $H^1\hookrightarrow L^6$ gives

For $u_{tt}-\Delta u+u^3=0$ in three spatial dimensions, conserved positive wave energy $E$ controls $\Vert \nabla u{\Vert }_2$. The Sobolev inequality gives $\Vert u^2\nabla u{\Vert }_2\lesssim \Vert \nabla u{\Vert }_2^2\Vert D^{2}u{\Vert }_2\lesssim E\Vert D^{2}u{\Vert }_2$. Differentiated wave energy estimates and the Gronwall inequality then give $\Vert u(t){\Vert }_{{\dot{H}}^2}+\Vert u_t(t){\Vert }_{{\dot{H}}^1}\le CD{e}^{CEt}$, where $D$ bounds the corresponding initial Sobolev norms.

In three spatial dimensions, with $Z$ the eleven translations, spatial rotations, boosts and scaling commutation vector fields for the wave equation, a sufficiently decaying smooth function satisfies $|u(t,x)|\le C(1+t+|x|{)}^{-1}(1+|t-|x||{)}^{-1/2}{\sum }_{|I|\le 2}\Vert Z^{I}u(t){\Vert }_2$. This weighted Sobolev inequality converts control of commuted wave energy into decay. It holds independently of any wave equation satisfied by $u$.

Set $N={\int }_V{P}^{2}dV$, $D={\int }_{{\mathbb{R}}^3}|\nabla P{|}^{2}dV$, $M={\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$, and $q={\max }_V|Q|$. The confined flow satisfies the impermeability condition $\mathbf{u}\cdot \mathbf{n}=0$, so $Q=0$ at $r=a$. Multiplying the radial magnetic induction scalar equation by $P$ and integrating, the advection term vanishes by the divergence theorem:The same integration by parts, using $\nabla \cdot \mathbf{B}=0$, gives

For diffusion, Green's first identity in the sphere gives ${\int }_{V}P\Delta P=-{\int }_V|\nabla P{|}^2+{\int }_{\partial V}P{\partial }_{r}P$. In the exterior, $\Delta P=0$ and the normal on its inner boundary is $-\hat{\mathbf{r}}$, soThe infinity term vanishes for the isolated decaying field. The insulating boundary condition for the radial magnetic scalar permits matching the two traces, and therefore the exterior contribution is essential: it turns the diffusive term into $-\eta D$. Altogether,Applying the pointwise Cauchy-Schwarz inequality and then its integral version,Thus a nonzero field with stationary or growing $N$ must satisfy the radial-flow dynamo energy bound:

The temporal qualification matters. For general time-dependent dynamo action, nondecay does not make $N^{\prime }\ge 0$ at every instant; the boxed bound applies at nondecreasing instants, to a marginal or growing mode, or in an appropriate long-time sense. A uniform strict violation, $q\sqrt{M/D}\le \eta -\epsilon$ with $\epsilon >0$, gives $N^{\prime }\le -2\epsilon D$. The whole-space Sobolev inequality and Holder inequality give $N\le |V{|}^{2/3}\Vert P{\Vert }_6^2\le C_{a}D$, forcing exponential decay of $N$. Alternatively, for a bounded statistically steady field, time averaging the energy balance and applying Cauchy-Schwarz once more gives $q_{\ast }^2\ge {\eta }^2⟨D⟩/⟨M⟩$, where $q_{\ast }={\sup }_{t}q(t)$. Sustained dynamo action cannot have a uniform negative dissipation gap.

As usual, the bounded set is assumed Lebesgue measurable and an $L^q$ exponent is positive; the conventional Banach-space assertion concerns $1\le q<2^{\ast }=2n/(n-2)$. No boundary regularity of $\Omega$ is needed. Choose a smooth compactly supported cutoff $\eta$ equal to one on a ball containing $\Omega$. Weak convergence and the Uniform boundedness principle give a common $H^1$ bound for $f_j$, hence also for $g_j=\eta f_j$. These functions have a common compact support.

We prove their local $L^2$ compactness directly. Translation and the fundamental theorem along line segments give, by density from smooth functions,For a compactly supported mollifier ${\rho }_{\epsilon }$, integration of this estimate givesAt fixed $\epsilon$, Hölder's inequality bounds $\Vert g_j\ast {\rho }_{\epsilon }{\Vert }_{\infty }$ and $\Vert \nabla (g_j\ast {\rho }_{\epsilon }){\Vert }_{\infty }$ uniformly by the $L^2$ norm of the corresponding smoothing kernels. Their supports lie in one fixed compact set, so the Arzelà-Ascoli theorem makes this smoothed family precompact in $L^2$. The uniform $O(\epsilon )$ approximation makes the original family totally bounded in $L^2$: choose a finite net for the smoothed family and add the approximation error. Completeness gives relative compactness.

Every strongly convergent subsequence of $g_j$ has limit $\eta f$, since multiplication by $\eta$ preserves weak $L^2$ convergence. Consequently the whole sequence converges strongly: otherwise a subsequence with distance at least a fixed positive number would have a further convergent subsequence and contradict that unique limit. In particular $\Vert {\chi }_{\Omega }(f_j-f){\Vert }_2\to 0$.

The critical Sobolev inequality supplies a uniform bound in $L^{2^{\ast }}$. For clarity, this estimate follows from part 2(a), not from compactness at the critical exponent. For a compactly supported smooth $v$, integrating by parts against the vector field $(x-y)/|x-y{|}^n$ givesIts distributional divergence is the sphere-area multiple of a point mass; applying the Hardy–Littlewood–Sobolev inequality with $\alpha =1,p=2$ yields $\Vert v{\Vert }_{2^{\ast }}\le C\Vert \nabla v{\Vert }_2$. Density extends this to $H^1({\mathbb{R}}^n)$.

For $2<q<2^{\ast }$ choose $0<\theta <1$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Hölder's inequality givesFor $1\le q<2$, use $\Vert v{\Vert }_{L^q(\Omega )}\le |\Omega {|}^{1/q-1/2}\Vert v{\Vert }_{L^2(\Omega )}$. The same integral estimate also works for $0<q<1$ if $L^q$ is understood with its usual quasi-norm. Thus ${\chi }_{\Omega }{f}_j\to {\chi }_{\Omega }f$ strongly for every positive $q<2^{\ast }$. We have never asserted that multiplying by the possibly nonsmooth indicator preserves $H^1$.

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.

For an integer $s\ge 0$, the Sobolev space iswhere $D^{\beta }$ is a weak derivative. For $p<\infty$ one may use the Sobolev norm $({\sum }_{|\beta |\le s}\Vert D^{\beta }u{\Vert }_p^p{)}^{1/p}$; for $p=\infty$ use the maximum of the finitely many essential-supremum norms.

For $1\le p<d$, the Sobolev inequality is $\Vert u{\Vert }_{p^{\ast }}\le C_{d,p}\Vert \nabla u{\Vert }_p$, with Sobolev conjugate exponent $p^{\ast }=dp/(d-p)$. For $d<p<\infty$, Morrey's inequality supplies a continuous representative satisfyingAt $p=\infty$ the representative is Lipschitz continuous. At the critical exponent $p=d>1$, first-order Sobolev regularity gives every finite $L^q$ embedding for $q\ge d$, with an inhomogeneous norm, but generally no $L^{\infty }$ embedding. The one-dimensional endpoint $W^{1,1}(\mathbb{R})\hookrightarrow L^{\infty }(\mathbb{R})$ is an exception.

For the proof of Morrey's inequality, start with a smooth $u$ and write $u_{B(x,r)}$ for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables givesThe last step is the Holder inequality; integrability of the kernel to power $p^{\prime }$ is exactly $(d-1)p^{\prime }<d$. For $r=|x-y|$, translate the averaging ball along the segment from $x$ to $y$. The fundamental theorem of calculus along a line segment and the Holder inequality giveCombining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with $r=1$, together with $|u_{B(x,1)}|\le |B_1{|}^{-1/p}\Vert u{\Vert }_p$, gives the supremum estimate. Density of smooth functions in a Sobolev space then gives a uniformly convergent sequence of smooth representatives, preserving both bounds. For $p=\infty$, mollification gives the Lipschitz version.

For the decay conclusion assume $d<p<\infty$. The representative is uniformly continuous. If $|u(x_j)|\ge 2\theta >0$ along points escaping to infinity, the Hölder bound gives a radius $\rho >0$, independent of $j$, on which $|u|\ge \theta$. A subsequence has disjoint radius-$\rho$ balls, each contributing at least ${\theta }^p|B_{\rho }|$ to $\Vert u{\Vert }_p^p$, a contradiction. This is uniformly continuous integrable functions vanish at infinity.

The finite-$p$ restriction is necessary. If the printed range includes $p=\infty$, its decay assertion is false: $u\equiv 1$ belongs to $W^{1,\infty }$ but does not tend to zero.

Work with real-valued functions. The screened sine-Gordon energy is well defined: $0\le 1-\cos u\le u^2/2$, and $fu$ is integrable by the Cauchy-Schwarz inequality. It is coercive, becauseA minimizing sequence is bounded in the Hilbert space $H^1$. Weak sequential compactness of bounded sequences in a reflexive Banach space supplies a weakly convergent subsequence. The squared H1 space norm is weakly lower semicontinuous, the source pairing is weakly continuous, and the nonlinear term is covered by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. The direct method in the calculus of variations therefore gives a minimizer $\varphi$.

Taking its first variation in any $v\in H^1$ givesThus the Euler-Lagrange equation isas a weak solution, equivalently in distributions when tested against smooth compactly supported functions. The derivative of the nonlinear term is justified by $|\sin \varphi |\le |\varphi |$ and the second-order remainder bound $|\cos |\le 1$.

Now $G=f-\sin \varphi \in L^2$, so the supplied elliptic regularity estimate puts $\varphi$ in $H^2$. Applying the Sobolev inequality to $\varphi$ and each first weak derivative gives $\varphi \in W^{1,6}({\mathbb{R}}^3)$. Morrey's inequality and uniformly continuous integrable functions vanish at infinity prove that its continuous representative tends to zero.

The final printed supremum estimate is false in general. The maximum bound for a monotone reaction term involves $h(s)=s+\sin s$, which is odd and strictly increasing: $h^{\prime }=1+\cos s\ge 0$, and its zeros are isolated. If $M=\Vert f{\Vert }_{\infty }<\infty$, a positive maximum $m$ of $\varphi$ satisfies $h(m)\le f\le M$; apply the same argument to $-\varphi$. The valid general estimate isThe bound by $M$ is valid if $M\le \pi$, but $h(s)$ can be smaller than $s$ for larger positive $s$.

For an explicit failure of the source-size bound for the screened sine-Gordon equation, take $a=3\pi /2$, $L=10$, $\varphi (x)=a{e}^{-|x{|}^2/L^2}$ and $f=-\Delta \varphi +\varphi +\sin \varphi$. These are smooth functions in the required spaces. With $q=|x{|}^2/L^2$,Hence $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert \varphi {\Vert }_{\infty }$. Moreover $V(s)=s^2/2+1-\cos s$ has $V^{\prime \prime }=1+\cos s\ge 0$, so the energy is convex and this critical point is a minimizer. The example therefore satisfies even the minimizing and $C^2$ hypotheses of the printed claim.

Put $p_j=2{\kappa }^j$, where $\kappa =n/(n-2)>1$. Since $p_j^2/(p_j-1)\le 2p_j$, part (ii), followed by a $p_j$-th root, gives the Moser iteration stepAfter $N$ steps,The geometric series and its differentiated form giveIn particular the exact convergent product isThe printed hint's equality to a single power of $2\kappa$ is not correct in general; its claimed finiteness is correct and the displayed expression supplies the correction.

On a finite-measure domain, Lp norms converge to the supremum norm for a bounded continuous function. Indeed, the upper bound is $\Vert u{\Vert }_p\le |\Omega {|}^{1/p}\Vert u{\Vert }_{\infty }$; for every $a<\Vert u{\Vert }_{\infty }$, the set $\{|u|>a\}$ has positive measure, giving $\Vert u{\Vert }_p\ge a|\{|u|>a\}{|}^{1/p}$. Let $N\to \infty$ in the iteration and use part (i), which gives $\lambda >0$. The Dirichlet eigenfunction supremum estimate isThe eigenvalue exponent is exactly $n/4$. The constant depends only on $n$: the supplied zero-boundary Sobolev inequality has a dimension-only constant, and every product factor above depends only on $\kappa$.

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.

Multiply the Dirichlet Laplacian eigenfunction equation by $u$ and use integration by parts. The zero Dirichlet boundary condition removes the boundary term and givesThe denominator is positive because the eigenfunction is nonzero. The numerator is also positive: if it were zero, the given Sobolev inequality for $u\in W_0^{1,2}(\Omega )$ would force $u=0$. Therefore the Dirichlet Laplacian eigenvalue has the positive Rayleigh quotientEvery nonzero Dirichlet eigenfunction in this problem has a positive eigenvalue.

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.

Here $w$ is the harmonic replacement of $u$ in $B_r=B(x_0,r)$, so $v=u-w\in H_0^1(B_r)$. Subtract the weak equations and test with $v$:Because $B_r\subset B(0,1)$, $|x_1|\le 1$. The Sobolev inequality $\Vert v{\Vert }_6\le C\Vert \nabla v{\Vert }_2$ and the Holder inequality giveAfter division and squaring,

After passing to a subsequence, weak compactness and the Rellich-Kondrachov compactness theorem giveFor fixed $w\in H^1(U)$, the Sobolev inequality gives $w\in L^6(U)$, soMeanwhile ${\partial }_j{u}_{n_k}\rightharpoonup {\partial }_{j}u$ in $L^2(U)$. Pairing this weak convergence with the strong convergence of the products, or equivalently using the weak-strong product convergence lemma, yields

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

The Holder inequality interpolates between $L^2(U)$ and $L^6(U)$:Taking cube roots and applying the three-dimensional Sobolev inequality $H^1(U)\hookrightarrow L^6(U)$ givesThis is the H1 L3 interpolation inequality in three dimensions.

In three dimensions, the Sobolev inequality gives $\Vert v{\Vert }_{L^6(U)}\le C_U\Vert v{\Vert }_{H_0^1(U)}$. Thereforeand hence

Using $w^3-{\tilde{w}}^3=(w-\tilde{w})(w^2+w\tilde{w}+{\tilde{w}}^2)$, the Holder inequality and the same Sobolev embedding give at each timeTaking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.