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Recall that $g_i\in L_{\mathrm{loc}}^1(\Omega )$ is the weak derivative $D_{i}u$ if $u\in L_{\mathrm{loc}}^1(\Omega )$ and $\int u{D}_i\varphi =-\int g_i\varphi$ for all $\varphi \in C_c^{\infty }(\Omega )$. A global weak derivative restricts to every open neighborhood simply by taking test functions supported there.

Conversely, suppose local weak derivatives exist near each point. local integrability of $u$ follows by a finite cover of each compact subset. On overlapping neighborhoods the two candidates for $D_{i}u$ have identical integrals against every smooth compactly supported test function. The fundamental lemma for locally integrable functions therefore makes them equal almost everywhere. Choose a countable local cover, possible in Euclidean space; the countably many exceptional overlap sets can all be discarded. The derivatives consequently patch to measurable $g_i\in L_{\mathrm{loc}}^1(\Omega )$.

For a test function $\varphi$, take a smooth partition of unity $({\chi }_j)$ subordinate to the local cover and locally finite on its support. Only finitely many terms occur there. Because ${\sum }_j{\chi }_j=1$ near the support,The derivatives of the partition sum to zero, explaining the first equality. Thus the patched functions are the global weak derivatives. This proves both directions of locality of weak differentiability.

By locality of weak differentiability, it is enough to work near the exceptional point; translate it to zero. Let $g_i$ denote the classical derivative off zero, assigning any value at zero. For $\varphi \in C_c^{\infty }(\Omega )$ choose a smooth cutoff ${\chi }_{\epsilon }$ which is zero on $B_{\epsilon }$, one outside $B_{2\epsilon }$, satisfies $|D{\chi }_{\epsilon }|\le C/\epsilon$, and has its transition region contained in $\Omega$. The product $\varphi {\chi }_{\epsilon }$ avoids the singular point, so classical integration by parts givesThe middle term has absolute value at mostwhen $n\ge 2$. The other terms tend to $\int u{D}_i\varphi$ and $-\int g_i\varphi$, since $u$ is locally integrable and $g_i\in L_{\mathrm{loc}}^1$. Therefore $\boxed{D_{i}u=g_i\text{ weakly on all of }\Omega }$. Its value at the single point is irrelevant. This is bounded isolated-point removability for weak first derivatives.

For $n=1$, take $u(x)=0$ for $x<0$ and $u(x)=1$ for $x>0$ on $(-1,1)$, with any value at zero. It is bounded and $C^1$ off zero, with classical derivative zero there. Butso its distributional derivative is a Dirac delta. If a locally integrable weak derivative existed, it would be zero off zero and hence zero almost everywhere, contradicting this identity. The step function is the required counterexample; the cutoff error in dimension one need not vanish.

Put $\delta =\mathrm{dist}({\Omega }^{\prime },\partial \Omega )>0$. Choose a smooth cutoff $0\le \eta \le 1$, equal to one on ${\Omega }^{\prime }$, supported within its $\delta /2$ neighborhood, with $|D\eta |\le C_n/\delta$. Such a cutoff is obtained by smoothing a distance cutoff; its support is relatively compact in $\Omega$. Let $0<|h|<\delta /8$, fix a coordinate $k$, and put $z={\Delta }_k^{h}u$. It lies in $W^{1,2}$ on the cutoff neighborhood because both translated copies of $u$ do. Extend ${\eta }^{2}z$ by zero, and use the test functionThe weak equation is $\int Du\cdot D\varphi =-\int f\varphi$. The discrete integration-by-parts identity $\int a{\Delta }_k^{-h}b=-\int ({\Delta }_k^{h}a)b$, valid by translation with these supports, turns it intoWrite $X=\Vert \eta Dz{\Vert }_2$, $U=\Vert z{\Vert }_{L^2(\mathrm{supp}\eta )}$, $A=\Vert D\eta {\Vert }_{\infty }$ and $F=\Vert f{\Vert }_2$. Expanding the left side gives $X^2$ plus a cross term of absolute value at most $2AUX$. Applying part (a) on the whole space to the zero extension of ${\eta }^{2}z$ bounds the right side byConsequentlywhere the last step is the elementary Young inequality. Part (a), on a slightly larger cutoff neighborhood if needed, gives $U\le \Vert D_{k}u{\Vert }_{L^2(\Omega )}$. We conclude thatuniformly as $h\to 0$.

To obtain actual second weak derivatives, choose $h_j\to 0$. For each $i,k$, the bounded sequence ${\Delta }_k^{h_j}{D}_{i}u$ has a weakly convergent subsequence in $L^2({\Omega }^{\prime })$, say to $g_{ik}$. For every smooth compactly supported $\psi$ in ${\Omega }^{\prime }$,Thus $g_{ik}=D_k{D}_{i}u$ is the second weak derivative. weak lower semicontinuity gives the same uniform bound for these derivatives. Adding the already known zeroth and first derivative norms and summing over the finitely many coordinates provesAll derivative existence and estimates have been established by the difference-quotient argument; no second derivatives were assumed in choosing the test. This proves the Interior second-derivative estimate for the Poisson equation in the stated setting.

We prove the Interior second-derivative estimate for the Poisson equation using difference quotients, rather than differentiating the $L^2$ forcing. Fix a coordinate $e_k$ and write ${\delta }_{h}v(x)=(v(x+h{e}_k)-v(x))/h$. Choose a smooth cutoff function $\eta$ equal to one on ${\Omega }^{\prime }$ with support compactly contained in $\Omega$, leaving room for all translations with sufficiently small $|h|$. We can arrange $|D\eta |\le C(n)/d$, where $d=\mathrm{dist}({\Omega }^{\prime },\partial \Omega )$.

The standard difference quotient lemmas give discrete integration by parts, $\int a{\delta }_{-h}b=-\int ({\delta }_{h}a)b$, and $\Vert {\delta }_{h}v{\Vert }_2\le \Vert D_{k}v{\Vert }_2$ on the relevant enlarged set. In the weak formulation $\int Du\cdot D\phi =-\int f\phi$, use the admissible test function $\phi =-{\delta }_{-h}({\eta }^2{\delta }_{h}u)$. Moving a difference quotient onto $Du$ givesSet $X=\Vert \eta D{\delta }_{h}u{\Vert }_2$ and $Y=\Vert D\eta {\delta }_{h}u{\Vert }_2$. The cross term has absolute value at most $2XY$. Since $0\le \eta \le 1$, the difference quotient lemma on the right givesThe Young inequality now implies $X^2\le C(Y^2+\Vert f{\Vert }_2^2)$. Moreover $Y\le C(n)d^{-1}\Vert Du{\Vert }_{L^2(\Omega )}$, independently of $h$. Hence, for every coordinate $k$,By the Sobolev characterization by bounded difference quotients, applied to each first weak derivative on smaller interior sets, these uniform bounds give all second weak derivatives. Their bounds pass to the weak limit as $h\to 0$. Thus $u\in W_{\mathrm{loc}}^{2,2}(\Omega )$ andNo derivative of $f$ is used, and no boundary condition is needed for this interior elliptic regularity argument.

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.

For every compactly supported smooth test function $\varphi$, integration by parts in the definition of the weak derivative givesBoth $\varphi$ and ${\partial }_{\ell }\varphi$ belong to $L^2({\mathbb{R}}^n)$, so the assumed weak limits implyThis is exactly the distributional characterization of ${\partial }_{\ell }g=h_{\ell }$. Since $g,h_1,\dots ,h_n$ are all in $L^2$, the defining condition of the H1 space is satisfied. ThusFor complex functions the same argument uses conjugated test functions in the $L^2$ pairing. No prior identification of an $H^1$ weak limit is needed.

Let $I=(-1,1)$ and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm $\Vert v{\Vert }_{H^2}^2=\Vert v{\Vert }_2^2+\Vert v^{\prime }{\Vert }_2^2+\Vert v^{\prime \prime }{\Vert }_2^2$. The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the $H^2$ closure of $C_c^{\infty }(I)$, so it is a closed Hilbert space. The derivative boundary conditions are essential, and $H_0^1(I)$ alone is insufficient.

Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace $a$ by its sesquilinear version, take the real part of the forcing term and use absolute squares in $J$; the same argument applies.

Its first variation is $J^{\prime }(u)[v]=a(u,v)-\ell (v)$. Thus stationary points satisfy $a(u,v)=\ell (v)$ for every $v\in H$, the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because $v=v^{\prime }=0$ at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution. This avoids assuming nonexistent classical derivatives of $p$ or $q$.

Subtract the two weak solution identities and test with their difference $w\in H^1(U)$. ThenA Sobolev function with zero weak gradient is constant on each connected component. One justification is to mollify locally: each mollification has zero gradient and is constant on its ball, and overlaps identify the constants; taking limits gives the original assertion. Since $U$ is connected, $w$ is one constant on $U$.

Conversely, adding a constant changes neither the weak derivative nor the weak identity. The solution is unique up to an additive constant.

For $I=(-a,a)$, the first-order Sobolev space isThe weak derivative $v=u^{\prime }$ is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm $\Vert u{\Vert }_{H^1(I)}^2=\Vert u{\Vert }_{L^2(I)}^2+\Vert u^{\prime }{\Vert }_{L^2(I)}^2$. Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.

Reflect $u$ across the two endpoints of $I$ into a slightly larger interval, giving an $H^1$ extension, and multiply by a smooth cutoff function equal to one near $\overline{I}$. Convolve the resulting function on $\mathbb{R}$ with a mollifier. The convolutions $u_{\epsilon }$ are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in $L^2$ proves both $u_{\epsilon }\to u$ and $u_{\epsilon }^{\prime }\to u^{\prime }$ in $L^2(I)$, hence convergence in the Sobolev norm.

To obtain convergence almost everywhere, select a sequence with $\Vert u_n-u{\Vert }_2\le 2^{-n}$. Then $\int {\sum }_n|u_n-u{|}^2<\infty$, so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in $H^1$ and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.

With only $u\in L^2(\Omega )$ available, use a distributional weak solution: for every test function $\eta \in C_c^{\infty }(\Omega )$ requireAll terms are meaningful because the smooth coefficients are bounded on the compact support of the test function and $u$ is locally integrable. This is exactly $\Delta u={\sum }_i{D}_i(b^{i}u)+cu+f$ as an identity of distributions. No first weak derivative of $u$ is assumed in this definition, and no boundary condition is being imposed.

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.

A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

Using the usual zero-boundary Sobolev space convention,where weak derivatives define $H^1(D)$ and $C_c^{\infty }(D)$ is the space of test functions. In the Gaussian free field convention, the same notation often denotes the Dirichlet energy space, the completion in the gradient norm alone. On bounded domains the Poincaré inequality makes the two definitions equivalent; on unbounded domains one must distinguish them. The following gradient-pairing argument applies in either setting whenever the energy completion is realized as weak functions.

Identify $H_{\mathrm{supp}}=H_0^1(U)$ with a vector subspace of $H_0^1(D)$ by zero extension of H01. Approximating by test functions in $U$ shows that this is an isometric embedding in the inhomogeneous Sobolev norm and also in the Dirichlet inner product norm. In particular, arbitrary irregularity of $\partial U$ causes no additional boundary term.

DefineThese are weakly harmonic Sobolev functions. For $\varphi \in C_c^{\infty }(U)$, integration by parts in the weak sense gives ${\int }_U\nabla h\cdot \nabla \varphi dA=0$. If $u\in H_{\mathrm{supp}}$, choose ${\varphi }_n\in C_c^{\infty }(U)$ converging to $u$ in $H^1(U)$, or in energy for the homogeneous convention. The Cauchy-Schwarz inequality givesHence the orthogonality of supported and harmonic Dirichlet functions isBoth are linear vector subspaces. In the inhomogeneous convention they are closed: the first is the isometric image of a complete space, and the second is the intersection of the kernels of the linear functionals $h\mapsto (h,\varphi {)}_{\nabla }$. No spanning assertion is needed. The PDF contains this orthogonality statement; the TeX has badly corrupted it into an assertion about openness.