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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.

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.

No: the conclusion need not survive without an additional restriction at infinity. A counterexample that keeps the functional finite is $\Omega ={\mathbb{R}}^n$ and $F(p)=\frac{1}{2}|p{|}^2$. The distinct functions $u\equiv 0$ and $v\equiv 1$ both lie in $C^2(\overline{\Omega })$ and have energy zero. Their boundary data agree because $\partial \Omega$ is empty. For any variation with finite gradient energy,so both are genuine critical points, not merely formal solutions with infinite functional value.

With nonempty boundary, the classical Euler-Lagrange equation also illustrates the missing condition: in dimension $n\ge 3$, on $\Omega =\{|x|>1\}$, both $0$ and $1-|x{|}^{2-n}$ are harmonic functions with zero boundary values. The latter has finite Dirichlet energy, because its gradient-energy integral is proportional to ${\int }_1^{\infty }{r}^{1-n}dr$, and is stationary under compactly supported variations. It approaches $1$ at infinity while the zero solution approaches $0$.

This is nonuniqueness of Dirichlet data without control at infinity. For a global variational formulation one must specify the admissible variations: the exterior example is not critical against every finite-energy zero-boundary variation, since varying it in its own direction changes its energy to first order. Prescribing an appropriate limit at infinity, or a variational space in which the difference is an admissible zero-boundary test function and constants are excluded, can restore uniqueness.

There is a qualification in the printed hypotheses: quadratic ellipticity does not bound a nonsymmetric coefficient matrix. The intended standard existence proof needs $a_{ij}\in L^{\infty }(\Omega )$, or more generally a bounded principal bilinear form. To see why this does not follow from the displayed inequalities, take $\Omega =(-1,1{)}^2$ andThen ${\zeta }^{T}A(x)\zeta =|\zeta {|}^2$ for every $x,\zeta$. However, choose $u_0,{\phi }_0\in C_c^{\infty }(\Omega )$ equal to $x_2,x_1$, respectively, on a small rectangle about the origin. There $(AD{u}_0)\cdot D{\phi }_0=k(x_1)$, whose positive part has infinite integral. Thus the standard principal pairing is not a finite Lebesgue integral even for these smooth test functions. This demonstrates the missing form hypothesis, rather than assuming that all unbounded skew coefficients prevent solvability.

Under the intended bounded-coefficient hypothesis the proof is as follows. On $H=H_0^1(\Omega )$ use the Hilbert space norm $\Vert Dw{\Vert }_2$, equivalent to the $H^1$ norm by the Poincaré inequality on a bounded domain. The form from part (i) is bounded, sinceIt is a coercive bilinear form even without symmetry:Here the nonpositive sign of $q$ is essential. With $u=\psi +w$, the required equation becomesThe Cauchy-Schwarz inequality, bounded coefficients and the Poincaré inequality show that $\ell$ is a bounded linear functional on $H$. The Lax-Milgram theorem states that a bounded coercive bilinear form on a real Hilbert space represents each bounded linear functional in this way with a unique $w\in H$. It therefore yields $u=\psi +w$ with the required boundary condition and weak equation. For completeness, if $u_1,u_2$ are solutions, $z=u_1-u_2\in H_0^1$ satisfies $B(z,z)=0$; coercivity and the Poincaré inequality imply $z=0$.

With the missing bounded-form hypothesis, the weak solution exists and is unique. For nonsymmetric matrices the printed quadratic inequalities omit the hypothesis needed for this standard formulation and proof. No maximum principle is used. In part (iii), symmetry makes the quadratic upper bound a bound on the whole matrix, so no such qualification is needed there.

First use weak compactness of bounded sets in the Hilbert space $L^2(\Omega )$ to pass to a subsequence with $u_k\rightharpoonup u$ globally in $L^2$. The weak lower semicontinuity of the Hilbert norm gives $\Vert u{\Vert }_{L^2(\Omega )}\le K$. This global step also applies if $\Omega$ is unbounded.

Choose a nested exhaustion $U_j⋐U_{j+1}⋐\Omega$ by bounded smooth open sets. Part (i) bounds $u_k$ in $H^1(U_j)$ for every $j$. By the Rellich-Kondrachov compactness theorem, $H^1(U_j)$ embeds compactly in $L^2(U_j)$. Weak compactness in $H^1(U_j)$ and a diagonal subsequence argument therefore give, for the same subsequence,The local limit agrees with the global weak $L^2$ limit. Using smooth exhaustion sets avoids imposing any unprinted boundary regularity on a general ${\Omega }^{\prime }$.

For a smooth compactly supported test function, the weak equation passes to the limit: the fixed bounded coefficients multiply fixed test functions, and local weak gradient convergence handles both the principal and drift terms. The zeroth-order term passes by local $L^2$ convergence. Thus $Lu=f$ weakly, $u\in H_{\mathrm{loc}}^1(\Omega )$, and the global bound already proved gives $u\in S_K$.

To improve convergence of the gradients, put $w_k=u_k-u$. Crucially the forcing is the same for all $k$, so $L{w}_k=0$. Given ${\Omega }^{\prime }⋐\Omega$, choose $j$ with $\overline{{\Omega }^{\prime }}\subset U_j$. Apply part (i) on $U_j$ to the homogeneous equation for $w_k$:This establishes strong local Sobolev compactness for a fixed elliptic equation and proves

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.

Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is $-u^{\prime \prime }+xu$, so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find $u\in V=H_0^1(0,1)$ such thatHere $V$ is the zero-boundary Sobolev space and can be equipped with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The interval Poincaré inequality states $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$, so this norm makes $V$ a Hilbert space and is equivalent to its usual $H^1$ Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, $u^{\prime \prime }=xu+1\in L^2$ gives $u\in H^2$ on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.

Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at $u$. This proves the minimization equivalence, including the sign of the linear term.

Restricting to a conforming finite element space $V_h\subset V$ gives a unique discrete solution and Galerkin orthogonality, $a(u-u_h,v_h)=0$ for every $v_h\in V_h$. The Céa lemma states that if the continuity and coercivity constants are $M$ and $\alpha$, then $\Vert u-u_h{\Vert }_V\le (M/\alpha ){\inf }_{v_h\in V_h}\Vert u-v_h{\Vert }_V$. For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.

Weak convergence of distributions means $⟨T_n,\phi ⟩\to ⟨T,\phi ⟩$ for every test function $\phi \in C_c^{\infty }(\Omega )$. The distributional derivative is defined byIt is linear and continuous: differentiation preserves compact support and bounds each test-function seminorm by one of higher derivative order, so composing it with the continuous functional $T$ again gives a distribution. Finally, for every fixed test function,Thus distributional differentiation preserves the stated convergence.

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.

Let $\varphi$ be a test function supported in $[-R,R]$. Symmetric cancellation givesThe integrand is bounded by $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, so the limit exists andThis is a continuous linear functional on every fixed-support test space, hence a distribution. It also specifies its local order as at most one.

For the boundary value, split the ordinary complex denominator:The real part paired with $\varphi$ isIt converges to the principal value by dominated convergence theorem, using the same bound $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. In the second part set $x=\epsilon t$:Here $\Vert \varphi {\Vert }_{\infty }/(1+t^2)$ is an integrable majorant. The correct Sokhotski–Plemelj formula is thereforeThe resulting functional is continuous, with bound $2R\Vert {\varphi }^{\prime }{\Vert }_{\infty }+\pi \Vert \varphi {\Vert }_{\infty }$, so it too is a distribution.

The printed PDF has the wrong sign on its delta term. For a real even nonnegative test function with $\varphi (0)>0$, the principal value is zero while the imaginary part of every regularized integral is negative. Its limit is $-\pi \varphi (0)$, disproving the printed plus sign directly. A plus sign instead belongs to the boundary value $(x-i0{)}^{-1}$.

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.

Fix a compact $K\subset \Omega$ and choose $\delta >0$ such that a closed $3\delta$-neighbourhood of $K$ is compactly contained in $\Omega$. The hypothesis gives a common bound $M$ on that neighbourhood. Choose a smooth radial mollifier ${\rho }_{\delta }$ supported in $B_{\delta }(0)$ with integral one. The mean value property for harmonic functions, integrated against its radial weights, givesin a neighbourhood of $K$. Differentiating this fixed-radius convolution gives the uniform interior estimateThus the sequence is equicontinuous there; taking its pointwise limit gives the same local modulus of continuity for $u$. A finite small-ball cover of $K$ now proves uniform convergence directly: choose the cover so both $u_j$ and $u$ vary by less than $\epsilon$ inside each ball, then use convergence at its finitely many centres to make $|u_j-u|<\epsilon$ there. The triangle inequality gives ${\sup }_K|u_j-u|<3\epsilon$ for large $j$.

Finally, pass to the limit in the reproducing convolution identity on smaller compact sets. It yields $u=u\ast {\rho }_{\delta }$, so $u$ is smooth locally. Also, for each compactly supported smooth test function $\varphi$, uniform convergence on its support givesHence its smooth Laplacian is zero. We have proved $u_j\to u$ uniformly on every compact subset and $u$ is harmonic. Using the continuous representatives in the pointwise hypothesis makes this conclusion pointwise, rather than only almost everywhere. This is compact convergence of locally bounded harmonic functions.

Inside the fan $u=x/t$, one has $u_t=-x/t^2$ and $u{u}_x=x/t^2$, while the constant regions also satisfy the equation. Across $x=0$ and $x=t$, the two traces of both $u$ and its flux $u^2/2$ agree. Integrating by parts separately in the three regions therefore cancels all interface terms and givesfor every smooth compactly supported test function. The initial term is correct because $u(\cdot ,t)\to 1_{\{x>0\}}$ in local $L^1$: the difference is supported in $(0,t)$ and its integral is at most $t$. Thus the fan is a weak solution of the Burgers equation with the required initial data.

Write $[u]=u_r-u_l$ and $[f]=u_r^2/2-u_l^2/2$. Splitting the test-function integral along $x=s(t)$ and integrating by parts gives a remaining interface contributionFor example the normal to the interface in $(x,t)$ coordinates is proportional to $(1,-s^{\prime })$, so the jump in the conserved spacetime flux $(u^2/2,u)$ has normal component $[f]-s^{\prime }[u]$. Since the traces of the test function are arbitrary, the interior weak equation holds if and only if $s^{\prime }[u]=[f]$. At a genuine jump $[u]\ne 0$, this isThis is the Rankine-Hugoniot condition. An initial-value weak formulation also requires the prescribed initial trace; the jump condition alone concerns the interior equation. The source's reference to “(1)” is to the displayed Burgers equation, not a different PDE.

The Gaussian free field is a random distribution, so its purported level curve cannot be defined by evaluating its height at every point. A coupling must instead be stated using conditional means and covariances. The SLE4 coupling with a Gaussian free field does exactly this.

Choose the Green-function normalizationA Zero-boundary Gaussian free field $H_0$ is characterized on real compactly supported smooth test functions byEquivalently its energy normalization is the Dirichlet inner product $(2\pi {)}^{-1}\int \nabla f\cdot \nabla gdA$. This fixes the otherwise convention-dependent height constant.

Set $\lambda =\pi /2$ andThe field $H$ has prescribed Dirichlet values $-\lambda$ on the negative half-line and $+\lambda$ on the positive half-line. Subtracting $m_0$ recovers the zero-Dirichlet field $H_0$. It is the shifted field $H$, rather than an unshifted zero-boundary field, whose zero-height interface has ordinary chordal ${\mathrm{SLE}}_4$ law.

Let $\eta$ be chordal ${\mathrm{SLE}}_4$ from $0$ to $\infty$, with Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-W_t)$ and $W_t=2{\beta }_t$. Write $D_t=\mathbb{H}\setminus \eta [0,t]$ and $f_t=g_t-W_t$. The coupling statement is:where, conditionally on the curve, $H_0^{D_t}$ is an independent zero-boundary field in the slit domain. The equality is for restrictions to test functions in that domain, and extends to suitable stopping times. Its covariance is $G_{D_t}(z,w)=G_{\mathbb{H}}(f_t(z),f_t(w))$ by conformal invariance. The two sides of the revealed slit have heights $-\lambda$ and $+\lambda$, in the order specified by their real images under $f_t$. This is the continuum meaning of a Gaussian free field level line.

The fundamental calculations explain why the parameter is four. For $Z_t=f_t(z)$, the Itô formula givesAt $\kappa =4$ the drift vanishes. ThusEach harmonic mean is a bounded martingale before approaching its point. Differentiating the explicit conformally transformed Green function gives the Loewner variation of the Dirichlet Green functionThereforeIn words, the variance learned from the evolving mean exactly equals the covariance lost when the slit is removed. For a Green kernel $cG$ the same calculation gives $\lambda =(\pi /2)\sqrt{c}$; using a different Green normalization changes the height constant, not the SLE parameter.

To construct the coupling, first sample the SLE curve. In the two components to its left and right, sample independent zero-boundary fields and add the corresponding constant heights. Extend these as distributions to obtain the candidate full field. The following martingale identity proves its marginal law rather than merely matching its first two moments.

For a real test function $\phi$, put $M_t=(m_t,\phi )$ andLocalization permits integration of the pointwise identities. They yield $d⟨M{⟩}_t=-d{V}_t$. Henceis a complex local martingale: the $-\frac{1}{2}d{V}_t$ drift cancels the $-\frac{1}{2}d⟨M{⟩}_t$ Itô correction. Since $V_t\ge 0$, $|{\mathcal{Z}}_t|\le 1$, making it a true martingale. At the complete-curve limit, $m_t$ becomes the constant height in each component and the remaining Green kernel becomes that of those components. Thus ${\mathcal{Z}}_{\infty }$ is the conditional characteristic function of the sampled candidate field. Taking expectations givesApplying this to every linear combination of test functions proves the entire Gaussian law, not only its covariance. The conditional version $\mathbb{E}({\mathcal{Z}}_{\infty }\mid {\mathcal{F}}_t)={\mathcal{Z}}_t$ proves the displayed conditional-field statement. Exhaustion by compact test supports justifies passage across the slit and the limiting distributional extensions.

This revealed curve is a local set of a Gaussian free field: conditionally on it, the remaining field is a zero-boundary field plus a specified harmonic function. The spatial Markov property is thereby preserved at random domains, a property not available for arbitrary field-dependent sets. One can strengthen the coupling to a curve measurable from the field. The proof explores compatible interfaces in small subdomains and uses the conditional boundary heights and monotonicity to show that two such interfaces for the same field cannot separate; a countable exhaustion gives uniqueness. This supplies a rigorous replacement for the informal phrase “draw the zero contour.” It does not assert that the distribution has pointwise values.

The coupling is useful in both directions: SLE martingales give exact conditional Gaussian free field data, while the Gaussian Markov structure explains the SLE domain Markov property and its distinguished parameter. With the above normalization, the height jump is $2\lambda =\pi$, and the corresponding interface is chordal SLE4.