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For an open set $U\subseteq {\mathbb{R}}^n$ and $1\le p\le \infty$, the first-order Sobolev space iswhere each $D_{i}u$ is a weak derivative. Its standard norm is $\Vert u{\Vert }_{W^{1,p}}=\Vert u{\Vert }_p+{\sum }_i\Vert D_{i}u{\Vert }_p$.

Every element of $W^{1,p}(a,b)$ for $1\le p<\infty$ has an absolutely continuous representative. Its ordinary derivative exists almost everywhere, equals its weak derivative, and belongs to $L^p(a,b)$.

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

Solved by gpt-5.6-sol high.

For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

Solved by gpt-5.6-sol high.

Let $g=Du\in L^p(0,1)$ be the weak derivative and defineThe fundamental theorem of calculus for Lebesgue integration makes $F$ an absolutely continuous function, differentiable almost everywhere, with $F^{\prime }=g$ almost everywhere. The distributional derivative of $u-F$ is zero. A locally integrable function with zero distributional derivative on a connected interval is equal almost everywhere to a constant $C$. Consequentlyis an absolutely continuous representative of $u$, and ${\tilde{u}}^{\prime }=g\in L^p(0,1)$ almost everywhere.

Solved by gpt-5.6-sol high.

Write points of $U$ as $(x,y)$ with $x>0$. For $h>0$, translate $u$ into the domain byContinuity of translations in $L^p$ applied to $u,D_{x}u,D_{y}u$ givesas $h\downarrow 0$. Choose a standard mollifier ${\rho }_{\epsilon }$ supported in a ball of radius $\epsilon <h/2$. For $(x,y)\in U$, the convolutiononly samples points with first coordinate greater than $-h$, so it is well-defined and smooth throughout $U$. The approximation-to-the-identity theorem, applied also to each weak derivative, allows $\epsilon (h)<h/2$ to be chosen so thatTaking $h=1/j$ and using the triangle inequality proves the density of smooth functions in a Sobolev space.

Solved by gpt-5.6-sol high.