First-order Sobolev space ($W^{1,p}(U)$)

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 $\Vert u{\Vert }_{W^{1,p}}=\Vert u{\Vert }_p+{\sum }_i\Vert D_{i}u{\Vert }_p$.

For a Sobolev space function, almost every line parallel to a coordinate axis gives an absolutely continuous restriction. An absolutely continuous function has derivative zero almost everywhere on each of its level sets. Slicing and Fubini's theorem give the assertion in higher dimension. If $w\in W^{2,p}$, apply it to $w$ and then to each component of its gradient to get $D^{2}w=0$ almost everywhere on $\{w=0\}$. This supplies the contact differentiation used in noncontact of ROF level boundaries.

On a connected open set, a Sobolev space function with zero weak gradient is constant almost everywhere. Local mollification gives constant smooth representatives on interior balls; overlaps and connectedness identify their constants. Without connectedness, the constant may differ between components. This explains the constant ambiguity in a Neumann Poisson problem.

The H1 space is the first-order Sobolev space whose elements and first weak derivatives belong to L2 space. It is a Hilbert space with inner product given by the sum of the L2 inner products of the functions and of their first weak derivatives.

For $f\in H^1(\mathbb{R})$, its difference quotient equals $D_{h}f={\int }_0^1{f}^{\prime }(\cdot +\theta h)d\theta$ in L2 space. Thus $\Vert D_{h}f{\Vert }_2\le \Vert f^{\prime }{\Vert }_2$, and translation of a function gives $D_{h}f\to f^{\prime }$ in L2 space. Conversely, bounded difference quotients have a weakly convergent subsequence as $h\to 0$. Against a test function, $\int D_{h}fv\to -\int f{v}^{\prime }$, so that weak limit is the weak derivative of $f$. It lies in L2 space, which is precisely the first-order Sobolev space condition.

The zero-boundary Sobolev space $H_0^1(U)$ is the closure of the space of test functions $C_c^{\infty }(U)$ in $H^1(U)$. Under standard regularity assumptions its elements have zero boundary trace.

On a nonempty bounded domain where the Poincaré inequality holds, no $a\in H_0^1(\Omega )$ can satisfy $a\ge a_{-}>0$ almost everywhere. For $g(s)=\min (\max (s,0),a_{-})$, Lipschitz truncation preserves zero-boundary Sobolev spaces, so $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$, whose zero gradient contradicts the Poincaré inequality. In a uniformly elliptic divergence-form elliptic operator, the homogeneous boundary condition belongs to the unknown and test functions, not to a coefficient bounded below by a positive constant.

If $v\in H_0^1(\Omega )$ and $g:\mathbb{R}\to \mathbb{R}$ is Lipschitz continuous with $g(0)=0$, then $g(v)\in H_0^1(\Omega )$. The Lipschitz version of the Sobolev chain rule controls its weak derivatives, and its boundary trace is $g(0)=0$. Equivalently, approximate $v$ by compactly supported smooth functions, apply the derivative bound to their compositions, and pass weakly in $H^1$ and strongly in $L^2$. The closed linear space $H_0^1$ is weakly closed. This allows positive-part and bounded truncations in weak formulations without losing the Dirichlet boundary condition.

On a domain $U\subseteq {\mathbb{R}}^n$, a common normalization of the Dirichlet inner product isThe Poincaré inequality makes its induced seminorm a norm on $H_0^1(U)$.

For a conformal bijection $f:D\to \tilde{D}$ and smooth $u,v$ of compact support on $\tilde{D}$, $\boxed{(u\circ f,v\circ f{)}_{\nabla ,D}=(u,v{)}_{\nabla ,\tilde{D}}}$. The Jacobian matrix of $f$ is a rotation times $|f^{\prime }|$. Consequently the chain rule introduces $|f^{\prime }{|}^2$ into the gradient pairing, while the change of variables formula introduces precisely the same factor into area. They cancel. The identity extends by completion to the Dirichlet energy spaces; it does not preserve the $L^2$ term of an inhomogeneous Sobolev norm.

For planar Gaussian free field theory, this is the completion of real test functions $C_c^{\infty }(D)$ in the norm induced by the Dirichlet inner product, $(f,g{)}_{\nabla }=(2\pi {)}^{-1}{\int }_D\nabla f\cdot \nabla g$. On a general unbounded domain this convention must be distinguished from completion in the inhomogeneous $H^1$ norm. A conformal map identifies its energy norm with that on the unit disc by conformal invariance.

Let $U\subseteq D$ be open. Regard $H_{\mathrm{supp}}=H_0^1(U)$ as functions on $D$ by zero extension of H01, and let $H_{\mathrm{harm}}$ consist of the weakly harmonic Sobolev functions on $U$ belonging to $H_0^1(D)$. For $h\in H_{\mathrm{harm}}$, the Dirichlet inner product $(h,\varphi {)}_{\nabla }$ vanishes for each test function supported in $U$. Approximate any $u\in H_{\mathrm{supp}}$ by those test functions in the gradient norm and use the Cauchy-Schwarz inequality. This gives $\boxed{(h,u{)}_{\nabla }=0}$. The proof also works with the inhomogeneous zero-boundary Sobolev space convention, and with a homogeneous completion realized as weak functions.

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 $L^p(a,b)$.

On $I=(-a,a)$, a first-order Sobolev space function has a continuous representative and obeysBound the average by the Cauchy-Schwarz inequality and average the identity $u(x)-u(y)={\int }_y^x{u}^{\prime }$. This also gives $[u{]}_{C^{0,1/2}}\le \Vert u^{\prime }{\Vert }_2$. A unit constant with the standard Sobolev norm cannot be asserted on arbitrary interval lengths: $u=1$ already disproves it when $2a<1$.

For every open set $U\subseteq {\mathbb{R}}^n$, smooth functions in $U$ are dense in $W^{1,p}(U)$ for $1\le p<\infty$. Near a flat boundary, one may first translate the function into the domain and then apply a mollifier.

By the definition of $W_0^{1,p}(U)$ as the closure of $C_c^{\infty }(U)$, extending a function by zero outside $U$ gives a bounded map into $W^{1,p}({\mathbb{R}}^n)$ and does not create a boundary distribution.

A Sobolev extension operator is a bounded linear map $E:W^{1,p}(U)\to W^{1,p}({\mathbb{R}}^n)$ whose restriction to $U$ is the original function. For a half-space, reflection across the boundary gives such an operator.

For $U=(0,\infty )\times {\mathbb{R}}^{n-1}$, the formula $Eu(x_1,x^{\prime })=u(|x_1|,x^{\prime })$ defines a bounded Sobolev extension operator. Its weak normal derivative changes sign across the boundary, while its tangential weak derivatives are reflected unchanged.

The odd reflection of a function on a half-space is $Eu(x^{\prime },x_n)=\mathrm{sgn}(x_n)u(x^{\prime },|x_n|)$. When $u$ has zero trace on the boundary, this extension preserves the relevant weak regularity and often converts a homogeneous boundary problem into an interior problem.

A function $f:[a,b]\to \mathbb{R}$ is absolutely continuous exactly when there is a function $g\in L^1(a,b)$ such thatThen $f$ is differentiable almost everywhere and $f^{\prime }=g$ almost everywhere.

For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.