Sobolev space

For real $s$, the $L^2$-based Sobolev space $H^s({\mathbb{R}}^n)$ consists of tempered distributions satisfying

The bounded angular function has ordinary derivatives $D_{j}u={\delta }_{1j}/r-x_1{x}_j/r^3$ away from zero. Their magnitudes are at most $2/r$, whose $p$-th power is integrable on the unit ball exactly below the dimension threshold in this estimate. To verify the weak derivative across zero, integrate by parts outside a ball of radius $\epsilon$ against a compactly supported smooth test. The boundary term has magnitude at most $C{\epsilon }^{n-1}$, tending to zero for $n>1$, while the derivative volume term converges by integrability. This proves the stated Sobolev membership with those weak derivatives. The dimensional condition distinguishes it from the one-dimensional sign function, whose jump produces a Dirac derivative.

Let $0<a<b<\infty$ and $1\le p<\infty$. A radial function in $W^{1,p}(\{a<|x|<b\})$ has a radial profile in $W^{1,p}(a,b)$; its first weak derivative is the profile of the radial derivative. To see the identity, sum the weak derivative tests against the components of $\varphi (r)x/r$ to obtain $\int g\varphi r^{n-1}=-\int \bar{u}(\varphi r^{n-1}{)}^{\prime }$. Substitute $\varphi =\zeta /r^{n-1}$ for arbitrary $\zeta \in C_c^{\infty }(a,b)$. The weighted Lp norms and $a>0$ give the displayed bound. Radiality of the radial derivative follows by applying a rotation to the weak derivative identity: if $u(Qx)=u(x)$, then $\nabla u(Qx)=Q\nabla u(x)$ almost everywhere, so the radial scalar product is invariant.

For $p>1$, the one-dimensional Sobolev representative obeys $|\bar{u}(r)-\bar{u}(s)|\le |r-s{|}^{1-1/p}\Vert {\bar{u}}^{\prime }{\Vert }_p$. The reverse triangle inequality transfers this to $u$, giving a continuous embedding of this radial Sobolev space into $C^{0,1-1/p}$ on the annulus. The two Hölder seminorms are equal: one inequality follows from $||x|-|y||\le |x-y|$, and the reverse follows by putting both points on the same ray. The restriction $a>0$ is essential for the unweighted profile bound.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

A discrete Sobolev norm combines weighted norms of nodal values and their finite differences. For a real sequence $W_0,\dots ,W_N$ with zero endpoints, the displayed expression is a norm: it is the Euclidean norm of one vector containing both weighted lists, and zero norm forces all nodal values to vanish. It approximates the Sobolev space $H^1$ norm. Endpoint weights or prescribed zero endpoints must be specified when calling a discrete expression a norm.

For arbitrary real $s,t$, the convolution of $u\in H^s(\mathbb{R})$ and $v\in H^t(\mathbb{R})$ has a canonical definition as a tempered distribution. With $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$, write $A=⟨\xi {⟩}^s\hat{u}$ and $B=⟨\xi {⟩}^t\hat{v}$, both in the L2 space. Then $AB\in L^1$ by Cauchy-Schwarz inequality, andThus the product defines a tempered distribution, and the displayed inverse Fourier transform defines $u\ast v$. This extends ordinary convolution of Schwartz functions by the convolution theorem and density in the Sobolev norm.

The result need not belong to any Sobolev space, even when both inputs belong to every Sobolev space. For $\hat{u}=\hat{v}={\xi }^{-1/3}{1}_{(0,1)}(\xi )$, each weighted squared input is integrable, but the squared product is ${\xi }^{-4/3}{1}_{(0,1)}(\xi )$, which is not locally integrable at zero. Every weight $(1+{\xi }^2{)}^r$ is bounded above and below there, proving the failure for all real $r$.

The zero-trace first-order Sobolev space is the closure of $C_c^{\infty }(\Omega )$ in the $H^1$ norm. On a bounded Lipschitz domain it is the kernel of the Sobolev trace operator. On an interval, its elements have absolutely continuous representatives with both endpoint values zero. The Poincaré inequality makes the derivative norm equivalent to the full Sobolev norm. This is the natural form domain for an elliptic problem with homogeneous Dirichlet boundary conditions.

For matrix-valued circle loops in $H^s$, $s>1/2$, use a submultiplicative algebra norm with contractive Fourier projections. If $\Vert X-I\Vert <1$, solve $Y_{+}=I+y_{+}$ by $y_{+}=-P_{>0}((X-I)(I+y_{+}))$, and solve $V_{-}=I+v_{-}$ by $v_{-}=-P_{<0}((I+v_{-})(X-I))$. Both are contractions. Then $X{Y}_{+}$ has only nonpositive modes, $V_{-}X$ has only nonnegative modes, and their compatible products equal a constant matrix. Invertibility of the Hardy compression $I+P_{\ge 0}(X-I)$ makes that constant invertible. This yields the stated normalized factors; uniqueness follows because the two Fourier subalgebras intersect in constants.

On a bounded Lipschitz domain, the Sobolev trace operator is the continuous linear extension to $H^1(\Omega )$ of the boundary restriction of smooth functions. Continuity and density make this extension unique; its existence is the Sobolev trace theorem. The kernel prescribes a homogeneous Dirichlet boundary condition. Vanishing Sobolev trace on only part of the boundary instead gives the form domain for a mixed boundary condition, with the natural Neumann boundary condition on the remainder. On an assembly of congruent tiles, taking a Sobolev trace commutes with a constant matrix acting on the tile restrictions. This lets a transplantation theorem be proved on quadratic form domains even when corner eigenfunctions have singular classical derivatives.

This theorem describes the continuous extension of the Trace operator. For $s>1/2$, restriction to the coordinate hyperplane extends uniquely to a bounded linear mapIn Fourier variables, Cauchy--Schwarz bounds the integral over the normal frequency because $(1+t^2{)}^{-s}$ is integrable exactly when $s>1/2$.

On a bounded smooth domain, the Sobolev trace theorem can be strengthened toFor smooth $u$, extend the outward unit normal to a smooth vector field $X$ near the closure of $U$ with $X\cdot \nu =1$ on its boundary. The divergence theorem gives ${\int }_{\partial U}{u}^2={\int }_U(\mathrm{div}X)u^2+2uX\cdot Du$, and the Cauchy-Schwarz inequality gives the estimate. Density extends it to $H^1(U)$. A bound containing only the product $\Vert u{\Vert }_2\Vert Du{\Vert }_2$ is false for nonzero constant functions. Equivalently, for every $\epsilon >0$, the Young inequality gives $\Vert Tu{\Vert }_2^2\le \epsilon \Vert Du{\Vert }_2^2+C_{\epsilon }\Vert u{\Vert }_2^2$.

For $1\le p<\infty$, boundary restriction on smooth functions extends uniquely to a bounded linear map $T:W^{1,p}({\mathbb{R}}_{+}^n)\to L^p({\mathbb{R}}^{n-1})$. The estimate follows by applying the fundamental theorem of calculus on normal line segments and then integrating over the boundary variables.

For $1\le p<\infty$, no bounded map from $L^p({\mathbb{R}}^n)$ to $L^p({\mathbb{R}}^{n-1})$ can agree with restriction on continuous functions. Indeed,has a fixed nonzero trace while $\Vert u_k{\Vert }_p=k^{-1/p}\Vert \varphi {\Vert }_p\Vert \psi {\Vert }_p\to 0$.

For $a<b$ and $0\le \theta \le 1$, the displayed estimate holds on Euclidean space and a torus with the Fourier definition of the Sobolev norm. In the squared norm, write the weighted squared Fourier coefficient as the product of its weight-$a$ value to the power $1-\theta$ and its weight-$b$ value to the power $\theta$. Holder inequality then proves the estimate. On a compact manifold, a finite coordinate cover and a partition of unity give the same estimate up to a constant. Young inequality implies the useful absorption form $\Vert u{\Vert }_{H^{s+1}}\le \epsilon \Vert u{\Vert }_{H^{s+2}}+C_{s,\epsilon }\Vert u{\Vert }_2$ for nonnegative $s$.

A domain has the uniform interior cone property if fixed-height, fixed-opening cones can be placed inside the domain at its points, with an allowed orientation depending on the point. This geometric regularity permits the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem. It is an interior condition, distinct from the exterior cone condition used in boundary regularity.

This is the closure of the compactly supported test functions $C_c^{\infty }(U)$ in the $H^2(U)$ norm. On a smooth bounded domain it consists precisely of the $H^2$ functions with zero value and zero normal derivative traces. Tangential first derivatives then vanish as well. The clamped Hessian identity makes the Laplacian norm equivalent to the full $H^2$ norm on this space, enabling a Lax-Milgram theorem treatment of the clamped biharmonic problem.

For $v\in H_0^2(-1,1)$, integrate its weak derivative from the left endpoint, where $v=v^{\prime }=0$. The Cauchy-Schwarz inequality gives $\Vert v^{\prime }{\Vert }_2\le 2\Vert v^{\prime \prime }{\Vert }_2$ and $\Vert v{\Vert }_2\le 2\Vert v^{\prime }{\Vert }_2$. Adding the three squared derivative norms gives the displayed bound. The constant is sufficient, rather than asserted sharp. Thus the second-derivative norm is equivalent to the full Sobolev norm on this clamped space.

For continuous real coefficients with $p>0$ and $q,r\ge 0$ on a bounded interval, use the clamped second-order Sobolev space and the displayed bilinear form. A forcing in $L^2$ defines $\ell (v)=\int fv$. The weak solution satisfies $a(u,v)=\ell (v)$ for every clamped test vector. Compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution; density extends that equation to the energy space. Continuous coefficients alone do not justify differentiating them classically. The clamped interval derivative norm bound gives coercivity, so Lax-Milgram theorem supplies a unique solution and the quadratic energy has that unique minimum.

For $u\in H_0^2(U)$, ${\sum }_{i,j}\Vert {\partial }_{ij}u{\Vert }_2^2=\Vert \Delta u{\Vert }_2^2$. Two integrations by parts prove this for compactly supported test functions, and $H^2$ density extends it to the clamped second-order Sobolev space. Together with the Poincare-Wirtinger inequality applied to the mean-zero first derivatives and the zero-boundary Poincaré inequality, it controls the full $H^2$ norm by $\Vert \Delta u{\Vert }_2$. The boundary conditions are essential to this exact identity on bounded domains.

A homogeneous Sobolev space controls derivatives without an independent low-frequency L2 norm: formally $\Vert f{\Vert }_{{\dot{H}}^s}^2=\int |\xi {|}^{2s}|\hat{f}(\xi ){|}^{2}d\xi$, with the normalization chosen according to the Fourier transform. Completions or quotient conventions must be specified when polynomial ambiguities occur. In three dimensions ${\dot{H}}^1$ is represented by $L^6$ functions with gradient in $L^2$, and $\Vert f{\Vert }_6\le C\Vert \nabla f{\Vert }_2$.

In three dimensions let $f\in L^{6/5}({\mathbb{R}}^3)$ have compact support. Let $v_R\in H_0^1(B_R)$ solve $-\Delta v_R=f$ and extend it by zero. There is a unique $v\in {\dot{H}}^1({\mathbb{R}}^3)$ with $\int \nabla v\cdot \nabla \varphi =\int f\varphi$ for all $\varphi \in {\dot{H}}^1$. Indeed Holder inequality and the homogeneous Sobolev inequality make the right side a continuous linear functional in the gradient norm, and the Riesz representation theorem applies to this Hilbert space.

The weak formulation says that $v_R$ is the orthogonal projection of $v$ onto the closed subspace consisting of zero extensions of $H_0^1(B_R)$. These subspaces increase with $R$ and their union contains every test function, which is dense by the definition of the homogeneous Sobolev space. The best-approximation property of orthogonal projection therefore gives $\Vert \nabla (v_R-v){\Vert }_2\to 0$. The homogeneous Sobolev inequality also gives $\Vert v_R-v{\Vert }_6\to 0$. This convergence does not generally extend to the inhomogeneous Sobolev space $H^1$, as the spherical mean of a Poisson solution outside its source shows for a source of nonzero total mass.

The first Gaussian Sobolev space consists of functions and their weak first derivatives square-integrable for standard Gaussian measure. If $f=\sum c_n{h}_n$ in the normalized Probabilists' Hermite polynomial basis, Gaussian integration by parts gives $⟨f^{\prime },h_{n-1}⟩=\sqrt{n}{c}_n$. Thus Parseval identity identifies it with $\sum (1+n)|c_n{|}^2<\infty$, and polynomial truncations converge in its norm. It is the form domain of the Gaussian number operator, with Gaussian Dirichlet energy $\sum n|c_n{|}^2$.

For $1\le p<\infty$, an inhomogeneous Sobolev space norm is $\Vert u{\Vert }_{W^{k,p}(D)}=({\sum }_{|\alpha |\le k}\Vert D^{\alpha }u{\Vert }_{L^p(D)}^p{)}^{1/p}$, using weak derivatives. In particular, $\Vert u{\Vert }_{H^1(D)}^2={\int }_D(|u{|}^2+|\nabla u{|}^2)$. The gradient seminorm alone defines a different completion, the Dirichlet energy space, unless a suitable Poincaré inequality makes the two norms equivalent on the chosen zero-boundary space.

The displayed inner product on the first-order Sobolev space induces its usual Sobolev norm. A continuous reaction-diffusion weak formulation with unit diffusion and unit reaction has this exact bilinear form, so its conforming Galerkin method gives a best approximation with constant one in the first-order Sobolev norm.

Multiplication by a test function is bounded on every real-index Sobolev space. Its Fourier transform acts by convolution with a Schwartz function; frequency-weight estimates and Young's convolution inequality give the norm bound.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.

If $u\in W^{1,p}(U)$, then after modification on a null set its restriction to almost every line parallel to a coordinate axis is absolutely continuous and satisfieswhenever the line segment lies in $U$.

For $u\in W^{1,q}((0,1{)}^2)$ with $q>1$, almost every coordinate-line slice has an absolutely continuous representative and is Hölder continuous with exponent $1-1/q$. Slice bounds need not be uniform. Global Hölder continuity follows from Morrey's inequality only when $q>2$, with exponent $1-2/q$; unbounded examples exist at and below $q=2$.

On a sufficiently regular domain in ${\mathbb{R}}^n$, $H^s$ is a Banach algebra when $s>n/2$:

In dimensions at least two a function in the Sobolev space $H^1$ can have finite norm while $\int w^2|\nabla w{|}^2=\infty$. Localize the following radial examples with a smooth cutoff function in an interior ball. For $n\ge 3$, set $w(r)=r^{-\alpha }$, $\alpha =(n-2)/3$. Its gradient energy has radial exponent $n-3-2\alpha >-1$, but its weighted gradient energy has exponent $n-3-4\alpha <-1$. For $n=2$, take $w(r)=(\log (1/r){)}^{1/3}$ near zero. Substituting $s=\log (1/r)$ makes the two gradient integrals proportional to ${\int }^{\infty }{s}^{-4/3}ds$ and ${\int }^{\infty }{s}^{-2/3}ds$, respectively. The first converges and the second diverges. In dimension one the Sobolev embedding theorem gives $H^1\subset L^{\infty }$, preventing this failure. These examples show why a nonnegative coefficient $w^2$ need not even define a bounded gradient form when only $w\in H^1$ is assumed.

For an integer $k\ge 0$ and sufficiently regular bounded functions, $\Vert fg{\Vert }_{H^k}\le C_k(\Vert f{\Vert }_{\infty }\Vert g{\Vert }_{H^k}+\Vert g{\Vert }_{\infty }\Vert f{\Vert }_{H^k})$. The Leibniz rule and derivative interpolation give the estimate. In a differentiated semilinear wave equation, this separates a controlled low-order coefficient from an arbitrarily high Sobolev norm, allowing propagation of smoothness after a fixed base norm has been controlled.

The periodic Sobolev space $H^s({\mathbb{T}}^d)$ consists of periodic distributions whose Fourier coefficients obeyIn one dimension, $H^1(\mathbb{T})$ embeds continuously into $L^{\infty }(\mathbb{T})$ and is a Banach algebra.

A distribution $u$ on an open set $X$ belongs to $H_{\mathrm{loc}}^s(X)$ when $\chi u\in H^s({\mathbb{R}}^n)$ for every test function $\chi \in C_c^{\infty }(X)$.

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.

For a boundary portion ${\Gamma }_D\subseteq \partial U$, the spaceis the kernel of the restricted trace map, hence a closed vector subspace of $H^1(U)$.

For prescribed boundary data $\phi$, the affine Sobolev spaceconsists of the $W^{1,p}$ functions whose trace equals the trace of $\phi$. It is the natural admissible set for a variational Dirichlet problem.

To solve a linear Dirichlet problem with boundary value $g$, choose a sufficiently regular extension $G$ whose trace is $g$ and write $v=w+G$. Then $w$ has zero trace, while moving the known term $LG$ to the right-hand side turns the equation $Lv=f$ into the homogeneous-boundary problem $Lw=f-LG$.

The Hardy operator is the averaging operatoron functions over the positive half-line or an interval beginning at zero.

For $1<p<\infty$, the Hardy operator is bounded on $L^p(0,1)$ with operator norm at most $p/(p-1)$.

If $u\in W^{1,p}(0,1)$ has zero trace at zero, thenIndeed, the one-dimensional Sobolev representative satisfies $u(x)={\int }_0^x{u}^{\prime }(t)dt$, so the claim is the Hardy averaging inequality applied to $u^{\prime }$.

A Poincare inequality controls a function by its derivatives after the constant ambiguity has been removed. For example, on a bounded connected Lipschitz domain,