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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.

A bounded function which is $C^1$ away from one point in dimension $n\ge 2$, and whose classical gradient is locally integrable across that point, has that gradient as its weak derivative. Cut a test function off within distance $\epsilon$ of the point. The error from differentiating the cutoff is bounded by $C\Vert u{\Vert }_{\infty }{\epsilon }^{n-1}$ and vanishes. The other terms converge by local integrability. In dimension one the Heaviside step function is a counterexample: its classical derivative off the point is zero, but its distributional derivative is a Dirac delta.

For $u\in H_0^2(U)$ on an open set $U$, the identity ${\sum }_{i,j}\Vert {\partial }_{ij}u{\Vert }_2^2=\Vert \Delta u{\Vert }_2^2$ follows by two integrations by parts for compactly supported test functions, followed by $H^2$ density. Thus the Laplacian and full Hessian have the same L2 norm on the clamped second-order Sobolev space.

If $U$ is bounded, the Poincaré inequality applies to $u$ and its first weak derivatives, all of which have zero-boundary approximations. It then controls the full $H^2$ Sobolev norm by $\Vert \Delta u{\Vert }_2$. The exact second-derivative identity holds without this boundedness hypothesis; the full inhomogeneous norm estimate needs the extra low-frequency control.

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 the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, use the compactly supported trial ${\psi }_{\gamma }=\cos (\gamma x)$ on $|x|\le \pi /(2\gamma )$ and zero elsewhere, with $\gamma >0$. It is continuous and belongs to the oscillator energy quadratic form domain: its weak derivative is square-integrable even though its ordinary derivative jumps at the endpoints. Exact integration givesThe minimum occurs at ${\gamma }^4=({\pi }^2-6)/12$, giving the displayed upper bound, approximately $1.13572$, compared with exact ground state value $1$. This illustrates why Rayleigh-Ritz variational principle trials need the energy form domain rather than the smaller operator domain.

For the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, the normalized compact trial function is ${\psi }_a=\sqrt{15/(16a^5)}(a^2-x^2)$ on $|x|\le a$ and zero elsewhere. Its kinetic and potential contributions to the energy quadratic form are $5/(2a^2)$ and $a^2/7$. Minimizing their sum gives $a^4=35/2$ and the displayed upper bound, approximately $1.19523$, compared with exact ground state eigenvalue $1$. The weak derivative is square-integrable because the trial is continuous at its endpoints. Jumps in the ordinary first derivative do not invalidate the Rayleigh-Ritz variational principle: it requires the energy form domain, rather than the smaller operator domain.

If a locally integrable function has weak derivatives represented by a continuous vector field $g$, it has a representative $U\in C^1$ with $\nabla U=g$. The assertion is local, so work on a ball compactly contained in the domain and regularize using a mollifier. The regularizations satisfy the segment identity $u_{\epsilon }(x)-u_{\epsilon }(y)={\int }_0^1\nabla u_{\epsilon }(y+t(x-y))\cdot (x-y)dt$. Averaging over a smaller ball in $y$ fixes the additive constant through local $L^1$ convergence. The gradients converge uniformly on compact sets, so the averaged identity gives local uniform convergence to $U$. Passing to the limit on each segment then gives $U(x+h)-U(x)-g(x)\cdot h=o(|h|)$. Continuity of $g$ proves that $U$ is continuously differentiable.

For $p>n$, the Hölder continuous function representative $u^{\ast }$ of $u\in W^{1,p}({\mathbb{R}}^n)$ has a classical Fréchet derivative at almost every point, equal to its weak derivative. At a point $x$ where the $p$-mean oscillation of $Du$ tends to zero, apply the Morrey inequality on a cube to $u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. On a cube of side comparable to $|y-x|$, its oscillation is at mostThe required points have full measure by the Lebesgue differentiation theorem.

For $\Delta v=\mathrm{div}F+G$ and $U\subset \subset V$, a cutoff function energy estimate gives $\Vert v{\Vert }_{H^1(U)}\le C(\Vert v{\Vert }_{L^2(V)}+\Vert F{\Vert }_{L^2(V)}+\Vert G{\Vert }_{L^2(V)})$. One derivative of forcing is permitted because integration by parts places it on the test function. This estimate allows a mollified distributional solution initially in $L^2$ to gain its first weak derivative.

A divergence-form elliptic operator is $Lu=D_i(a^{ij}{D}_{j}u)$, with bounded measurable coefficients whose symmetric part has a uniform positive lower bound. Its weak formulation involves only first weak derivatives, so differentiability of the coefficients is unnecessary.

An elliptic boundary value problem couples an elliptic differential equation in a domain to specified boundary conditions. An important example is the Poisson equationHere the Dirichlet boundary data fix the boundary value. Prescribing outward flux instead gives a Neumann boundary-value problem, which has different compatibility and normalization requirements. Thus the differential expression alone does not specify the problem.

For zero Dirichlet boundary conditions, multiply the Poisson equation by a smooth compactly supported test function and apply integration by parts:This weak formulation requires only first weak derivatives, so it makes sense for $u\in H_0^1(\Omega )$ and $f\in L^2(\Omega )$. On a bounded domain, the Poincaré inequality makes its bilinear form coercive, and the Lax-Milgram theorem gives a unique weak solution. The right side also extends to suitable forcing in the continuous dual space of $H_0^1(\Omega )$.^[ref]

The energy method explains the value of this formulation: subtracting two solutions and testing with their difference proves uniqueness. Elliptic regularity asks whether additional smoothness of coefficients, forcing and boundary upgrades the weak solution to a classical one. A Schauder estimate quantifies this gain in Hölder spaces; rough coefficients instead lead to results such as the De Giorgi-Nash-Moser theorem.

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

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

If $u\in W_{\mathrm{loc}}^{1,2}$, then $Du=0$ almost everywhere on every level set $\{u=t\}$. Choose smooth truncations ${\Phi }_{\epsilon }(s)$ with $|{\Phi }_{\epsilon }|\le 2\epsilon$, derivative one near zero, bounded derivative, and derivative zero outside $(-2\epsilon ,2\epsilon )$. The Sobolev chain rule gives $D{\Phi }_{\epsilon }(u-t)\to 1_{\{u=t\}}Du$ in local $L^2$ by dominated convergence theorem, while ${\Phi }_{\epsilon }(u-t)\to 0$. The limiting weak derivative is therefore zero.

The inhomogeneous $H^1$ norm on a domain $D\subseteq {\mathbb{R}}^n$ iswhere the derivatives are weak derivatives. It measures both the size of the function and its first-order variation. The underlying Sobolev space consists of the $L^2$ functions whose first weak derivatives also lie in $L^2$.

For $u(x)=x$ on $(0,1)$, the squared norm is $1/3+1=4/3$, so the norm is $2/\sqrt{3}$. A nonzero constant has zero derivative but positive $H^1$ norm. Thus the gradient alone defines a seminorm on the full space, not the displayed norm.

On a suitable zero-boundary space over a bounded domain, a Poincaré inequality controls the $L^2$ term by the gradient and makes the gradient norm equivalent to the full norm. The boundary or normalization condition is essential for this replacement.

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.

Integration by parts transfers a derivative from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible antiderivative. The integral is then replaced by a boundary term minus another integral.

For real continuously differentiable functions $u,v$ on $[a,b]$, the precise definite integral formula isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[ref]

For example, choose $u=x$ and $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$, since the exponential function is its own derivative:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ and area one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the area under either factor separately.

The same choice repeatedly lowers the degree of a polynomial when integrating a polynomial times an exponential function, while choosing $u=\ln x$ gives $\int \ln xdx=x\ln x-x+C$ for $x>0$. More broadly, moving a derivative onto a convenient test function is the mechanism behind a weak derivative. Always evaluate the boundary term: it disappears only when the relevant endpoint values justify doing so, often through a boundary condition.

For a function smooth on either side of a joining point $a$, integration by parts on the two intervals adds a Dirac mass whose coefficient is the right trace minus the left trace. If the one-sided derivatives are locally integrable, a first weak derivative can be represented by a locally integrable function across the join only when the jump vanishes. Derivative jumps themselves do not obstruct first-order Sobolev membership: a continuous piecewise Lipschitz function still has a bounded weak first derivative.

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.

A locally integrable function has weak derivatives on an open set exactly when it has them near every point. Local derivatives agree almost everywhere on overlaps because they define the same distributional derivative. A countable local cover therefore gives globally defined, locally integrable derivative functions. A smooth partition of unity subordinate to the cover splits each compactly supported test function; summing the local integration identities yields the global weak-derivative identity.