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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.

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 $u^{\prime }$.

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 $Lw=f-LG$.

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 $\Vert Tu{\Vert }_2^2\le \epsilon \Vert Du{\Vert }_2^2+C_{\epsilon }\Vert u{\Vert }_2^2$.

Let $I=(-1,1)$ and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm $\Vert v{\Vert }_{H^2}^2=\Vert v{\Vert }_2^2+\Vert v^{\prime }{\Vert }_2^2+\Vert v^{\prime \prime }{\Vert }_2^2$. The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the $H^2$ closure of $C_c^{\infty }(I)$, so it is a closed Hilbert space. The derivative boundary conditions are essential, and $H_0^1(I)$ alone is insufficient.

Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace $a$ by its sesquilinear version, take the real part of the forcing term and use absolute squares in $J$; the same argument applies.

Its first variation is $J^{\prime }(u)[v]=a(u,v)-\ell (v)$. Thus stationary points satisfy $a(u,v)=\ell (v)$ for every $v\in H$, the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because $v=v^{\prime }=0$ at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution. This avoids assuming nonexistent classical derivatives of $p$ or $q$.

The clamped second-order Sobolev space is $H_0^2(U)={\overline{C_c^{\infty }(U)}}^{H^2}$. On a smooth bounded domain the Sobolev trace theorem characterizes it by zero value and zero normal derivative on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.

For a smooth weak solution, compactly supported test functions and two integrations by parts giveso ${\Delta }^{2}u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u={\partial }_{n}u=0$ on $\partial U$. Thus it is a classical solution of the clamped biharmonic problem.

Conversely, a $C^4(\overline{U})$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported test function, two integrations by parts give ${\int }_U\Delta u\Delta v={\int }_{U}fv$. Both sides are continuous for the $H^2$ norm, so the defining density of $C_c^{\infty }(U)$ in $H_0^2(U)$ extends this equality to every required test. The two notions agree under the stated smoothness.

Use the interior $\Omega =\{1/2<|x|<2\}$ for the Sobolev space notation; the closed shell in the PDF specifies its boundary. Put $A(x)=\mathrm{diag}(1,1,g(|x|))$. A weak solution has zero boundary trace in the sense of the Sobolev trace theorem, equivalently $u\in H_0^1(\Omega )$, and satisfiesThe minus sign is necessary because the operator in the paper is positive divergence, not its negative. One can first test with $C_c^{\infty }$ functions and then use density in the zero-boundary Sobolev space. Smoothness of $g$ makes its restriction to the compact radial interval bounded, so both sides define continuous functionals on this space. Merely belonging to $H^1$ without the zero trace would omit the boundary condition.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives

Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the Sobolev trace theorem. The weak Robin problem for a uniformly elliptic operator uses all of $H^1(U)$ as the test space, since a Robin boundary condition does not require a zero trace. DefineThe weak formulation isFor a classical solution, integration by parts produces the boundary term $-{\int }_{\partial U}(\sum a^{ij}{u}_{x_i}{\nu }_j)v$. Substituting the Robin boundary condition replaces it by ${\int }_{\partial U}\beta uv$, giving the displayed identity.

Conversely, if $u\in C^2(\overline{U})$ satisfies the identity, tests $v\in C_c^{\infty }(U)$ first show $(L+\lambda )u=f$ in the sense of distributions. Subtracting this interior equality after integration by parts givesEvery smooth boundary function is the trace of a smooth function on $\overline{U}$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood almost everywhere.

If the Poincare inequality with a boundary trace failed, after normalization there would be $u_k\in H^1(U)$ withAs in part 1(c)(i), a subsequence converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to a constant function $u$. The Sobolev trace theorem is a bounded linear map, so the traces converge weakly while their norms tend to zero; hence the trace of $u$ is zero. A constant with zero trace is zero, contradicting $\Vert u{\Vert }_2=1$. Consequently

The zero-boundary Sobolev space isHere $H^1(\Omega )$ consists of the $L^2$ functions whose first weak derivatives also belong to $L^2$, with normOn a sufficiently regular domain, the closure condition is equivalent to having zero boundary trace.

For $s>\frac{1}{2}$, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

Take first $u$ in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,The Cauchy-Schwarz inequality, with weights $1+|{\xi }^{\prime }{|}^2+{\xi }_n^2$, givesAfter the substitution ${\xi }_n=(1+|{\xi }^{\prime }{|}^2{)}^{1/2}t$, the second factor iswhere finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|{\xi }^{\prime }{|}^2{)}^{s-1/2}$ and integrating in ${\xi }^{\prime }$ yieldsDensity of the Schwartz space in the Sobolev space completes the unique bounded extension.

The one-dimensional Sobolev representative of $u$ is absolutely continuous, and its zero trace givesConsequently $u(x)/x=A(u^{\prime })(x)$ for the Hardy operator. Applying the Hardy averaging inequality to $u^{\prime }$ gives the Hardy inequality on an interval:

The Sobolev trace theorem makes evaluation at zero a continuous linear map $T_0:H^1(0,1)\to \mathbb{R}$. Henceis a closed vector subspace of the Hilbert space $H^1(0,1)$. Every closed vector subspace of a Hilbert space is complete with the restricted inner product, so $H$ is a Hilbert space with the standard $H^1$ inner product.

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

Suppose the claimed Poincare inequality with a partial Dirichlet boundary fails. There are $v_k\in V$ withAfter the normalization $w_k=v_k/\Vert v_k{\Vert }_{L^2(U)}$,Thus $(w_k)$ is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrashov compactness theorem and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The Sobolev space with a partial Dirichlet condition $V$ is a closed vector subspace, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a constant function. Its trace vanishes on the positive-measure set ${\Gamma }_1$, so that constant is zero. This contradictsTherefore some $C_P$ satisfiesSince the reverse bound $\Vert \nabla v{\Vert }_2\le \Vert v{\Vert }_{H^1}$ is immediate,The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on ${\Gamma }_1$ is zero.

The restricted trace map is continuous, and $V$ is its kernel, so $V$ is a closed vector subspace of the Hilbert space $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient normis equivalent to that norm, so it is complete as well. It comes from the inner productConsequently $(V,\Vert \cdot {\Vert }_V)$ is a Hilbert space.

On a bounded connected Lipschitz domain $U$,for every $u\in H^1(U)$. If this failed, the Rellich-Kondrachov compactness theorem would produce a normalized strong $L^2$ limit with zero gradient, hence a constant function; continuity of the Sobolev trace theorem would force that constant to vanish, contradicting its unit norm.