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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 $u^{\prime }$.

By the extreme value theorem, $p_0={\min }_{[-1,1]}p$ is strictly positive, and $p,q,r$ are bounded. The Cauchy-Schwarz inequality therefore givesFor completeness, the clamped interval derivative norm bound follows from $v^{\prime }(-1)=v(-1)=0$ and the one-dimensional Sobolev representative: integrate $v^{\prime \prime }$ to recover $v^{\prime }$, and then $v^{\prime }$ to recover $v$. Applying Cauchy-Schwarz and integrating over an interval of length two givesConsequently $\Vert v{\Vert }_{H^2}^2\le 21\Vert v^{\prime \prime }{\Vert }_2^2$. Nonnegativity of $q,r$ yields the precise coercive bilinear form boundThe Lax-Milgram theorem states that a bounded bilinear form on a real Hilbert space satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ with $c>0$ represents every bounded linear functional uniquely: there is exactly one $u$ with $a(u,v)=\ell (v)$ for all $v$. All of its hypotheses have now been verified on $H$.

This weak solution is the unique global minimum, rather than merely a stationary point. The Ritz-Galerkin equivalence for a symmetric coercive form here is the exact identityIt is strictly positive for every nonzero $w\in H$. Conversely, a minimum has zero first variation in every direction, so it satisfies the same weak equation. Density of compactly supported test functions and boundedness of $a,\ell$ extend the distributional equation back to all $H$ test vectors. There is exactly one minimum, and it is exactly the clamped weak solution. No additional coefficient smoothness or classical fourth derivative is needed.

A one-dimensional Sobolev representative is absolutely continuous. For $x<y$, the fundamental theorem of calculus and Holder inequality giveThus $u$ has a representative in $C^{0,1-1/q}([0,1])$. The qualification about representatives matters because a Sobolev space element is an almost-everywhere equivalence class.

In two dimensions, the Sobolev fundamental theorem of calculus on lines and Fubini's theorem imply that almost every horizontal and vertical slice belongs to $W^{1,q}(0,1)$ and has this one-dimensional Hölder continuity. The slice seminorm depends on the slice; this does not give one uniform pointwise estimate on the square. For $q>2$, Morrey's inequality additionally gives a globally Hölder continuous representative of exponent $1-2/q$. For $1<q\le 2$, global continuity need not hold. For example, with a smooth cutoff around an interior point, $u(x)=|x-x_0{|}^{-a}$ lies in $W^{1,q}$ when $0<a<2/q-1$, but is unbounded. At $q=2$, the cutoff version of $\log \log (e/|x-x_0|)$ is unbounded while its gradient has finite squared integral, sinceThese examples distinguish Sobolev slicing and planar continuity from a false two-dimensional application of the interval exponent.

Put $\Omega =(0,1{)}^2$. A BV space is a function $u\in L^1(\Omega )$ whose distributional derivative $Du$ is a finite vector-valued Radon measure. Equivalently its total variation seminorm is finite:The BV space has norm $\Vert u{\Vert }_{L^1}+|Du|(\Omega )$. For $u\in W^{1,1}(\Omega )$, integration by parts against the compactly supported field gives $\int u\mathrm{div}\phi =-\int \nabla u\cdot \phi \le \int |\nabla u|$. Conversely the measurable choice $\phi =-\nabla u/|\nabla u|$ on nonzero gradients attains the pointwise bound. Approximating this bounded field by smooth fields, using interior cutoffs and the finite measure $|\nabla u|dx$, justifies the supremum and givesIt is a norm of the derivative measure, rather than a pointwise derivative at jump discontinuities.

There is a genuine mismatch in the printed definition of the next functional. Its constraints on ${\phi }_0$ and $\phi$ are independent. Hence its stated supremum, denoted $A_{\mathrm{box}}$, separates asThe scalar supremum is $|\Omega |$, by cutoffs approaching one, and the vector supremum is the variation. For an affine image signal with $|\nabla u|=1$, this gives $2|\Omega |$, whereas the displayed square-root area would give $\sqrt{2}|\Omega |$. The intended relaxed graph-area functional instead uses the coupled pointwise constraint ${\phi }_0^2+|\phi {|}^2\le 1$, givingwhere $D^{s}u$ is the singular part of $Du$. Both readings have a minimizer, but their equations are different.

Here is the direct method in the calculus of variations for either reading. Let $A_{\bullet }$ be the literal $A_{\mathrm{box}}$ or the corrected $A$, and define the energy on $BV(\Omega )\cap L^2(\Omega )$, assigning infinity elsewhere. A minimizing sequence has bounded energy by comparison with $u=0$. Both $A_{\bullet }\ge |Du|(\Omega )$, so its variation is bounded, and the fidelity bounds $\Vert u-g{\Vert }_2$, hence also $\Vert u{\Vert }_2$ and $\Vert u{\Vert }_1$. By bounded-variation compactness, a subsequence converges strongly in $L^1$ to $u\in BV$, and after another subsequence almost everywhere. Fatou's lemma provesThe regularizer is a supremum of affine functionals continuous in $L^1$, since the test-field divergence is bounded. It is therefore lower semicontinuous. Combining the two lower bounds proves existence of a minimizer. In fact the convex regularizer and the strictly convex squared fidelity make the minimizer unique up to null sets. This does not assert that the minimizer must belong to $W^{1,1}$.

For the intended graph area, conditionally assume that the minimizer is in $W^{1,1}$. For $\eta \in C_c^{\infty }(\Omega )$, differentiate at $u+t\eta$. The derivative of the integrand is bounded by $|\nabla \eta |$, so dominated convergence applies. The weak equation isthat is,This is the graph-area Euler-Lagrange equation. Compactly supported variations impose no boundary condition in this statement.

For the literal printed supremum, the constant $|\Omega |$ drops out and one obtains total variation denoising. Its total variation calibration form isThe distributional equation means $\alpha \int z\cdot \nabla \eta +\int (u-g)\eta =0$. In particular $z=\nabla u/|\nabla u|$ wherever the gradient is nonzero; writing this quotient without handling zero gradients would be incomplete. Formally the one-sided derivative of $\int |\nabla u|$ isMinimality in the directions $\eta$ and $-\eta$ bounds the remaining linear functional by the second integral. The Hahn-Banach theorem extends it on that zero-gradient set to a bounded vector field of magnitude at most one, furnishing $z$ and the displayed weak equation. Thus the literal definition has a nonsmooth subgradient equation, not the square-root equation above.

Interpret the given velocity regularity in the stated Sobolev space sense. Repeated integration by parts, or the Fourier transform of a derivative in distributions, givesThe usual one-dimensional one-dimensional Sobolev representative or a smooth approximation justifies this identity without imposing extra decay of classical derivatives at specific boundary points. The transform of an integrable function has absolute value at most its $L^1$ norm. ThereforeSetting $\xi =kt$ and taking the supremum gives the required uniform weighted bound. The $k=0$ term on the left is zero; there is no division by that frequency in this argument.

For $I=(-a,a)$, the first-order Sobolev space isThe weak derivative $v=u^{\prime }$ is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm $\Vert u{\Vert }_{H^1(I)}^2=\Vert u{\Vert }_{L^2(I)}^2+\Vert u^{\prime }{\Vert }_{L^2(I)}^2$. Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.

With the standard Sobolev norm, the printed constant one is not valid for arbitrary $a$. The constant function $u=1$ has supremum norm one and $H^1$ norm $\sqrt{2a}$, which is smaller when $a<1/2$.

For the continuous one-dimensional Sobolev representative, let $\bar{u}=(2a{)}^{-1}{\int }_{I}u$. The Cauchy-Schwarz inequality gives $|\bar{u}|\le (2a{)}^{-1/2}\Vert u{\Vert }_2$. Using the preceding Hölder seminorm estimate,Consequently the corrected interval Sobolev supremum estimate isOne may first prove this for smooth approximations and pass on the full-measure convergence set. Averaging differences also handles complex-valued functions without assuming that a function takes its complex average at some point. At the fixed interval used below the constant is absolute, which suffices for the intended interior estimate.

The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincaré inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

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: