= Solution
A <one-dimensional Sobolev representative> is absolutely continuous. For $x<y$, the <fundamental theorem of calculus> and <Holder inequality> give
$$
|u(y)-u(x)|\leq\int_x^y|u'(s)|ds
\leq\|u'\|_{L^q(0,1)}|y-x|^{1-1/q}.
$$
Thus \b[$u$ 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 $1<q\leq2$, 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, since
$$
\int_0^\varepsilon\frac{dr}{r\log^2(e/r)}<\infty.
$$
These 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:
$$
\boxed{|Du|(\Omega)=\sup_{\substack{\varphi\in C_c^1(\Omega;\mathbb R^2)\\|\varphi(x)|\leq1}}
\int_\Omega u\,\operatorname{div}\varphi\,dx.}
$$
The <BV space> has <norm> $\|u\|_{L^1}+|Du|(\Omega)$. For $u\in W^{1,1}(\Omega)$, <integration by parts> against the compactly supported field gives $\int u\operatorname{div}\varphi=-\int\nabla u\cdot\varphi\leq\int|\nabla u|$. Conversely the measurable choice $\varphi=-\nabla u/|\nabla u|$ 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 gives
$$
\boxed{|Du|(\Omega)=\int_\Omega|\nabla u|dx.}
$$
It 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 $\varphi_0$ and $\varphi$ are independent. Hence its stated <supremum>, denoted $A_{\mathrm{box}}$, separates as
$$
\boxed{A_{\mathrm{box}}(u)=|\Omega|+|Du|(\Omega).}
$$
The 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 $\sqrt2|\Omega|$. The intended <relaxed graph-area functional> instead uses the coupled pointwise constraint $\varphi_0^2+|\varphi|^2\leq1$, giving
$$
A(u)=\int_\Omega\sqrt{1+|\nabla u|^2}\,dx+|D^su|(\Omega),
$$
where $D^su$ 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 $A_\bullet\geq|Du|(\Omega)$, so its variation is bounded, and the fidelity bounds $\|u-g\|_2$, hence also $\|u\|_2$ and $\|u\|_1$. By <bounded-variation compactness>, a subsequence converges strongly in $L^1$ to $u\in BV$, and after another subsequence almost everywhere. <Fatou's lemma> proves
$$
\int_\Omega(u-g)^2\leq\liminf_j\int_\Omega(u_j-g)^2.
$$
The 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 \b[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 is
$$
\alpha\int_\Omega\frac{\nabla u\cdot\nabla\eta}{\sqrt{1+|\nabla u|^2}}dx
+\int_\Omega(u-g)\eta\,dx=0,
$$
that is,
$$
\boxed{u-g-\alpha\operatorname{div}\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=0
\quad\hbox{in }\mathcal D'(\Omega).}
$$
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 is
$$
\boxed{u-g-\alpha\operatorname{div}z=0,\qquad
|z|\leq1,\qquad z\cdot\nabla u=|\nabla u|\ \hbox{a.e.}}
$$
The 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 $\int|\nabla u|$ is
$$
\int_{\{\nabla u\ne0\}}\frac{\nabla u}{|\nabla u|}\cdot\nabla\eta
+\int_{\{\nabla u=0\}}|\nabla\eta|.
$$
Minimality 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.
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