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A continuous function at $x_0$ satisfies the following quantified condition:The choice of $\delta$ may depend on $x_0$ and $\epsilon$.

For the bounded example, takeThis is a continuous function on its domain and $|f(x)|\le 1-x<1$. Along $x_n=(\pi /2+2\pi n{)}^{-1}$ its values tend to $1$, while along $y_n=(3\pi /2+2\pi n{)}^{-1}$ its values tend to $-1$. Hence the supremum is $1$ and the infimum is $-1$, and neither is attained. “Upper and lower bound” here means the least upper and greatest lower bounds.

For the nonnegative function on the whole real line, if $f\equiv 0$ the conclusion is immediate. Otherwise choose $x_{\ast }$ with $m=f(x_{\ast })>0$. The limits at both infinities give $R>|x_{\ast }|$ such that $f(x)<m/2$ whenever $|x|>R$. By the extreme value theorem, a continuous function on the compact interval $[-R,R]$ has a maximum $M=f(\alpha )$ there. Since $M\ge m$, it also exceeds every value outside this interval. Thus $f$ is bounded above and attains its global maximum. Nonnegativity ensures that either the function is zero or such a positive comparison value exists; without it, a function such as $-e^{-x^2}$ would have an unattained supremum of zero.

For fixed-design nonparametric regression, let $x_i=i/n$ and $K_h(u)=h^{-1}K(u/h)$. The degree-$p$ local polynomial estimator takes the fitted intercept ${\hat{m}}_h(x;p)={\hat{\beta }}_0$, whereUse a nonnegative regression kernel and an invertible local polynomial Gram matrix to obtain a unique fit. For $p=0$, solving the single weighted least squares normal equation gives the local constant estimator:At the interior point $x_0$, use the standard unit-integral regression kernel convention ${\mu }_0(K)=1$. Otherwise the leading variance below is $R(K)/(nh{\mu }_0(K{)}^2)$; unlike a kernel density estimator, the local constant estimator itself is unchanged by rescaling $K$.

The local constant estimator is a linear estimator in nonparametric regression with weights $W_i=K_h(x_i-x_0)/(n{s}_{0,h})$. The independent unit-variance errors implyTo control the bias of an estimator, differentiability at this one point is enough. WriteSymmetry makes ${\mu }_1(K)=0$, and the supplied moment approximation gives $s_{1,h}=o(h)$. Therefore the linear contribution to the bias of an estimator is $m^{\prime }(x_0)s_{1,h}/s_{0,h}=o(h)$. The remaining contribution has absolute value at most $\eta (h)h$, because only $|x_i-x_0|\le h$ receive positive weight. This is the interior first-order bias cancellation for local constant regression, and gives squared bias of an estimator $o(h^2)$. Combining with the variance provesFor fixed $\epsilon >0$, take $h_{\epsilon }={\epsilon }^{-1/3}{n}^{-1/3}$. It satisfies both asymptotic smoothing bandwidth conditions, and the displayed remainder, multiplied by $n^{2/3}$, tends to zero for this fixed $\epsilon$. ThusLetting $\epsilon \downarrow 0$ proves the optimized scaled pointwise error tends to zero. The order of these limits matters: no remainder uniform in $\epsilon$ has been assumed.

For the lower bound over the whole function class, the absence of a uniform bound on derivatives permits a particularly direct argument. It also avoids assuming a normal distribution for errors when the model only supplies their first two moments. Fix $n$ and $x_0$. Choose a smooth bump function $b$ on the interval, with $b(x_0)=1$, whose support contains no design point except possibly $x_0$. Such a bump exists because the design is finite; at an endpoint use the restriction of a smooth bump on the real line. Every $m_a=ab$, $a\in \mathbb{R}$, belongs to the differentiability class.

If $x_0$ is not a design point, all observation means are zero for every $a$. The Le Cam two-point lemma applied to $a=\pm A$ has total variation distance zero and gives a lower bound tending to infinity with $A$. In fact, the supremum of the mean squared error is infinite at every such $x_0$.

If $x_0$ is a design point, only its response depends on $a$. The resulting location family is the single observation $Y=a+ϵ$; all remaining observations are independent of $a$ and give no extra information. The single-observation location minimax bound isHere is a proof covering discrete as well as continuous errors. Give $a$ the prior distribution $U[-A,A]$, independently of the noise. Put $p_M=\mathbb{P}(|ϵ|\le M)$ and $v_M=\mathrm{Var}(ϵ\mid |ϵ|\le M)$. Consider an easier experiment in which an oracle also reveals whether $|ϵ|\le M$, and reveals $a$ itself on the complementary event. On the event $|ϵ|\le M$ and $|Y|\le A-M$, every possible truncated noise value places $a=Y-ϵ$ within $[-A,A]$. The flat prior distribution then makes the Bayesian posterior noise law exactly its truncated original law. The smallest conditional squared-error loss is its conditional variance $v_M$, attained by the posterior mean.

The probability of that event is at least $p_M(1-2M/A)$ for $A>2M$, by restricting further to $|a|\le A-2M$. Thus every rule's supremum risk function is at least the original Bayes risk, which is at least the easier experiment's Bayes risk, and therefore at least $p_M(1-2M/A)v_M$. Let $A\to \infty$ and then $M\to \infty$. Finite second moment and zero mean give $p_M{v}_M\to 1$. The rule $T(Y)=Y$ has constant mean squared error one, proving the equality.

Consequently, at every $x_0\in [0,1]$, the unrestricted differentiability class satisfies the stronger conclusionThus $c=1$ works for every $n\ge 1$. There is no conflict with the preceding fixed-function limit: the worst functions, including arbitrarily narrow bumps, can vary with $n$. This is the pointwise versus uniform risk distinction. A smoothness ball with a common derivative bound would pose a different minimax risk problem.

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.