Two energy estimates

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

We continue the proof by contradiction of Lipschitz sheets and a lexicographic maximum: \(f:S\to\mathbb R^k\) satisfies (2.1) there, and \(K\), \(A\), \(B\) are defined by its sheets. Near a good point where the derivative columns of a sheet \(F\) have squared lengths close to \((A,B)\), we add one coordinate \(re_{j,i}\) at a fresh fine level \(j\) and obtain new sheets \(F_1,\dots,F_N\), each within \(Dr\) of \(F\) on its strips. This lesson shows that their derivatives are close to that of \(F\) in mean square, in a rectangle \(Q\) tiled by period blocks of the strips at level \(j\) (Proposition 2.1) [OpenAI-L, Section 4]. Two features of the strips are used: a horizontal strip of colour \(i\) crosses the whole rectangle, so \(F_i-F\) can be integrated along its full width; and the vertical strips of level \(j\) are \(n\) times wider than the horizontal ones, so that integrating across one of them already gives an error of order \(1/n\). The horizontal estimate comes with the factor \(n^2\), which the lesson Crossings and the Lang–Plaut problem needs because the horizontal lines it uses are \(n\) times longer than the vertical ones.

We use Lemmas 1.1, 1.2, 3.1 and 4.1 of Lipschitz sheets and a lexicographic maximum, the set \(S\), the strips and the sheets of Doubling spaces and coloured strips, and Fubini's theorem for bounded measurable functions on rectangles (core course Measure and Integration, Fremlin, Measure Theory, Volume 2, 251N and 252B).

1. Choices

The map \(f\), the numbers \(k\), \(D\), \(A\), \(B\) and the set \(K\) are fixed. Fix an integer \[ N>(1+2D)^k.\tag{1.1} \] Let \(n\) be a positive integer. All the following choices depend on \(n\).

A near-extremal sheet. Since \((A,B)\in K\) lies in the closure of the pairs given by good points, there are a sheet \(F=F_w\) and a good point \(p_0\) of \(F\) with \(u=F_x(p_0)\), \(v=F_y(p_0)\) and \[ \begin{gathered} 0\le A-\|u\|^2<n^{-4},\\ \bigl|B-\|v\|^2\bigr|<n^{-4}; \end{gathered}\tag{1.2} \] the first difference is nonnegative by the definition of \(A\). Since \(p_0\) is good and \(\Omega_w\) is open, there is a closed square \(Q^0\) of side \(l\), centred at \(p_0\) and contained in \(\Omega_w\), satisfying the bound (1.3): \[ \begin{aligned} &\bigl\langle\|F_x-u\|^2+\|F_y-v\|^2\bigr\rangle_{Q^0}\\ &\quad<n^{-8}. \end{aligned} \]

A fresh level. Every pair of positive integers occurs infinitely often in the sequence \((N_j,W_j)\), so there is a level \(j\), larger than every level where \(w\) is nonzero, with \[ \begin{gathered} (N_j,W_j)=(N,n),\\ Nnr_j<\tfrac1{100}\,l\,n^{-8}. \end{gathered}\tag{1.4} \] Put \(r=r_j\). The horizontal strips of level \(j\) have height \(r\) and their colours repeat with period \(Nr\); the vertical strips have width \(nr\) and their colours repeat with period \(Nnr\). A period block is a rectangle \([aNnr,(a+1)Nnr]\times[bNr,(b+1)Nr]\), \(a,b\in\mathbb Z\). It contains exactly one horizontal strip (a row) and one vertical strip (a column) of each colour.

The rectangle. Let \(Q\) be the union of the period blocks contained in \(Q^0\), a closed rectangle with sides \(L_x\) and \(L_y\). Trimming \(Q^0\) to the grid of blocks loses less than one period at each end, so \(L_x\ge l-2Nnr\ge l/2\) and \(L_y\ge l/2\). Hence \(|Q|\ge|Q^0|/4\), and with (1.3) and (1.4), \[ \begin{aligned} &\bigl\langle\|F_x-u\|^2+\|F_y-v\|^2\bigr\rangle_Q\\ &\quad <4n^{-8},\\ &\frac r{L_x},\ \frac r{L_y}<\frac{n^{-9}}{50N}. \end{aligned}\tag{1.5} \]

The new sheets. For \(1\le i\le N\), let \(F_i\) be the sheet of the sequence obtained from \(w\) by putting \(re_{j,i}\) at level \(j\). Its domain is \(\Omega_w\cap U_{j,i}\), which contains the rows and the columns of colour \(i\) inside \(Q^0\). The two points of \(S\) behind \(F_i(p)\) and \(F(p)\) are at distance \(r\), so by (2.1) of Lipschitz sheets and a lexicographic maximum \[ \begin{gathered} \|F_i-F\|\le Dr\\ \text{on }\Omega_w\cap U_{j,i}. \end{gathered}\tag{1.6} \]

The selected fields. Off the boundaries of the strips (a null set), define on \(Q\): \(G_H^x=(F_i)_x\), where \(i\) is the colour of the row containing the point; and \(G_V^x=(F_i)_x\), \(G_V^y=(F_i)_y\), where \(i\) is the colour of the column containing the point. Good points of the finitely many sheets \(F,F_1,\dots,F_N\) have full measure, so almost everywhere on \(Q\) \[ \begin{gathered} \|G_H^x\|^2\le A,\\ \bigl(\|G_V^x\|^2,\|G_V^y\|^2\bigr)\in K. \end{gathered}\tag{1.7} \]

2. The estimates

Proposition 2.1 (two energy estimates; OpenAI). With the choices of Section 1, averages over \(Q\) satisfy \[ \begin{aligned} &n^2\bigl\langle\|G_H^x-F_x\|^2\bigr\rangle\\ &\quad \le2(1+4D)n^{-2}\\ &\quad +\frac{4D^2}{25N}\,n^{-7}+8n^{-6}, \end{aligned}\tag{2.1} \] \[ \begin{gathered} \bigl\langle\|G_V^y-F_y\|^2\bigr\rangle\longrightarrow0\\ (n\to\infty). \end{gathered}\tag{2.2} \]

Proof. The row field. A row of colour \(i\) meets \(Q\) in a horizontal band across the whole width of \(Q\). By Lemma 1.2 of Lipschitz sheets and a lexicographic maximum, applied to the Lipschitz map \(F_i-F\) on \(\Omega_w\cap U_{j,i}\), for almost every height \(y\) in the band, \[ \begin{aligned} &\Bigl\|\int_{x_L}^{x_R}\bigl((F_i)_x-F_x\bigr)(x,y)\,dx\Bigr\|\\ &=\|(F_i-F)(x_R,y)-(F_i-F)(x_L,y)\|\\ &\le2Dr, \end{aligned} \] where \(x_L,x_R\) are the ends of \(Q\); the end points lie in the domain because \(Q\subseteq\Omega_w\) and \(y\) lies in the open row, and (1.6) applies. The integrand is \(G_H^x-F_x\) almost everywhere in the band. Integrating over the heights and dividing by \(|Q|=L_xL_y\), \(\|\langle G_H^x-F_x\rangle\|\le2Dr/L_x\). By the Cauchy–Schwarz inequality and (1.5), \(\|\langle F_x\rangle-u\|\le\langle\|F_x-u\|^2\rangle^{1/2}<2n^{-4}\), so \[ \begin{aligned} &\|\langle G_H^x\rangle-u\|\\ &\quad \le\frac{2Dr}{L_x}+2n^{-4}. \end{aligned}\tag{2.3} \] Lemma 4.1 of that lesson, with \(\langle\|G_H^x\|^2\rangle\le A\) from (1.7), (1.2) and \(\|u\|\le D\), gives \[ \begin{aligned} &\bigl\langle\|G_H^x-u\|^2\bigr\rangle\le A-\|u\|^2\\ &\quad +2D\|\langle G_H^x\rangle-u\|\\ &\quad \le(1+4D)n^{-4}+\frac{4D^2r}{L_x}. \end{aligned} \] Since \(\|G_H^x-F_x\|^2\) is at most \(2\|G_H^x-u\|^2+2\|F_x-u\|^2\), (1.5) gives \[ \begin{aligned} &\bigl\langle\|G_H^x-F_x\|^2\bigr\rangle\le2(1+4D)n^{-4}\\ &\quad +\frac{8D^2r}{L_x}+8n^{-8}, \end{aligned} \] and multiplying by \(n^2\) and using \(r/L_x<n^{-9}/(50N)\) gives (2.1).

The column field, first column. A column of colour \(i\) meets \(Q\) in a vertical band of width \(nr\), and \(Q\) is the union of \(L_x/(nr)\) such bands. For almost every height \(y\), Lemma 1.2 of that lesson applies to \(F_i-F\) on the open segment crossing the band at height \(y\), whose end points lie on the edges of the band; the continuous extension to the end points still satisfies (1.6), by continuity. So the integral of \(G_V^x-F_x\) across each band at height \(y\) has norm at most \(2Dr\). Adding over the \(L_x/(nr)\) bands and dividing by \(L_x\) gives at most \(2D/n\) for almost every \(y\). Hence, as above, \[ \begin{aligned} &\|\langle G_V^x\rangle-u\|\\ &\quad \le\frac{2D}n+2n^{-4}. \end{aligned}\tag{2.4} \] Let \(\eta_n=\langle A-\|G_V^x\|^2\rangle\), which is nonnegative by (1.7). By Lemma 4.1 of that lesson, whose left-hand side is nonnegative, \(\langle\|G_V^x\|^2\rangle\ge\|u\|^2-2\|u\|\,\|\langle G_V^x\rangle-u\|\), so \[ \begin{aligned} 0&\le\eta_n\le A-\|u\|^2+2D\|\langle G_V^x\rangle-u\|\\ &\le(1+4D)n^{-4}+\frac{4D^2}n\longrightarrow0. \end{aligned} \] Apply Lemma 3.1(b) of that lesson to the pairs \((\|G_V^x\|^2,\|G_V^y\|^2)\), which lie in \(K\) almost everywhere by (1.7), on \(Q\) with the normalized area measure: \[ \limsup_{n\to\infty}\bigl\langle\|G_V^y\|^2\bigr\rangle\le B.\tag{2.5} \]

The column field, second column. A column of colour \(i\) crosses \(Q\) from bottom to top. As for the rows, for almost every \(x\) in it the integral of \((F_i)_y-F_y\) along the vertical segment through \(Q\) has norm at most \(2Dr\), so \(\|\langle G_V^y-F_y\rangle\|\le2Dr/L_y\) and \[ \begin{aligned} &\|\langle G_V^y\rangle-v\|\\ &\quad \le\frac{2Dr}{L_y}+2n^{-4}\longrightarrow0. \end{aligned}\tag{2.6} \] By Lemma 4.1 of that lesson and \(\|v\|\le D\), \[ \begin{aligned} 0&\le\bigl\langle\|G_V^y-v\|^2\bigr\rangle\le\bigl\langle\|G_V^y\|^2\bigr\rangle\\ &\quad -\|v\|^2+2D\|\langle G_V^y\rangle-v\|. \end{aligned} \] By (2.5), (1.2) and (2.6), the right-hand side has \(\limsup\) at most \(B-B+0=0\). So \(\langle\|G_V^y-v\|^2\rangle\to0\), and with \(\langle\|F_y-v\|^2\rangle<4n^{-8}\) this gives (2.2). \(\square\)

The vertical estimate has no rate: it comes from the compactness argument of Lemma 3.1 of Lipschitz sheets and a lexicographic maximum, and it does not need the vectors \(v\) to converge. The order of the maximization in the definition of \((A,B)\) is what makes it work: the first column of the column field is forced towards the largest possible squared length \(A\), and then its second column cannot have a larger mean squared length than \(B\).

3. Exercises

Exercise 3.1 (easy). Derive the inequalities \(L_x,L_y\ge l/2\) and \(r/L_x<n^{-9}/(50N)\) from (1.4).

Exercise 3.2 (easy). Show that a period block contains exactly one row and one column of each colour, and that \(Q\) contains at least one period block.

Exercise 3.3 (medium). Show that (2.1) remains true, with a right-hand side tending to \(0\), when \(n^2\) is replaced by \(n^{2+\gamma}\) for a fixed \(0\le\gamma<2\). Which term limits \(\gamma\)?

Exercise 3.4 (medium). In the estimate (2.4), the integral was taken across each column separately. Explain why integrating \(G_V^x-F_x\) along the whole width of \(Q\) at once would not give a useful bound.

4. Solutions

3.1. \(2Nnr<\frac1{50}ln^{-8}\le l/2\), so \(L_x\ge l-2Nnr\ge l/2\), and likewise \(L_y\ge l-2Nr\ge l/2\). Then \(r/L_x\le2r/l\), which is less than \(\frac2{100}\,n^{-8}/(Nn)=n^{-9}/(50N)\), and the same for \(L_y\).

3.2. The rows of level \(j\) in \([bNr,(b+1)Nr]\) are those with \(q=bN,\dots,bN+N-1\), whose colours run through all residues; similarly for columns. \(Q^0\) has side \(l>2Nnr\), so it contains a full period of the grid in each direction.

3.3. Multiplying the bound before (2.1) by \(n^{2+\gamma}\) gives \(2(1+4D)n^{\gamma-2}+\frac{4D^2}{25N}n^{\gamma-7}+8n^{\gamma-6}\). The first term, which comes from \(A-\|u\|^2<n^{-4}\) and \(\|\langle F_x\rangle-u\|<2n^{-4}\), needs \(\gamma<2\).

3.4. Along a horizontal line, \(G_V^x\) switches between the derivatives of different sheets \(F_i\) at the edges of the columns. Only within one column is it the derivative of a single map \(F_i-F\), whose increments are bounded by \(2Dr\) through (1.6).

References