Two-by-two upper triangular matrices and Morrey’s conjecture

Abstract

It is shown that every homogeneous gradient Young measure supported on matrices of the form $\begin{pmatrix} a_{1,1} &{} \cdots &{} a_{1,n-1} &{} a_{1,n} \\ 0 &{} \cdots &{} 0 &{} a_{2,n} \end{pmatrix}$ is a laminate. This is used to prove the same result on the 3-dimensional nonlinear submanifold of $\mathbb {M}^{2 \times 2}$ defined by $\det X = 0$ and $X_{12}>0$.

Appendix: The diagonal case

This section contains one particular generalisation of Müller’s result from the $2 \times 2$ diagonal matrices $\mathbb {M}^{2 \times 2}_{{{\mathrm{diag}}}}$ to the subspace

$$\begin{aligned} \mathbb {M}^{2 \times n}_{{{\mathrm{diag}}}} := \left\{ \begin{pmatrix} a_{1,1} &{} \cdots &{} a_{1,n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} a_{2,n} \end{pmatrix} \in \mathbb {M}^{2 \times n} \right\} , \quad n \ge 2. \end{aligned}$$

This is used to prove the result for $\mathbb {M}^{2 \times n}_{{{\mathrm{tri}}}}$. The proof has only minor modifications from the one in [12], but is included for convenience. As in [8], define the elements of the Haar system in $L^2(\mathbb {R}^n)$ by

$$\begin{aligned} h_Q^{(\epsilon )}(x) = \prod _{j=1}^n h_{I_j}^{\epsilon _j}(x_j), \quad \text { for } x \in \mathbb {R}^n, \end{aligned}$$

where $\epsilon \in \{0,1\}^n \setminus \{(0, \ldots , 0)\}$, $Q= I_1 \times \cdots \times I_n$ is a dyadic cube in $\mathbb {R}^n$, the $I_j$’s are dyadic intervals of equal size and the convention $0^0=0$ is assumed. A dyadic interval is always of the form $\left[ k \cdot 2^{-j}, (k+1)\cdot 2^{-j} \right)$ with $j,k \in \mathbb {Z}$. For a dyadic interval $I=[a,b)$, $h_I$ is defined by

$$\begin{aligned} h_I(x) = h_{[0,1)}\left( \frac{ x-a}{b-a} \right) \quad \text { for } x \in \mathbb {R}, \end{aligned}$$

where

$$\begin{aligned} h_{[0,1)} = \chi _{\left[ 0, \frac{1}{2}\right) } - \chi _{\left[ \frac{1}{2}, 1\right) }. \end{aligned}$$

For $j \in \mathbb {Z}$ and $k \in \mathbb {Z}^n$, the notation $h^{(\epsilon )}_{j,k}=h^{(\epsilon )}_Q$ will be used, where

$$\begin{aligned} Q =Q_{j,k} = \bigg [ \frac{ k_1}{2^j},\frac{ k_1+ 1}{2^j} \bigg ) \times \cdots \times \bigg [ \frac{ k_n}{2^j}, \frac{ k_n+ 1}{2^j} \bigg ). \end{aligned}$$

The standard basis vectors in $\mathbb {R}^n$ or $\{0,1\}^n$ will be denoted by $e_j$. The Riesz transform $R_j$ on $L^2(\mathbb {R}^n)$ is defined through multiplication on the Fourier side by $-i \xi _j/|\xi |$. In [8, Theorem 2.1] and [12, Theorem 5] it was shown that if $\epsilon \in \{0,1\}^n$ satisfies $\epsilon _j =1$, then there is a constant C such that

$$\begin{aligned} \left\| P^{(\epsilon )}u \right\| _2 \le C \Vert u\Vert _2^{1/2} \Vert R_j u \Vert _2^{1/2} \quad \text { for all } u \in L^2(\mathbb {R}^n), \end{aligned}$$

(4)

where $\epsilon$ is fixed and $P^{(\epsilon )}$ is the projection onto the closed span of the set

$$\begin{aligned} \left\{ h_Q^{(\epsilon )} : Q \subseteq \mathbb {R}^n \text { is a dyadic cube } \right\} . \end{aligned}$$

Lemma 1

If $f: \mathbb {M}^{2 \times n} \rightarrow \mathbb {R}$ is rank-one convex with $f(0)=0$, and if $u_1, \ldots , u_{n-1}, v_n$ have finite expansions in the Haar system

$$\begin{aligned} u_i = \sum _{ \epsilon _n =0} \sum _{j=J}^K \sum _{k \in \mathbb {Z}^n} a_{j,k, i}^{(\epsilon )} h_{j,k}^{(\epsilon )} \quad \text { for } 1 \le i \le n-1, \text { and } \quad \quad v_n = \sum _{j=J}^K \sum _{k \in \mathbb {Z}^n} b_{j,k} h_{j,k}^{(e_n)}, \end{aligned}$$

so that $a_{j,k, i}^{(\epsilon )}=b_{j,k}=0$ whenever |k| is sufficiently large, then

$$\begin{aligned} \int _{\mathbb {R}^n} f\begin{pmatrix} u_1 &{} \cdots &{} u_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} v_n \end{pmatrix} \, dx \ge 0. \end{aligned}$$

Proof

The assumption that $a_{j,k, i}^{(\epsilon )}=b_{j,k}=0$ for |k| sufficiently large means the integral converges absolutely. Let

$$\begin{aligned} \widetilde{u}_i = \sum _{ \epsilon _n = 0} \sum _{j=J}^{K-1} \sum _{k \in \mathbb {Z}^n} a_{j,k, i}^{(\epsilon )} h_{j,k}^{(\epsilon )} \quad \text { for } 1 \le i \le n-1, \text { and let } \quad \widetilde{v}_n = \sum _{j=J}^{K-1} \sum _{k \in \mathbb {Z}^n} b_{j,k} h_{j,k}^{(e_n)}. \end{aligned}$$

Then on $Q_{K,k}$, for any $k \in \mathbb {Z}^n$,

$$\begin{aligned} u_i' := u_i - \widetilde{u}_i =\sum _{ \epsilon _n = 0} a_{K,k, i}^{(\epsilon )} h_{K,k}^{(\epsilon )}, \quad v_n' := v_n -\widetilde{v}_n = b_{K,k}h_{K,k}^{(e_n)}, \end{aligned}$$

and

$$\begin{aligned}&\int _{Q_{K,k}} f\begin{pmatrix} u_1 &{} \cdots &{} u_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} v_n \end{pmatrix} \, dx \\&\quad = \int _{Q_{K,k}} f\begin{pmatrix} \widetilde{u}_1+ u_1' &{} \cdots &{} \widetilde{u}_{n-1}+ u_{n-1}' &{} 0 \\ 0 &{} \cdots &{} 0 &{} \widetilde{v}_n + v_n' \end{pmatrix} \, dx_1 \, \cdots \, dx_n. \end{aligned}$$

The bottom row is constant in $x_1, \ldots , x_{n-1}$ on $Q_{K,k}$, and so the function is convex for the integration with respect to $x_1, \ldots , x_{n-1}$. The terms $\widetilde{u}_i$ and $\widetilde{v}_n$ are constant on $Q_{K,k}$, and the $x_1, \ldots , x_{n-1}$ integral of $u_i'$ over the $(n-1)$-dimensional dyadic cube inside $Q_{K,k}$ is zero (for any $x_n$). Hence applying Jensen’s inequality for convex functions gives

$$\begin{aligned}&\int _{Q_{K,k}} f\begin{pmatrix} u_1 &{} \cdots &{} u_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} v_n \end{pmatrix} \, dx \\&\quad \ge \int _{Q_{K,k}} f\begin{pmatrix} \widetilde{u}_1 &{} \cdots &{} \widetilde{u}_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} \widetilde{v}_n + v_n' \end{pmatrix} \, dx_1 \, \cdots \, dx_n. \end{aligned}$$

Applying Jensen’s inequality similarly to the integration in $x_n$, and summing over all $k \in \mathbb {Z}^n$ gives

$$\begin{aligned} \int _{\mathbb {R}^n} f\begin{pmatrix} u_1 &{} \cdots &{} u_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} v_n \end{pmatrix} \, dx \ge \int _{\mathbb {R}^n} f\begin{pmatrix} \widetilde{u}_1 &{} \cdots &{} \widetilde{u}_{n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} \widetilde{v}_n \end{pmatrix} \, dx. \end{aligned}$$

By induction this proves the lemma. $\square$

Proof of Theorem 2

Let $\mu$ be a homogeneous gradient Young measure supported in $\mathbb {M}^{2 \times n}_{{{\mathrm{diag}}}}$, and let $f: \mathbb {M}^{2 \times n} \rightarrow \mathbb {R}$ be a rank-one convex function. It is required to show that

$$\begin{aligned} \int f \, d\mu \ge f( \overline{\mu }). \end{aligned}$$

Without loss of generality it may be assumed that $\overline{\mu }=0$ and that $f(0)=0$. After replacing f by an extension of f which is equal to f on $({{\mathrm{supp}}}\mu )^{co}$, it can also be assumed that there is a constant C with

$$\begin{aligned} |f(X)| \le C(1+|X|^2) \quad \text { for all } X \in \mathbb {M}^{2 \times n}. \end{aligned}$$

(5)

Let $\varOmega \subseteq \mathbb {R}^n$ be the open unit cube. By the characterisation of gradient Young measures [15, Theorem 8.16] there is a sequence $\phi ^{(j)}=(\phi ^{(j)}_1,\phi ^{(j)}_2)$ in $W^{1, \infty }(\varOmega , \mathbb {R}^2)$ whose gradients generate $\mu$, which means that

$$\begin{aligned} \lim _{j \rightarrow \infty } \int _{\varOmega } \eta (x) g\left( \nabla \phi ^{(j)}(x) \right) \, dx = \int g \, d\mu \cdot \int _{\varOmega } \eta (x) \, dx, \end{aligned}$$

(6)

for any continuous g and for all $\eta \in L^1(\varOmega )$. In particular $\nabla \phi ^{(j)} \rightarrow 0$ weakly in $L^2(\varOmega , \mathbb {M}^{2 \times n})$. By the sharp version of the Zhang truncation theorem (see [13, Corollary 3]) it may be assumed that

$$\begin{aligned} \left\| {{\mathrm{dist}}}\left( \nabla \phi ^{(j)}, \mathbb {M}^{2 \times n}_{{{\mathrm{diag}}}} \right) \right\| _{\infty } \rightarrow 0. \end{aligned}$$

(7)

As in Lemma 8.3 of [15], after multiplying the sequence by cutoff functions and diagonalising in such a way as to not affect (7), it can additionally be assumed that $\phi ^{(j)} \in W_0^{1, \infty }(\varOmega , \mathbb {R}^2)$ (the choice $p=\infty$ is not that important, any large enough p would work).

Let $P_1: L^2(\mathbb {R}^n) \rightarrow L^2(\mathbb {R}^n)$ be the projection onto the closed span of

$$\begin{aligned} \left\{ h_Q^{(\epsilon )} : Q \subseteq \mathbb {R}^n \text { is a dyadic cube and } \epsilon _n = 0 \right\} , \end{aligned}$$

and let $P_2: L^2(\mathbb {R}^n) \rightarrow L^2(\mathbb {R}^n)$ be the projection onto the closed span of

$$\begin{aligned} \left\{ h_Q^{(\epsilon )} : Q \subseteq \mathbb {R}^n \text { is a dyadic cube and } \epsilon = e_n \right\} . \end{aligned}$$

Write $w^{(j)} = \nabla \phi ^{(j)}$, so that by (7) and the fact that $R_i \partial _j \phi = R_j \partial _i \phi$,

$$\begin{aligned} \left\| R_n w^{(j)}_{1,1} \right\| _2, \ldots , \left\| R_nw^{(j)}_{1,n-1}\right\| _2 \rightarrow 0, \quad \left\| R_1w^{(j)}_{2,n}\right\| _2, \ldots , \left\| R_{n-1}w^{(j)}_{2,n}\right\| _2 \rightarrow 0. \end{aligned}$$

Hence by (4) and orthogonality

$$\begin{aligned} \left\| w^{(j)}_{1,1}-P_1 w^{(j)}_{1,1} \right\| _2, \ldots ,\left\| w^{(j)}_{1,n-1}-P_1 w^{(j)}_{1,n-1}\right\| _2 \rightarrow 0, \quad \left\| w^{(j)}_{2,n}-P_2 w^{(j)}_{2,n} \right\| _2 \rightarrow 0. \end{aligned}$$

(8)

The function f is separately convex since it is rank-one convex. Hence by the quadratic growth of f in (5) (see Observation 2.3 in [10]), there exists a constant K such that

$$\begin{aligned} |f(X)-f(Y)| \le K(1+|X|+|Y|)|X-Y| \text { for all } X,Y \in \mathbb {M}^{2 \times n}. \end{aligned}$$

(9)

Hence applying (6) with $\eta = \chi _{\varOmega }$ gives

$$\begin{aligned} \int f \, d\mu&= \lim _{j \rightarrow \infty } \int _{\varOmega } f\left( w^{(j)} \right) \, dx \nonumber \\&= \lim _{j \rightarrow \infty } \int _{\varOmega } f\begin{pmatrix} P_1w^{(j)}_{11} &{} \cdots &{} P_1w^{(j)}_{1,n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} P_2w^{(j)}_{2,n} \end{pmatrix} \, dx, \end{aligned}$$

(10)

by (7), (8), (9) and the Cauchy–Schwarz inequality. The functions $w^{(j)}$ are supported in $\overline{\varOmega }$ and satisfy $\int _{\varOmega } w^{(j)} \, dx = 0$ by the definition of weak derivative. Hence the $L^2(\mathbb {R}^n)$ inner product satisfies $\left\langle w^{(j)}, h_Q^{(\epsilon )} \right\rangle = 0$ whenever Q is a dyadic cube not contained in $\overline{\varOmega }$. This implies that $P_1w^{(j)}$ and $P_2w^{(j)}$ are supported in $\overline{\varOmega }$. The integrand in (10) therefore vanishes outside $\overline{\varOmega }$, and so

$$\begin{aligned} \int f\, d\mu = \lim _{j \rightarrow \infty } \int _{\mathbb {R}^n} f\begin{pmatrix} P_1w^{(j)}_{11} &{} \cdots &{} P_1w^{(j)}_{1,n-1} &{} 0 \\ 0 &{} \cdots &{} 0 &{} P_2w^{(j)}_{2,n} \end{pmatrix} \, dx \ge 0 \end{aligned}$$

by (9) and Lemma 1. This finishes the proof. $\square$