Extremal Functions for Morrey’s Inequality

Abstract

We give a qualitative description of extremals for Morrey’s inequality. Our theory is based on exploiting the invariances of this inequality, studying the equation satisfied by extremals, and the observation that extremals are optimal for a related convex minimization problem.

Appendices

A Finite Chain Lemma

We will first pursue the Finite Chain Lemma for $n=2$. To this end, we will need to recall a few facts about isosceles triangles. Suppose $a>0$ and consider an isosceles triangle with two sides equal to $1+a$ and another equal to a. Using the law of cosines we find that the angle $\theta (a)$ between the two sides of length $1+a$ satisfies (Fig. 5)

$$\begin{aligned} \cos (\theta (a))=1-\frac{1}{2}\left( \frac{a}{1+a}\right) ^2. \end{aligned}$$

It is also easy to check that $0<\theta (a)<\pi /3$ and that $\theta (a)$ is increasing in a. In particular, $\theta (1)\leqq \theta (a)$ for $a\ge 1$. It will also be useful to note

$$\begin{aligned} \theta (1)=\cos ^{-1}\left( \frac{7}{8}\right) >\frac{\pi }{7}. \end{aligned}$$

With these observations, we can derive the following assertion:

Lemma A.1

Assume $a\geqq 1$ and $x,y\in \mathbb {R}^2$ with

$$\begin{aligned} |y|= |x|=1+a. \end{aligned}$$

If

$$\begin{aligned} |y-x|>a, \end{aligned}$$

there are $z_1,\ldots , z_m\in \mathbb {R}^2$ with $m\in \{1,\ldots , 6\}$ such that

$$\begin{aligned}&|z_1|=\cdots =|z_m|=1+a,\nonumber \\&|x-z_1|=|z_1-z_2|=\cdots =|z_{m-1}-z_m|=a, \end{aligned}$$

(A.1)

and

$$\begin{aligned} \left| y-z_m\right| \leqq a. \end{aligned}$$

Fig. 5

This figure on the left is a schematic of the assertion made in Lemma A.1. This corresponds to the case where three points $z_1,z_2,z_3$ are needed to form a finite chain linking x to y on the circle centered at 0 of radius $1+a$. The enlarged triangle on the right shows how $\theta (a)$ is defined and how it is related to the figure on the left

Proof

First assume $x=(1+a)e_1$. Note that

$$\begin{aligned} y=(1+a)(\cos \vartheta ,\sin \vartheta ) \end{aligned}$$

for some $\vartheta \in (\theta (a),2\pi -\theta (a))$. If $\vartheta \in (\theta (a),\pi ]$, we set

$$\begin{aligned} z_j=(1+a)(\cos (j\theta (a)),\sin (j\theta (a))) \end{aligned}$$

$j=1,\ldots , 6$. By the definition of $\theta (a)$, (A.1) holds.

As $7\theta (a)>\pi$, there is $m\in \{1,\ldots , 6\}$ such that

$$\begin{aligned} m\theta (a)\leqq \vartheta \leqq (m+1)\theta (a). \end{aligned}$$

It follows that

$$\begin{aligned} |y-z_m|^2&=(1+a)^22\left[ 1-\cos (m\theta (a)-\vartheta )\right] \\&\leqq (1+a)^22\left[ 1-\cos (\theta (a))\right] \\&=a^2. \end{aligned}$$

If $\vartheta \in (\pi ,2\pi -\theta (a))$, we consider the reflection Ry of y about the x-axis. In particular, we can reason as in the case when $\vartheta \in (\theta (a),\pi ]$ for Ry and obtain points $z_1,\ldots , z_m\in \mathbb {R}^2$. In this case, $Rz_1,\ldots , Rz_m$ satisfy the conclusion of this lemma. For a general x that is not necessarily equal to $(1+a)e_1$, we can find a rotation O of $\mathbb {R}^2$ so that $Ox=(1+a)e_1$. Then we can prove the assertion for $(1+a)e_1$ and Oy to get points $z_1,\ldots , z_m\in \mathbb {R}^2$ satisfying the conclusion of the lemma as we argued above. Then $O^{-1}z_1,\ldots , O^{-1}z_m$ satisfy the conclusion of this lemma for the given x and y. $\quad \square$

Corollary A.2

Assume $s\geqq t>0$ and $x,y\in \mathbb {R}^2$ with

$$\begin{aligned} |y|= |x|=t+s. \end{aligned}$$

If

$$\begin{aligned} |y-x|>s, \end{aligned}$$

there are $z_1,\ldots , z_m\in \mathbb {R}^2$ with $m\in \{1,\ldots , 6\}$ such that

$$\begin{aligned}&|z_1|=\cdots =|z_m|=t+s, \end{aligned}$$

(A.2)

$$\begin{aligned}&|x-z_1|=|z_1-z_2|=\cdots =|z_{m-1}-z_m|=s, \end{aligned}$$

(A.3)

and

$$\begin{aligned} \left| y-z_m\right| \leqq s. \end{aligned}$$

(A.4)

Proof

We can apply the previous lemma with $a=s/t\ge 1$ and $x/t, y/t\in \mathbb {R}^2$.

$\square$

We are of course interested in the scenario where |x| is not necessarily equal to |y|. Fortunately, we can just add an additional point to obtain an analogous statement. We also will need to use the following elementary fact: if $x,y\in \mathbb {R}^2$ with $|y|\geqq |x|>0$, then

$$\begin{aligned} \left| |x|\frac{y}{|y|}-x\right| \leqq |x-y|. \end{aligned}$$

(A.5)

Corollary A.3

Assume $s\geqq t>0$ and $x,y\in \mathbb {R}^2$ with

$$\begin{aligned} |y|\geqq |x|=t+s. \end{aligned}$$

Suppose

$$\begin{aligned} \left| |x|\frac{y}{|y|}-x\right| >s. \end{aligned}$$

• (i) Then there are $z_1,\ldots , z_{m}\in \mathbb {R}^2$ with $m\in \{1,\ldots , 7\}$ such that

  $$\begin{aligned} |z_1|=\cdots =|z_{m}|=t+s, \end{aligned}$$

  and

  $$\begin{aligned} |x-z_1|, \ldots ,|z_i-z_{i+1}|,\ldots ,|z_{m}-y|\leqq |y-x|. \end{aligned}$$

  (A.6)

• (ii) Furthermore,

  $$\begin{aligned} {\left\{ \begin{array}{ll} B_{t_0/2}\left( \frac{x+z_1}{2}\right) &{}\text {with}\quad t_0=|x-z_1|\\ \qquad \vdots &{} \\ B_{t_i/2}\left( \frac{z_i+z_{i+1}}{2}\right) &{}\text {with}\quad t_i=|z_i-z_{i+1}|\\ \qquad \vdots &{} \\ B_{t_{m}/2}\left( \frac{z_{m}+y}{2}\right) &{}\text {with}\quad t_{m}=|z_{m}-y| \end{array}\right. } \end{aligned}$$

  (A.7)

  are all subsets of $\mathbb {R}^2{\setminus } B_t(0)$.

Proof

(i) Corollary A.2 applied to x and $|x|\frac{y}{|y|}$ give at most six points $z_1,\ldots , z_{m-1}$ such that (A.2) holds. We also observe that by (A.3) and inequality (A.5)

$$\begin{aligned} |x-z_1|=|z_1-z_2|=\cdots =|z_{m-2}-z_{m-1}| =s<\left| |x|\frac{y}{|y|}-x\right| \leqq |y-x|. \end{aligned}$$

Set

$$\begin{aligned} z_{m}:=|x|\frac{y}{|y|}, \end{aligned}$$

and note by conclusion (A.4) of the previous corollary that $|z_{m-1}-z_m|\leqq s\leqq |y-x|$. Moreover, $|z_{m}|=|x|=t+s$. As

$$\begin{aligned} |z_{m}-y|=\left| |x|\frac{y}{|y|}-y\right| =|y|-|x|\leqq |y-x|, \end{aligned}$$

we have verified each inequality in (A.6).

(ii) Since $|x|=s+t$, $B_s(x)\subset \mathbb {R}^2{\setminus } B_t(0)$. Moreover, if $z\in B_{s/2}\left( \frac{x+z_1}{2}\right)$ then

$$\begin{aligned} |z-x|\leqq \left| z- \frac{x+z_1}{2}\right| + \left| \frac{x+z_1}{2} -x\right| \leqq \frac{s}{2}+\frac{|x-z_1|}{2}=s. \end{aligned}$$

Thus, $B_{s/2}\left( \frac{x+z_1}{2}\right) \subset B_s(x)\subset \mathbb {R}^2{\setminus } B_t(0)$ with $s=|x-z_1|$. The inclusions for $i=1,\ldots , m-1$ in (A.7) follow similarly.

As for the case $i=m$, set $t_m:=|z_{m}-x|$. Note that the closest point in $B_{t_m/2}\left( \frac{z_{m}+y}{2}\right)$ to the origin is $z_{m}=|x|\frac{y}{|y|}$. Thus for any $z\in B_{t_m/2}\left( \frac{z_{m}+y}{2}\right)$, $|z|\geqq ||x|\frac{y}{|y|}|=|x|>t$. Hence, $B_{t_m/2}\left( \frac{z_{m}+y}{2}\right) \subset \mathbb {R}^2{\setminus } B_t(0)$ and we conclude (A.7) (Fig. 6). $\quad \square$

Fig. 6

This diagram shows how we can adapt our proof of Lemma A.1 to the case where $|y|>|x|$. We do so by simply adding another point $z_4=(|x|/|y|)y$ to the chain we obtained by linking x and (|x|/|y|)y

It turns out that we can easily generalize the ideas we developed for $n=2$ to all $n\geqq 3$. The main insight is that for any two-dimensional subspace of $\mathbb {R}^n$ containing x and y, we can find a chain of points linking x to y that belong to this subspace. In particular, we can accomplish this task by applying Corollary A.3.

Corollary A.4

Suppose $n\geqq 2$. Assume $s\geqq t>0$ and $x,y\in \mathbb {R}^n$ with

$$\begin{aligned} |y|\geqq |x|=t+s. \end{aligned}$$

Suppose that

$$\begin{aligned} \left| |x|\frac{y}{|y|}-x\right| >s. \end{aligned}$$

• (i) Then there are $z_1,\ldots , z_{m}\in \mathbb {R}^n$ with $m\in \{1,\ldots , 7\}$ such that

  $$\begin{aligned} |z_1|=\cdots =|z_{m}|=t+s, \end{aligned}$$

  and

  $$\begin{aligned} |x-z_1|, \ldots ,|z_i-z_{i+1}|,\ldots ,|z_{m}-y|\leqq |y-x|. \end{aligned}$$

• (ii) Furthermore,

  $$\begin{aligned} {\left\{ \begin{array}{ll} B_{t_0/2}\left( \frac{x+z_1}{2}\right) &{}\text {with}\quad t_0=|x-z_1|\\ \qquad \vdots &{} \\ B_{t_i/2}\left( \frac{z_i+z_{i+1}}{2}\right) &{}\text {with}\quad t_i=|z_i-z_{i+1}|\\ \qquad \vdots &{} \\ B_{t_{m}/2}\left( \frac{z_{m}+y}{2}\right) &{}\text {with}\quad t_{m}=|z_{m}-y| \end{array}\right. } \end{aligned}$$

  are all subsets of $\mathbb {R}^n{\setminus } B_t(0)$.

Proof

Choose a two-dimensional subspace in $\mathbb {R}^n$ that includes x and y. We can then apply Corollary A.3 to obtain $z_1,\ldots , z_{m}\in \mathbb {R}^n$ that belong to this subspace and check that these points satisfy the desired conclusions. We leave the details to the reader. $\quad \square$

Proof of the Finite Chain Lemma

Without loss of generality, we may assume $|y|\geqq |x|$. Set $S:=|x|-R$, and note $S\geqq R$ with

$$\begin{aligned} |y|\geqq |x|=R+S. \end{aligned}$$

We will verify the claim by considering the following two cases:

• (i) $\displaystyle \left| |x|\frac{y}{|y|}-x\right| \leqq S$

• (ii) $\displaystyle \left| |x|\frac{y}{|y|}-x\right| > S$

Case $\underline{(i)}$: As $|x|=R+S$, $B_S(x)\subset \mathbb {R}^n{\setminus } B_R(0)$. And by adapting our the proof of (A.7), we find

$$\begin{aligned} B_{r/2}\left( \frac{x+(|x|/|y|)y}{2}\right) \subset B_S(x) \end{aligned}$$

for $\displaystyle r=\left| |x|\frac{y}{|y|}-x\right|$ and

$$\begin{aligned} B_{s/2}\left( \frac{y+(|x|/|y|)y}{2}\right) \subset \mathbb {R}^n{\setminus } B_R(0) \end{aligned}$$

for $s=\displaystyle \left| |x|\frac{y}{|y|}-y\right| =|y|-|x|$. We also have $r\leqq |x-y|$ by inequality A.5 and $s\leqq |y-x|$ by the triangle inequality. Therefore, the claim holds for $m=1$ and

$$\begin{aligned} z_1=|x|\frac{y}{|y|}. \end{aligned}$$

Case $\underline{(ii)}$: We can apply Corollary A.4 with $s=S$ and $t=R$ to obtain an $m=\{1,\ldots ,7\}$ and a finite sequence $z_1,\ldots , z_m\in \mathbb {R}^n{\setminus } B_{2R}(0)$ which satisfy (6.6) and (6.7). $\quad \square$

B Coordinate Gradient Descent

In this section, we will change notation and use (x, y) to denote a point in $\mathbb {R}^2$. For a given $\ell >1$, we seek to approximate minimizers of the two dimensional integral

$$\begin{aligned} \int ^\ell _{-\ell }\int ^{\ell }_{-\ell }|Dv(x,y)|^pdxdy \end{aligned}$$

among functions $v\in W^{1,p}([-\ell ,\ell ]^2)$ which satisfy

$$\begin{aligned} v(0,1)= 1\quad \text {and}\quad v(0,-1)=-1. \end{aligned}$$

(B.1)

It can be shown that a unique minimizer $u_\ell \in W^{1,p}([-\ell ,\ell ]^2)$ exists and that $u_\ell$ is p-harmonic and thus continuously differentiable in $(-\ell ,\ell )^2{\setminus }\{(0,\pm 1)\}$. Moreover, as $\ell \rightarrow \infty$, $u_\ell$ converges locally uniformly to an extremal of Morrey’s inequality in $\mathbb {R}^2$ which satisfies (B.1). Therefore, our numerical approximation for $u_\ell$ will in turn serve as an approximation for the corresponding extremal.

To this end, we suppose that $\ell \in \mathbb {N}$ and divide the interval $[-\ell ,\ell ]$ into $N-1$ evenly spaced sub-intervals of length

$$\begin{aligned} h=\frac{2\ell }{N-1}. \end{aligned}$$

Along the x-axis, we will label the endpoints of these intervals with

$$\begin{aligned} x_i=-\ell +(i-1)h \end{aligned}$$

for $i=1,\ldots , N$ and along the y-axis we will use the labels

$$\begin{aligned} y_j=-\ell +(j-1)h \end{aligned}$$

for $j=1,\ldots , N$.

Fig. 7

A numerically computed approximation for an extremal of Morrey’s inequality with $n=2$ and $p=4$. Here $\ell =6$ and $k=10$ (so that $N=121$), $\tau =10^{-10}$, and this approximation was obtained after $10^8$ iterations. Our initial guess was $v^0_{i,j}=w(x_i,y_j)$, where $w(x,y)=c \ln [(x^2+(y-1)^2+10^{-2})/(x^2+(y+1)^2+10^{-2})]$ and c is chosen to ensure $w(0,1)=1$ and $w(0,-1)=-1$

Assuming that $v:[-\ell ,\ell ]^2\rightarrow \mathbb {R}$ is continuously differentiable,

$$\begin{aligned}&\int ^\ell _{-\ell }\int ^{\ell }_{-\ell }|Dv(x,y)|^pdxdy\\&\quad \approx \sum ^{N-1}_{i,j=1}|Dv(x_i,y_j)|^ph^2\\&\quad =\sum ^{N-1}_{i,j=1}\left( v_{x}(x_{i},y_{j})^2+v_{y}(x_{i},y_{j})^2\right) ^{p/2}h^2\\&\quad \approx \sum ^{N-1}_{i,j=1}\left( \left( \frac{v(x_{i}+h,y_{j})-v(x_{i},y_{j})}{h}\right) ^2 +\left( \frac{v(x_i,y_j+h)-v(x_i,y_i)}{h}\right) ^2\right) ^{p/2}h^2\\&\quad =\sum ^{N-1}_{i,j=1}\left( \left( \frac{v(x_{i+1},y_j)-v(x_i,y_i)}{h}\right) ^2 +\left( \frac{v(x_i,y_{j+1})-v(x_i,y_i)}{h}\right) ^2\right) ^{p/2}h^2\\&\quad = h^{2-p}\sum ^{N-1}_{i,j=1}\left( \left( v(x_{i+1},y_j)-v(x_i,y_i)\right) ^2 +\left( v(x_i,y_{j+1})-v(x_i,y_i\right) ^2\right) ^{p/2}\\&\quad =h^{2-p}\sum ^{N-1}_{i,j=1}\left( \left( v_{i+1,j}-v_{i,j}\right) ^2 +\left( v_{i,j+1}-v_{i,j}\right) ^2\right) ^{p/2}. \end{aligned}$$

Here we have written

$$\begin{aligned} v_{i,j}=v(x_i,y_j). \end{aligned}$$

We now suppose that N is of the form

$$\begin{aligned} N=2\ell k+1 \end{aligned}$$

for some $k\in \mathbb {N}$; this assumption is equivalent to $h=1/k$. We can then attempt to minimize

$$\begin{aligned} E(v):=\sum ^{N-1}_{i,j=1}\left( \left( v_{i+1,j}-v_{i,j}\right) ^2 +\left( v_{i,j+1}-v_{i,j}\right) ^2\right) ^{p/2} \end{aligned}$$

among the $N^2-1$ variables

$$\begin{aligned} v=\left( \begin{array}{ccccc} v_{1,1} &{} v_{1,2}&{}\dots &{} v_{1,N-1}&{} v_{1,N}\\ v_{2,1} &{} v_{2,2}&{}\dots &{} v_{2,N-1}&{} v_{2,N}\\ \vdots &{} \vdots &{} \ddots &{} \vdots &{} \vdots \\ v_{N-1,1} &{} v_{N-1,2} &{} \dots &{} v_{N-1,N-1} &{}v_{N-1,N} \\ v_{N,1} &{} v_{N,2} &{} \dots &{} v_{N,N-1} &{} \end{array}\right) , \end{aligned}$$

which satisfy

$$\begin{aligned} v_{\ell k+1, (\ell +1)k+1}=1\quad \text {and}\quad v_{\ell k+1, (\ell -1)k+1}=-1. \end{aligned}$$

(B.2)

These constraints are natural as $x_{\ell k+1}=0, y_{(\ell +1)k+1}=1$, and $y_{(\ell -1)k+1}=-1$.

We are now in position to use coordinate gradient descent to minimize E. That is, we choose an initial guess $v^0=(v^0_{i,j})$ which satisfies (B.2), select a small parameter $\tau >0$, and then run the iteration scheme

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle v^{m}_{i,j}=v^{m-1}_{i,j} -\tau \frac{\partial E(v^{m-1})}{\partial v_{i,j}},\quad &{}(i,j) \ne (\ell k+1, (\ell \pm 1)k+1)\\ \\ \displaystyle v^{m}_{i,j}=v^{m-1}_{i,j},\quad &{}(i,j)=(\ell k+1, (\ell \pm 1)k+1) \end{array}\right. } \end{aligned}$$

for $m\in \mathbb {N}$. After we perform this scheme for large number of iterates $m=1,\ldots , M$, for $\tau$ small and k sufficiently large, we can use $v^M_{i,j}$ as an approximation for $u_\ell (x_i,y_j)$. This is what we did to produce the graphs in Figs. 1 and 7 and the contour plot in Fig. 3.