Inequality of polynomial on neighbour quadrilaterals

Question

Let $\mathbb{Q}_p$ (in $\mathbb{R}^2$) denotes the vector space of polynomials of degree $\le p$ in each variable $x_1,x_2$. Let $v_h\in \mathbb{Q}_p(K\cup K')$, where $K,K'\subset \mathbb{R}^2$ are two quadrilateral of $diam\sim h$, $h>0$ fixed, both affine to the square $(0,1)^2$, which either share a vertex either a side (note that we are in the finite element context).

It holds $$ \|v_h\|_{L^\infty(K)}\le C \|v_h\|_{L^\infty(K')}. $$

What does $C>0$ depend upon? What happens if $\operatorname{dist}(K,K')\sim C_1 h$?

Answer 1

Before I start, I'll like to point out:

The coefficient $C$ is invariant under global affine transformations of the plane. In particular, $C$ doesn't depend on $h$ explicitly. If you place $K$ and $K'$ at a distance $C_1 h$ apart and scale $K$, $K'$ and their separation by same amount, same $C$ follows.

With this in mind, we can transform the question as follows:

• Pick a fixed integer $p > 0$.
• Let $Y \subset \mathbb{R}[x,y]$ be the set of polynomials in $x,y$ whose degree in both $x, y$ are at most $p$.
• Let $K'$ be the square $[-1,1]^2$ and $K$ be another square of side $2$.

• For any $f \in Y$, let $\|f\|_K$ and $\|f\|_{K'}$ be the sup-norm of $f$ over $K$ and $K'$.

  • $\|f\|_K = \sup\limits_{(x,y) \in K} |f(x,y)|$
  • $\|f\|_{K'} = \sup\limits_{(x,y) \in K'} |f(x,y)|$

• Let $Y_1 = \{ f \in Y \;|\; \|f\|_{K'} \le 1 \}$

The questions become what are the coefficient $C$ to make following works:

$$\|f\|_K \le C \|f\|_{K'}, \;\forall f \in Y \quad\iff\quad \|f\|_K \le C,\;\forall f \in Y_1$$ This is equivalent to $\sup\limits_{f\in Y_1} \|f\|_K\;\le\; C$. Independent of where $K$ is located, what we need is compute/estimate $\sup\limits_{f\in Y_1} \|f\|_K$.

To my surprise, it is possible to compute this value.

To do this, we need to look at a related 1-dim problem.

• Let $X \subset \mathbb{R}[x]$ be the set of polynomials in $x$ with degree at most $p$.

• For $f \in X$, let $\|f\|_I$ be the sup-norm of $f$ over $I = [-1,1]$.

  • $\|f\|_I = \sup\limits_{x\in I}|f(x)|$.

• Let $X_1 = \big\{ f \in X : \|f\|_I \le 1 \big\}$

The new problem is given $t \in \mathbb{R}$, what is $\Delta(t) \stackrel{def}{=} \sup\limits_{f \in X_1}|f(t)|$?

Let's consider what happens when $t > 1$.

For any $f \in X_1$ and $\epsilon > 0$, consider the function: $$g(x) = (1+\epsilon)T_p(x) - f(x)$$

Let $x_k = \cos\frac{k\pi}{p}$ for $k = 0,1,\ldots,p$. Since $T_p(x)$ reaches $+1$ and $-1$ alternatively at these $p+1$ points, we find $$g(x_0) > 0, g(x_1) < 0, g(x_2) > 0,\ldots$$

This forces $g(x)$ to have at least $p$ roots $\lambda_1,\ldots,\lambda_p$ with $\lambda_i \in (x_{i},x_{i-1})$. Since $g(x)$ is a polynomial of degree at most $p$, these are all the roots of $g(x)$. $g(x)$ can be factorized as $A\prod_{i=1}^p(x - \lambda_i)$ for some constant $A$. Since $g(x_0) > 0$ and $x_0 > \lambda_i$ for all $i$, we find $A > 0$.

Since $t > 1 = x_0$, this leads to

$$(1+\epsilon)T_p(t) - f(t) = A\prod_{i=1}^p(t - x_i) > 0 \implies (1+\epsilon)T_p(t) > f(t) $$ Apply a similar argument to $-f(x)$, we can deduce $(1+\epsilon)T_p(t) > |f(t)|$.

Since this is true for all $f \in X_1$, we get $(1+\epsilon)T_p(t) \ge \sup_{f\in X_1}|f(t)|$

Since this is true for all $\epsilon > 0$, we obtain $T_p(t) \ge \sup_{f\in X_1}|f(t)|$

Since the bound $T_p(t)$ can be reached by taking $f(x)$ to $T_p(x)$, we have

$$\Delta(t) = \sup_{f\in X_1}|f(t)| = T_p(t)\quad\text{ for } t > 1$$

It is clear $\Delta(t) = 1$ for $|t| \le 1$ and $\Delta(t)$ is an even function in $t$. This implies

$$\Delta(t) = \max(1,|T_p(t)|)$$

With this 1-dim result at hand, we can ask a similar 2-dim question:

What is $\Delta(x,y) \stackrel{def}{=} \sup\limits_{f\in Y_1}|f(x,y)|$?

1. If $(x,y) \in K' = I^2$, $\Delta(x,y)$ is trivially $1$.

2. If one of $x,y \in I$, say $y \in I$, then for any $f\in Y_1$, $f(\cdot,y) \in X_1$. By above 1-dim result, it is easy to deduce $\Delta(x,y) = \Delta(x)$. Similarly, $\Delta(x,y) = \Delta(y)$ when $x \in I$.

3. When both $x, y \not\in I$, we can apply result in step 2. to deduce $|f(x,s)| \le \Delta(x)$ for all $f \in Y_1, s \in I$. Apply the 1-dim result to $h \in X_1$ where $h(s) = \frac{f(x,s)}{\Delta(x)}$, we can deduce $$\sup_{f\in Y_1}\left(\frac{f(x,y)}{\Delta(x')}\right) = \sup_{f\in Y_1} h(y) \le \sup_{h\in X_1} h(y) = \Delta(y)$$ This leads to $\Delta(x,y) = \sup\limits_{f\in Y_1}f(x,y) \le \Delta(x)\Delta(y)$. Since the bound on RHS is achieved by $f(s,t) = T_p(s)T_p(t)$, we can conclude $\Delta(x,y) = \Delta(x)\Delta(y)$ in this case.

Combine all these, we have

$$\Delta(x,y) = \Delta(x)\Delta(y) = \max(1,|T_p(x)|)\max(1,|T_p(y)|)$$ Armed with this function, it is easy to see the target "sup" can be computed by following formula.

$$\bbox[border:1px solid blue;padding: 1em]{C_{optimal} = \sup_{f\in Y_1} \|f\|_K = \max_{(x,y)\in K}\Delta(x)\Delta(y)}$$

For example,

1. When $K$ is sharing an edge with $K'$, say $K = [1,3]\times[-1,1]$, the largest $\Delta(x,y)$ in $K$ is achieved at $(3,\cdot)$. In this case, $C_{optimal} = \Delta(3) = T_p(3)$.

2. When $K$ is axis aligned and sharing an vertex with $K'$, say $K = [1,3]^2$, the largest $\Delta(x,y)$ in $K$ is achieved at $(3,3)$. In this case, $C_{optimal} = \Delta(3,3) = \Delta(3)^2 = T_p(3)^2$.

For other configuration where $K$ is separated by a distance $C_1h$ from $K'$, one can compute the optimal $C$ in a similar manner.

Let's say $K$ and $K'$ are related by a translation of distance $C_1h$. Under the affine transformation which maps $K'$ to $I^2$, the center of $K$ will be located at a distance $c = \sqrt{8}C_1$ from the origin. Let's say the center is $(c\cos\theta,c\sin\theta)$ for some $\theta \in [0,\frac{\pi}{2}]$. Over $K$, $\Delta(x,y)$ will be maximized at the vertex $( 1 + c\cos\theta, 1 + c\sin\theta )$. In this case $C_{optimal} = T_p(1+c\cos\theta)T_p(1+c\sin\theta)$.

When $K$ is not axis aligned with $K'$, it is not immediately obvious which point in $K$ will maximize $\Delta(x,y)$. I'll leave the fun (or mess?) to you.