Let $\Omega\subset\mathbb{R}^n$ be open and bounded. Suppose also that $\Omega$ satisfies the exterior sphere condition at $x_0$ and let $B=B_R(y)$ be a ball such that $B\cap\overline{\Omega}=x_0$. Let $L$ be the following linear differential operator: \begin{equation} Lu(x)=a_{ij}(x)D_{ij}u(x)+b_{i}(x)D_{i}u(x)+c(x)u(x)=f(x)\quad a_{ij}(x)=a_{ji}(x). \end{equation}

So $A(x)$ is a real $n\times n$ symmetric matrix at each $x\in\Omega$. I would like to show that if $\vert A(x)\cdot(x-y)\vert\geq\delta>0$ for all $x\in N\cap\Omega$ of some neighbourhood, $N$ of $x_0$, then: \begin{equation} (x-y)^T A\cdot(x-y)\geq \lambda \vert (x-y)\vert^2\quad\forall\ x\in N\cap\Omega \end{equation} where $\lambda>0$.

This is the claim in Gillbarg-Trudinger's Second Order Elliptic PDE book on page 117 (3rd edition).

I thought I'd try to argue by contradiction. So if there is some $x_{\ast}\in N\cap\Omega$ such that: \begin{equation} \vert A(x_{\ast})\cdot(x_{\ast}-y)\vert\geq\delta>0\quad\text{but}\quad (x_{\ast}-y)\cdot A(x_{\ast})\cdot(x-y)=0, \end{equation}

then it must be because the $(x_{\ast}-y)$ is perpendicular to $A(x_{\ast})\cdot(x_{\ast}-y)=0$. I'm not sure where to go from here.

  • $\begingroup$ Certainly this cannot be true as stated, since $w$ and $v$ may be orthogonal, in which case $w\cdot v=0$. $\endgroup$ – rondo9 Mar 2 '13 at 4:59


Ok so first simply note that since $A(x)$ is real and symmetric it can be diagonalized. We may therefore assume that $A(x)$ is diagonal- this is a very common trick in elliptic PDE and is equivalent to rotating the co-ordinate system. Then we simply calculate that:

$A(x)(x-y)$ is the vector of length $n$ where the $i$-th component is equal to $\sum_j a_{ij}(x)(x_j-y_j)$ but since $A$ is diagonal this is just $a_{ii}(x_i-y_i)$. Therefore,

$|A(x)(x-y)|^2 = \sum_i a_{ii}^2(x_i - y_i)^2 \ge \delta^2 >0$.

Therefore there is at least one $k$ so that

$a_{kk}^2 (x_k - y_k)^2 \ge \frac{\delta^2}{n} >0$. (2)

Now as soon as he writes your desired inequality Trudinger then assumes that the coefficients $a_{ij}$ are bounded. It is not clear if he assumes they are bounded to derive your inequality- but since he assumes they are bounded from then on from a practical viewpoint we may assume the $a_{ij}$ are bounded by some $C$. I thought about the proof if they are not bounded but I am not sure how to do it. Then we may use the bound to divide both sides of (2) through by $a_{kk}>0$ to derive:

$a_{kk} (x_k-y_k)^2 \ge \frac{\delta^2}{n C} >0$. (3)

We now write out a formula for

$(x-y)^T A(x) (x-y)= \sum_{ij} a_{ij}(x)(x_i-y_i)(x_j-y_j)$

but since $A$ is diagonal this becomes:

$(x-y)^T A(x) (x-y) = \sum_i a_{ii}(x_i-y_i)^2$.

But since $a_{ii}>0$ for all $i$ by assumption we get, from (3),

$(x-y)^T A(x) (x-y) \ge a_{kk}(x_k-y_k)^2 \ge \delta >0$ (4)

for $x \in \mathcal{N}$ and some new $\delta$. Now $|x-y|^2 \le 4R^2+diam(\Omega)^2$ because $x$ is in $\Omega$ and $y$ is in the enclosing ball and these two regions touch, so if we pick $\lambda>0$ small enough so that:

$\lambda (4 R^2+diam(\Omega)^2) \le \delta$ for the $\delta$ in (4) we are done.

  • $\begingroup$ Ok- I have changed this proof a few times to get it right but I think it is finally correct now. $\endgroup$ – James Mar 12 '13 at 13:06
  • $\begingroup$ The $a_{ii}$ are all strictly positive because (after diagonalization) they are the eigenvalues of $A$. $\endgroup$ – James Mar 12 '13 at 13:19
  • $\begingroup$ Sure thanks, also I think that (assuming my comment on the exterior sphere condition is correct) $\vert x-y\vert^2\leq MR^2$ for some $M$ not necessarily $4$ but the neighbourhood is only so far away from $y$. $\endgroup$ – Nirav Mar 12 '13 at 13:34
  • $\begingroup$ Yes- I already changed that. $\endgroup$ – James Mar 12 '13 at 13:35
  • $\begingroup$ Actually the bound on the $a_{ij}$ is necessary. Otherwise suppose that $(x_0)_1=y_1$ and $a_{11}(x)=(x_1-y_1)^{-1.5}$ wit all other $a_{ij}=0$. Then we calculate that $|A(x)(x-y)|^2 = a_{11}^2(x_1-y_1)^2 = (x_1-y_1)^{-1} \ge \delta^2 >0$ which is bounded below since $|x-y|$ is bounded above. But we calculate: $(x-y)^T A(x)(x-y) = a_{11}(x_1-y_1)^2 = (x_1-y_1)^{\frac{1}{2}} \rightarrow 0$ as $x \rightarrow x_0$. Since $|x-y|^2$ can be assumed bounded below in the neighbourhood $\mathcal{N}$ this means $(x-y)^T A(x)(x-y) \ge \lambda|x-y|^2$ is impossible. $\endgroup$ – James Mar 12 '13 at 15:09

The inequality $w\cdot Aw\geqslant\lambda |w|^2$ for a $\lambda>0$ can hold if and only if $A$ is positive definite.

Indeed, if so is $A$, then take $\lambda$ the smallest eigenvalue, and if $A$ is not positive definite, take $w\neq 0$ such that $w^tAw\leqslant 0$.

  • $\begingroup$ I guess what I am trying to ask is how do we use the fact that $\vert Aw\vert\geq\delta$ to show that $w\cdot Aw\geq\lambda\vert w\vert^2$. $\endgroup$ – Nirav Mar 2 '13 at 12:40
  • $\begingroup$ So, because the latter inequality is homogeneous in $w$ (that is, if it holds for $w$, it will work for $tw$ for real number $t$). $\endgroup$ – Davide Giraudo Mar 2 '13 at 12:41

Actually the thing you are trying to prove is not true. There must be some extra conditions in Trudinger. As a counter example suppose that $\Omega$ is the unit ball and the point $x_0$ is the north pole of the ball. Then we can take the enclosing sphere as a ball of radius $1+\epsilon$ for any $\epsilon >0$ and the point $y$ will just be $(-\epsilon, 0 , \cdots, 0)$ if we choose the $x_1$ axis pointing up to the north pole. Then we may suppose that $A(x_0)$ is the matrix given by $a_{11}=-1$ and all other entries are equal to $0$ since this matrix is real and symmetric and there are no other conditions given in your question.

Then $(x_0-y) = (1+\epsilon, 0, \cdots, 0)$ and so we have

$A(x_0)(x_0-y) = (-1-\epsilon, 0, \cdots, 0)$ so that:

$|A(x_0)(x_0-y)| = 1+\epsilon \ge \frac{1}{2} >0$. Then if the coefficients $a_{ij}(x)$ are continuous in $x$ if we make our neighbourhood $\mathcal{N}$ small enough (depending on the continuity of the $a_{ij}$) we then get

$|A(x)(x-y)| \ge \frac{1}{2} >0$

for all $x \in \mathcal{N}$. However, we now calculate:

$(x_0-y)^T A(x_0) (x_0-y) = -1 <0$

so again by continuity if we pick $\mathcal{N}$ small enough we have:

$(x-y)^T A(x) (x-y) <0$ in $\mathcal{N}$ so the result is not true.

In conclusion there must be some extra condition on the matrix $A$.

Also, the answer above by Davide is incorrect. You do not need the matrix $A(x)$ to be positive definite for $x \in \mathcal{N}$ as you do not need that inequality in Davides answer to be true for all $w$ (which would require $A$ to be positive definite) but only for a small subset of $w$ given by $(x-y)$ as $x$ varies inside $\mathcal{N}$.

  • $\begingroup$ OK I found my copy of Trudinger. The extra condition is that $A(x)$ is strictly positive- in the sense that all its eigenvalues are strictly greater than zero for $x \in \Omega$ but maybe not for $x \in \partial \Omega$ so maybe not for $x_0$. I will try to get back to you with a proof. $\endgroup$ – James Mar 12 '13 at 10:36
  • $\begingroup$ By the enclosing sphere do you mean the $B$ in my original post? I am fairly sure that $y\notin\overline{\Omega}$ and that $B$ is the ball that works for the exterior sphere condition. $\endgroup$ – Nirav Mar 12 '13 at 12:19
  • $\begingroup$ So this would mean that we still have $L$ as elliptic but not necessarily strictly elliptic, right? $\endgroup$ – Nirav Mar 12 '13 at 12:22
  • $\begingroup$ Yes I mean the B in your post. Well actually $y \in \overline{\Omega}$ is possible. Since when you say $B \cap \overline{\Omega} = x_0$ what I think you really mean is $\partial B \cap \overline{\Omega}= x_0$. $\endgroup$ – James Mar 12 '13 at 13:05
  • $\begingroup$ Yes thats right- which I should have guessed based on the title of your post. $\endgroup$ – James Mar 12 '13 at 13:06

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