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I'm given the following question (from Davies' book - Spectral theory of differential operators): Use the theorem (*) below to prove that if $\Omega$ is a convex region in $ \mathbb{R}^2 $ , then: $ \frac{1}{4} \int _\Omega \frac{|f|^2}{d^2} d^2 x \leq \int _\Omega | \bigtriangledown f | ^2 d^2 x $ for all $f \in C_c ^\infty (\Omega ) $ . Deduce that $ Spec(H) \subseteq [\lambda, \infty ) $ , where $ \lambda^{-1} := 4Inradius(\Omega)^2 $.

Theorem : Let $\Omega $ be a region in $ \mathbb{R}^2 $ and let $f \in C_c ^\infty (\Omega ) $. Then : $ \frac{1}{2} \int _\Omega \frac{|f|^2}{m(x)^2} d^2 x \leq \int _\Omega | \bigtriangledown f | ^2 d^2 x $ for all $f \in C_c ^\infty (\Omega ) $ where the pseudodistance $m(x)$ is defined by $ \frac{1}{m(x)^2 } := \frac{1}{2\pi } \int_{-\pi}^{\pi} \frac{d \Theta }{d_\Theta (x)^2 }$ and $ d_\Theta : \Omega \to (0,+\infty ] $ is defined by $ d_\Theta (x) = min \{ |s| : x+se^{i \Theta} \notin \Omega \} $ .

It's also known that if $\Omega $ is regular ( i.e.- there exists a constant $c$ such that $d(x)\leq m(x) \leq cd(x) $ ) , then $ Spec (H) \leq [ \frac{1}{2c^2 Inradius^2}, \infty ) $ .

How can I use these facts in order to answer Davies' question?

Any help will be greatfully acknowledged !

Thanks in advance

$ d(x):=min \{ |x-y| : y \notin \Omega \} $ .

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Sorry ! I added the definition of this $d(x)$ to my first message. Hope you'll be able to help Thanks ! – joshua Jul 13 '12 at 13:46
Another question: should the definition of the pseudodistance be $\frac{1}{m(x)^2} := \ldots$? You gave it with $\frac{1}{m(x)}$ which I think the unit is not right. – Willie Wong Jul 13 '12 at 13:52
You're indeed right again...Sorry ! – joshua Jul 13 '12 at 14:51
up vote 1 down vote accepted

To use the theorem and the fact, it suffices to prove that $\Omega$ is regular, and that $d(x) \leq m(x) \leq \sqrt{2} d(x)$. The first inequality is evident (since the average must be greater than the minimum). It suffices to prove the second inequality, which we re-write equivalently as

$$ \frac{1}{2\pi}\int_0^{2\pi}\frac{1}{d_\theta(x)^2} \mathrm{d}\theta = \frac{1}{m^2(x)} \geq \frac{1}{2d^2(x)} \tag{*} $$

Now, since $\Omega$ is assumed to be convex, its boundary "lies to one side of the tangent line(s)" (I use plural since a convex boundary need not be differentiable, but can have multiple supporting tangent lines). Furthermore, if $\theta_0$ is such that $d_{\theta_0}(x) = \inf_\theta d_\theta(x) = d(x)$ we must have that the boundary is a differentiable function of $\theta$ there, and that the tangent line is perpendicular to the direction $\theta_0$ (this is elementary convex analysis [see below the cut]).

We thus have [see below second cut] that $d_\theta(0) \leq d_{\theta_0}(x) \sec (\theta - \theta_0) $ by convexity. This implies that

$$ \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{d_\theta(x)^2}\mathrm{d}\theta \geq \frac{1}{2\pi}\frac{1}{d(x)^2} \int_0^{2\pi} \cos^2(\theta - \theta_0) \mathrm{d}\theta = \frac{1}{2d(x)^2} $$

which is precisely (*).

If $\Omega$ is convex, it has non-vanishing tangent support. That is for every point $x\in \partial\Omega$ there exists at least one line $\ell_x$ through $x$ such that $\ell_x$ does not intersect the interior of $\Omega$.

Now let $y$ be a point in the interior of $\Omega$. Look at the line $xy$. Suppose $\ell_x$ is not orthogonal to $xy$. Then you can choose a point $z$ on $\ell_x$ such that $|z-y|$ is less than $|x-y|$ strictly (using that the angle $\angle zyx$ is acute). Since along the line $zy$, the distance from $y$ to the boundary is $\leq |z-y|$ (by virtue of $\ell_x$ not intersecting the interior of $\Omega$), we arrive at the conclusion

If there exists a supporting line $\ell_x$ which is not perpendicular to $xy$, the direction $xy$ cannot minimise $d_\theta(y)$.

The contrapositive of that statement states that if $x$ is the point such that the direction $xy$ minimises $d_\theta(y)$, there exists a supporting line (automatically true by convexity) and that supporting line can only be the perpendicular to $xy$. This tells you that the function defining the boundary of $\Omega$ has a unique tangent at $x$, and hence is differentiable there.

Recall that the formula in radial coordinates for a line $\ell$ which comes closest to the origin at the point $(r_0,\theta_0)$ is precisely $$ r = r_0 \sec (\theta - \theta_0) $$ where $\theta \in (\theta_0 - \pi/2, \theta_0+\pi/2)$.

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Dear @Willie: Thanks a lot for your detailed answer. I understood most of it, until the part of the convex analysis: Why we must have that the boundary is a differentiable function of $ \theta $ there? In addition, I will be glad if you'll be able to detail a bit how did you get that $ d_\theta(0) \leq d_{\theta_0}(x) \sec (\theta - \theta_0) $ (I guess it has something to do with properties of triangles). Hope you'll be able to help me Thanks a lot again! – joshua Jul 13 '12 at 22:01
@joshua: see edit. – Willie Wong Jul 16 '12 at 8:10
Thanks a lot Willie Wong ! I appreciate it ! – joshua Jul 16 '12 at 11:31

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