From: Bjørn Rustad Date: Tue, 28 Oct 2014 16:34:08 +0000 (+0100) Subject: More theorems X-Git-Url: http://git.rustad.me/?a=commitdiff_plain;h=bac97117ae309902f46deeac6bee283640ca954d;p=master More theorems --- diff --git a/theory.tex b/theory.tex index 4019a30..44adc3d 100644 --- a/theory.tex +++ b/theory.tex @@ -353,6 +353,7 @@ motions, meaning combinations of translations and rotations. \end{equation} where $\#(l_{\phi, \rho} \cap C)$ is the number of times the line $l_{\phi, \rho}$ intersects the curve $C$. + \label{thm:euclidean_cauchy_crofton} \end{theorem} This elegant formula is very useful when we later will discretize our energy function. The set of lines $\mathcal{L}$ is then discretized @@ -364,24 +365,49 @@ product of two vectors in a point $p$ is calculated as $\langle a, b\rangle_M = \langle a, M(p) b \rangle$. The length of a curve $\gamma$ parametrized by some parameter $t$ then becomes \begin{equation} - \abs{\gamma}_M = \int_\gamma \langle \dot{\gamma}, - M\big(\gamma(t)\big) \, \dot{\gamma} \rangle \, dt + \abs{\gamma}_M = \int_\gamma \sqrt{\langle \dot{\gamma}, + M\big(\gamma(t)\big) \, \dot{\gamma} \rangle} \, dt + \label{eq:riemannian_length} \end{equation} We will now prove a Cauchy--Crofton formula in this case where we have a metric tensor in each point. \begin{theorem}[The Riemannian Cauchy--Crofton formula] - In the case where the scalar product in each point depends on a - metric tensor $M(x)$ varying continuously over our space, the - Cauchy--Crofton formula becomes \fixme{assuming some stuff about - stuff} + Assume that our space $\Omega$ is equipped with a continous metric + tensor $M(x)$, whose eigenvalues are bounded $0 < k \leq + \lambda_2 \leq \lambda_1 \leq K < \infty$ for all $x \in \Omega$. + The Cauchy--Crofton formula for a differentiable curve $C$ of finite + length then becomes \begin{equation} \abs{C}_M = \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu \right)^{\sfrac{3}{2}}} \, d\mathcal{L}(l_{\nu, \rho}). + \label{eq:riemannian_cauchy_crofton} \end{equation} + \label{thm:riemannian_cauchy_crofton} \end{theorem} -\begin{proof} +Before proving this we present an important result from measure theory +that we will need. +\begin{theorem}[The Dominated Convergence theorem] + Let $\{ f_n \}$ be a sequence of real-valued measurable functions on + a complete measure space $(S, \Sigma, \mu)$. Suppose that the + sequence converges pointwise to a function $f$ and is dominated by + some integrable function $g$ in the sense that + \begin{equation} + \abs{f_n(x)} \leq g(x) + \end{equation} + for all $n$ and almost all $x \in S$. Then $f$ is integrable and + \begin{equation} + \lim_{n \to \infty} \int_S \abs{f_n - f} \, d\mu = 0 + \end{equation} + which further implies that + \begin{equation} + \lim_{n \to \infty} \int_S f_n \, d\mu = \int_S f \, d\mu. + \end{equation} +\end{theorem} +For a proof and further background on measure theory and Lebesgue +integration theory see for example \fixme{ref} +\begin{proof}[Proof of the Riemannian Cauchy--Crofton formula] Assume first that our space is equipped with at constant metric tensor $M$. The length of our curve using this tensor can be calculated by transforming the curve and applying the Euclidean @@ -488,31 +514,35 @@ metric tensor in each point. = \frac{\sigma_1^2 \sigma_2^2 }{\left(\sigma_1^2 \nu_x^2 + \sigma_2^2 \nu_y^2\right)^{\sfrac{3}{2}}} - = \frac{\det \Sigma^2}{(\nu^T \, \Sigma^2 \, + = \frac{\det \Sigma^2}{(\nu^T \cdot \Sigma^2 \cdot \nu)^{\sfrac{3}{2}}}. \end{equation} We are interested in the Jacobian of the whole transformation, so all that is left to do is insert $U^T l_{\nu, \rho}$ to obtain \begin{equation} \abs{J_M(l_{\nu, \rho})} = \abs{J_{\Sigma^2}(U^T l_{\nu, \rho})} - = \frac{\det M}{\left(\nu^T U \Sigma^2 U^T \nu\right)^{\sfrac{3}{2}}} - = \frac{\det M}{\left(\nu^T M \nu\right)^{\sfrac{3}{2}}} + = \frac{\det M}{\left(\nu^T \cdot U \Sigma^2 U^T \cdot + \nu\right)^{\sfrac{3}{2}}} + = \frac{\det M}{\left(\nu^T \cdot M \cdot \nu\right)^{\sfrac{3}{2}}} \end{equation} We have now proven that for a constant metric tensor $M$, the length of the differentiable curve $C$ with regards to this tensor can be calculated as \begin{equation} - \abs{C}_M = \int_C \langle \dot{C}, M \dot{C} \rangle \, dt + \abs{C}_M = \int_C \sqrt{\langle \dot{C}, M \dot{C} \rangle} \, + dt = \int_\mathcal{L} \# (l_{\nu, \rho} \cap C) - \frac{\det M}{\left(\nu^T M \nu \right)^{\sfrac{3}{2}}} \, - d\mathcal{L} (l_{\nu, \rho}) + \frac{\det M}{\left(\nu^T \cdot M \cdot \nu + \right)^{\sfrac{3}{2}}} \, d\mathcal{L} (l_{\nu, \rho}) + \label{eq:riemannian_const_m} \end{equation} - We now argue that this also holds for a non-constant but continuous - metric tensor $M(x)$. By partitioning the domain into disjoint sets - $U_i$ such that $\Omega = \cup U_i$, we make a piecewise constant - approximation $M_\pi(x)$ such that if $x \in U_i$ then $M_\pi(x) = - M(x_i)$ for some fixed $x_i \in U_i$. We then approximate the right - side of \fixme{ref} by + We now argue that the similar formula in + \eqref{eq:riemannian_cauchy_crofton} holds for a non-constant but + continuous metric tensor $M(x)$. By partitioning the domain into + disjoint sets $U_i$ such that $\Omega = \cup_i U_i$, we make a + piecewise constant approximation $M_\pi(x)$ such that if $x \in U_i$ + then $M_\pi(x) = M(x_i)$ for some fixed $x_i \in U_i$. We then + approximate \eqref{eq:riemannian_const_m} by \begin{equation} \abs{C}_{M_\pi} = \sum_i \int_\mathcal{L} \#(l_{\nu, \rho} \cap C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho}) @@ -521,15 +551,15 @@ metric tensor in each point. where $w_i$ is the weight-function used in the set $U_i$ which can be written \begin{equation} - w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T M(x_i) \nu + w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T \cdot M(x_i) \cdot \nu \right)^{\sfrac{3}{2}}}. \end{equation} We further simplify the approximation by introducing the global weight-function $w_\pi(\nu, x)$ which is equal to $w_i(\nu)$ when $x \in U_i$. It can be written \begin{equation} - w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \nu - \right)^{\sfrac{3}{2}}}. + w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T \cdot M_\pi(x) + \cdot \nu \right)^{\sfrac{3}{2}}}. \label{eq:wpi_def} \end{equation} Using this weight in \eqref{eq:mpi_approx} we can get rid of the sum @@ -542,13 +572,21 @@ metric tensor in each point. U_i} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) \\ &= \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}). + \label{eq:riemannian_segments} \end{align} As our partition $\pi$ is refined the weight $w_\pi(x)$ converges - pointwise to the continously varying weight $w(x)$. Now it only - remains to show that the integral converges, and that the left side - converges to the left side of \fixme{ref}. + pointwise to the continously varying weight + \begin{equation} + w(\nu, x) = \frac{\det M(x)}{\left( \nu^T \cdot M(x) \cdot \nu + \right)^{\sfrac{3}{2}}} + \end{equation} + found in \eqref{eq:riemannian_cauchy_crofton}. Now it only remains + to show that the left and right hand side of + \eqref{eq:riemannian_segments} converges to the left and right hand + side of \eqref{eq:riemannian_cauchy_crofton}. - Recall from \fixme{ref} that the left hand side is calculated as + Recall from \eqref{eq:riemannian_length} that the left hand side is + calculated as \begin{equation} \abs{C}_{M_\pi} = \int_C \abs{\dot{C}}_{M_\pi} \, dt @@ -556,61 +594,37 @@ metric tensor in each point. \, dt. \end{equation} We know that $M_\pi(x)$ converges pointwise to $M(x)$, and thus - $\abs{\dot{C}}_{M_\pi}$ converges pointwise to $\abs{\dot{C}}_M$. - The largest eigenvalue of $M$ is equal to 1 and we can parametrize - $C$ by its arclength parameter such that $\abs{\dot{C}} = 1$. Thus - the integrand is bounded and we can apply Lebesgue's dominated - convergence theorem to see that $\abs{C}_{M_\pi} \to - \abs{C}_{M}$. + $\abs{\dot{C}(x)}_{M_\pi}$ converges pointwise to + $\abs{\dot{C}(x)}_M$. We have assumed bounds on the eigenvalues + of $M(x)$, and we can parametrize $C$ by its arc length such that + $\abs{\dot{C}(x)} = 1$, and thus the integrand is bounded and we can + apply Lebesgue's dominated convergence theorem to see that + $\abs{C}_{M_\pi} \to \abs{C}_M$. We apply the same theorem to show that the right hand side of - \fixme{ref} converges, but showing that $\sum_x w_\pi(\nu, x)$ is - bounded is a bit more involved. Recall the definition of $w_\pi$ in - \eqref{eq:wpi_def}. The numerator is equal to $\sigma_1^2 - \sigma_2^2$ and thus from the construction in \fixme{ref} bounded - from above by $1$. + \eqref{eq:riemannian_segments} converges. Recall the definition of + $w_\pi$ in \eqref{eq:wpi_def}. The numerator is equal to $\sigma_1^2 + \sigma_2^2 = \lambda_1 \lambda_2$ and is by assumption bounded from + above by $K^2$. - Next we need to bound $\nu^T M_\pi(x) \nu$ away from zero. According - to the Rayleigh principle \fixme{sigma squared is a bit meh?} - \begin{equation} - \sigma_2^2 = \min_{\norm{\xi} = 1} \sqrt{\xi^T M_\pi(x) \xi} - \end{equation} - and thus $\nu^T M_\pi(x) \nu \geq \sigma_2^2$ where $\sigma_2^2$ is - the smallest eigenvalue of $M_\pi(x)$. Recalling the construction in - \fixme{ref} we know that - \begin{align} - \sigma_2^2 &= \left(1 + \frac{(s_1 - - s_2)^2}{\gamma^2}\right)^{-1} \geq \left(1 + - \frac{s_1^2}{\gamma^2}\right)^{-1}, - \end{align} - where $s_1$ and $s_2$ are the largest and smallest eigenvalue of our - structure tensor respectively. Bounding $s_1$ from above would then - imply $\sigma_2^2 \geq K > 0$. + \fixme{sigma squared meh?} - The structure tensor is constructed as follows: + Next we need to bound $\nu^T M_\pi(x) \nu$ away from zero. According + to the Rayleigh principle \begin{equation} - S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla - \tilde{f}_{\sigma} \right)\right)(x). + \lambda_2 = \min_{\norm{\xi} = 1} \sqrt{\xi^T M_\pi(x) \xi} \end{equation} - This is a smooth continuous map from $\bar{\Omega}$ to - $\mathbb{R}^{2\times 2}$. As we can see in \fixme{ref} the - eigenvalues are the roots of a monic polynomial and thus depend - continuously on the coefficients of the polynomial, which in turn - are continuous functions of the elements in the structure tensor - $S(x)$. - - This proves that our weight function $w_\pi$ is bounded from above, - but the sum + and thus $\nu^T M_\pi(x) \nu \geq \lambda_2 \geq k$. The weight + function $w_\pi$ is then bounded, but not neccesarily the sum \begin{equation} - \sum_{x \in l_{\nu, \rho} \cap C} w_\pi(\nu, x) + \sum_{\mathclap{x \in l_{\nu, \rho} \cap C}} w_\pi(\nu, x). \end{equation} - might be infinite. However, if this happens for a set of lines with - measure greater than zero, the Euclidean Cauchy--Crofton formula in - \fixme{ref} implies that the curve is infinitely long. - - Assuming that our curve has finite length we can therefore bound the - integrand of \fixme{ref} and conclude using Lebesgue's dominated - convergence theorem \fixme{ref} that + However, if the $l_{\nu, \rho} \cap C$ is infinite for a set of + lines with measure greater than zero, the Euclidean Cauchy--Crofton + formula in Theorem \ref{thm:euclidean_cauchy_crofton} implies that + the curve is infinitely long \fixme{which we have assumed it is + not?} As the sum is also bounded we can apply Lebesgue's dominated + convergence theorem again and conclude that \begin{equation} \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) @@ -618,7 +632,6 @@ metric tensor in each point. \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} w(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) \end{equation} - \fixme{introduce $w$ somewhere} which--as both sides of the equality has been shown to converge--leaves us with what we wanted to prove \begin{equation} @@ -630,6 +643,39 @@ metric tensor in each point. \end{equation} \end{proof} +\section{Yeeep} + +Here we finish up the consideration of the contiuous problem, woop. + +The structure tensor is constructed as described in \fixme{ref} +\begin{equation} + S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla + \tilde{f}_{\sigma} \right)\right)(x), +\end{equation} +where $\tilde{f}$ is the symmetric extension of our input image $f$ to +$\mathbb{R}^2$. +Because of the convolutions with the Gaussian function, this is a smooth +continuous map from $\bar{\Omega}$ to $\mathbb{R}^{2\times 2}$. As we +can see in \fixme{ref} the eigenvalues are the roots of a monic +polynomial and thus depend continuously on the coefficients of the +polynomial, which in turn are continuous functions of the elements in +the structure tensor $S(x)$. The extreme value theorem \fixme{ref} +states that a continuous real-valued function on a nonempty compact +space is bounded above. Thus the eigenvalues $s_1$ and $s_2$ of $S(x)$ +are bounded from above and by construction the smallest eigenvalue of +our metric tensor $M(x)$ is bounded away from zero as +\begin{equation} + \lambda_2 = \left(1 + + \frac{(s_1 - s_2)^2}{\gamma^2}\right)^{-1} \geq \left(1 + + \frac{s_1^2}{\gamma^2}\right)^{-1} \geq Q > 0. +\end{equation} +Hence, our metric tensor $M(x)$ fulfills all the assumptions of Theorem +\ref{thm:riemannian_cauchy_crofton}. \fixme{did we argue for +continuous?} + +And here we write something about the perimeter not being the length of +the boundary and all that stuff. We really tie the room together. + \chapter{Discrete formulation} This is where we discretize! It will also be an important chapter, as