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Lemma 5.5.6. Let \( S \) be a finite set of places of \( K \) containing all archimedean places. For \( n \in {\mathbb{Z}}_{ \geq 1} \), there exists a cyclic extension \( L \) of \( K \) contained in \( K\left( {\zeta }_{N}\right) \) for some \( N \geq 1 \) such that \( \left\lbrack {{\bar{L}}_{w} : {K}_{v}}\right\rbr... | Proof. We can and do assume \( n \) is even, and write \( n = \mathop{\prod }\limits_{i}{p}_{i}^{{e}_{i}} \) . For each odd \( {p}_{i} \), consider \( {\widetilde{L}}_{i} \mathrel{\text{:=}} \mathop{\lim }\limits_{{ \rightarrow m}}K{\left( {\zeta }_{{p}_{i}^{m}}\right) }_{{p}_{i}} \), where \( K{\left( {\zeta }_{{p}_{i... | Yes |
Proposition 5.5.7. The map \( {\operatorname{inv}}_{K} = \mathop{\sum }\limits_{v}{\operatorname{inv}}_{v} : {H}^{2}\left( {\operatorname{Gal}\left( {\bar{K}/K}\right) ,{\bar{K}}^{ \times }}\right) \rightarrow \mathbb{Q}/\mathbb{Z} \) is trivial. | Proof. Let \( \beta \in {H}^{2}\left( {\operatorname{Gal}\left( {\bar{K}/K}\right) ,{\bar{K}}^{ \times }}\right) \), and \( S \) be the set of places such that \( {\operatorname{inv}}_{v}\left( \beta \right) \neq 0 \) . Let \( n \) be the least common multiple of the orders of the elements \( {\operatorname{inv}}_{v}\l... | Yes |
Corollary 5.5.8. \( {\Phi }_{K}\left( a\right) = 1 \) for all \( a \in {K}^{ \times } \) . | Proof. Let \( L \) be a finite abelian extension of \( K \) . We have \( \chi \left( {{\Phi }_{L/K}\left( a\right) }\right) = \mathop{\sum }\limits_{v}{\operatorname{inv}}_{v}\left( {a \cup \delta \left( \chi \right) }\right) = \) 0 for any character \( \chi : \operatorname{Gal}\left( {L/K}\right) \rightarrow \mathbb{Q... | Yes |
Proposition 5.5.9. Let \( L/K \) be a finite cyclic extension, then \( {\widetilde{\operatorname{inv}}}_{L/K} \) induces an isomorphism \( {\operatorname{inv}}_{L/K} : {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathbb{C}}_{L}}\right) \overset{ \sim }{ \rightarrow }\frac{1}{\left| \operatorname{Gal}\left( L... | Proof. The exact sequence \( 1 \rightarrow {L}^{ \times } \rightarrow {I}_{L} \rightarrow {\mathbb{C}}_{L} \rightarrow 1 \) induces\n\n\[ 1 \rightarrow {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) \rightarrow {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{I}_{L}}\right) \rightarr... | Yes |
Lemma 5.5.11. Let \( L \supset E \supset K \) be finite Galois extensions, then the following diagram commutes | Proof. Let \( v \) be a place of \( K \), and \( w \) be a place of \( L \) dividing \( v \) . For each place \( u \mid v \) of \( E \), there exists \( {\sigma }_{u} \in \operatorname{Gal}\left( {L/K}\right) \) such that \( {\sigma }_{u}\left( w\right) \mid u \) . Recall \( {\sigma }_{u} \) induces an isomorphism \( {... | Yes |
Corollary 5.5.13. The morphism \( \mathop{\lim }\limits_{{ \rightarrow L}}{H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{I}_{L}}\right) \rightarrow \mathop{\lim }\limits_{{ \rightarrow L}}{H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathbb{C}}_{L}}\right) \) is surjective, and \( {\operatorname{inv}}... | Proof. We have by Corollary 5.5.10 (where the injection is induced by inflation)\n\n\[ \mathbb{Q}/\mathbb{Z}\xrightarrow[ \sim ]{\mathop{\operatorname{inv}}\limits_{K}^{{-1}}}\mathop{\lim }\limits_{{L \in {\mathcal{E}}_{K}}}{H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathbb{C}}_{L}}\right) \rightarrow \math... | Yes |
Theorem 5.5.14. Let \( K \) be a number field. Then \( \left( {{\mathbb{C}}_{\bar{K}} \mathrel{\text{:=}} \mathop{\lim }\limits_{{ \rightarrow L}}{\mathbb{C}}_{L},\text{inv}}\right) \) is a class formation. Moreover, the induced reciprocity map \( {\mathbb{C}}_{K} \rightarrow \operatorname{Gal}{\left( \bar{K}/K\right) ... | Proof. Let \( L \) be a finite extension of \( K \) . We have \( {H}^{1}\left( {\operatorname{Gal}\left( {\bar{K}/L}\right) ,{\mathbb{C}}_{L}}\right) = 1 \), and \( {\operatorname{inv}}_{L} \) : \( {H}^{2}\left( {\operatorname{Gal}\left( {\bar{K}/L}\right) ,{\mathbb{C}}_{L}}\right) \overset{ \sim }{ \rightarrow }\mathb... | Yes |
Lemma 5.5.15. We have an isomorphism of topological groups \( {\mathbb{C}}_{K} \cong {I}_{K}^{1}/{K}^{ \times } \times {\mathbb{R}}_{ > 0} \) . | Proof. We have an exact sequence \( 1 \rightarrow {I}_{K}^{1}/{K}^{ \times } \rightarrow {\mathbb{C}}_{K}\xrightarrow[]{{\left| \cdot \right| }_{{I}_{K}}}{\mathbb{R}}_{ > 0} \rightarrow 1 \) . Let \( v \) be an archimedean place of \( K \), then \( {K}_{v}\overset{{\left| \cdot \right| }_{v}}{ \rightarrow }{\mathbb{R}}... | Yes |
Theorem 5.6.1. There exists a modulus \( \mathfrak{m} \) divisible exactly by all ramified primes of \( K \) in \( L/K \) such that the induced map \( {I}_{K}\left( \mathfrak{m}\right) \rightarrow {J}_{K}\left( \mathfrak{m}\right) \rightarrow \operatorname{Gal}\left( {L/K}\right) \) coincides with \( {\Phi }_{K} \) . M... | Proof. Let \( \mathfrak{m} = {\mathfrak{m}}_{\infty }{\mathfrak{m}}^{\infty } \) be a modulus divisible exactly by all ramified primes of \( K \) in \( L/K \) such that \( {W}_{K}\left( \mathfrak{m}\right) \subset {N}_{L/K}\left( {I}_{L}\right) \) . Then we have a surjective map \( {I}_{K}\left( \mathfrak{m}\right) /\l... | Yes |
Proposition 5.6.2. \( H \) is the maximal unramified abelian extension of \( K \) . | Proof. By the local-global compatibility of class field theory, we have for any place \( v \) of \( K \) :  We see \( {\rho }_{{H}_{w}/{K}_{v}}\left( {\mathcal{O}}_{v}^{ \times }\right) = 1 \) hence \( {H}_{w}/{K}_{v} ... | Yes |
Theorem 5.6.3. Let \( H \) be the Hilbert class field of \( K \) . For any fractional ideal \( \mathfrak{a} \) in \( K \) , \( \mathfrak{a}{\mathcal{O}}_{H} \) is principal in \( H \) . | Proof. Let \( {H}^{\prime } \) be the Hilbert class field of \( H \), then we have a commutative diagram \n\nThe theorem amounts to say that the top map is trivial, which follows from the following lemma.\n\nLemma 5.6.... | No |
Lemma 5.6.4. Let \( G \) be a finite group, \( {G}^{\prime } \) be the commutator subgroup. Then the transfer map \( {G}^{\mathrm{{ab}}} \rightarrow {G}^{\prime \mathrm{{ab}}} \) is trivial. | Proof. See Theorem VI.7.6 of Algebraic number theory by Neukirch. | No |
Example 1.1.1. Delimit accurately (can we?) the ozone regions of different Dobson units in the picture and then draw conclusions about overall concentrations and trends. | Let us look at the most latest picture at 1990 in Fig. 1.1(b). The result by thresholding is in Fig. 1.1(c). The result by using \ | No |
Example 1.4.1. How the human perceive process and store the visual information? | We do not have a clear understanding how the human perceive, process and store the visual information. We do not even know how the human measures internally the image visual quality and discrimination. | No |
Given an matrix, what is the perceived information by human? | Do the following exercise:\n\n- Read and display a image file as a two dimensional function. The example matlab script file is here matlab display example.\n\nBoth presentation contain exactly the same information, but for a human observer it is very difficult to find a correspondence between, and without the second, i... | No |
The distance between two pixels in a digital image is a significant quantitative measure. The distance between points with co-ordinates \( \left( {i, j}\right) \) and \( \left( {h, k}\right) \) may be defined in several different ways: The Euclidean distance \( {D}_{E} \) is defined by \[ {D}_{E}\left\lbrack {\left( {i... | The advantage of the Euclidean distance is the fact that it is intuitively obvious. The disadvantages are costly calculation due to the square root, and its not-integer value. | No |
Example 2.3.1. The following figure shows three digital lines with \( {45}^{o} \) and \( - {45}^{o} \) slope. | - If 4-connectivity is used, the lines are not contiguous at each of their points.\n- An even worse conflict with intuitive understanding of line properties is: two perpendicular lines do intersect in one case (upper right intersection) and do not intersect in another case (lower left), as they do not have any common p... | No |
Example 2.3.3. (Connectivity paradox).\n\n\n\nFigure 2.3: Connectivity paradox on a discrete grid.\n\n- If we assume 4-connectivity, the figure contains four separate contiguous regions \( A, B, C \) and \( D \) . \( A... | - One possible solution to contiguity paradox is to treat objects using 4-neighborhoods and background using 8-neighborhoods (or vice versa).\n\n- More exact treatment of digital contiguity paradox and their solution for binary images and images with more brightness levels can be found in [Pavlidis, 1977].\n\n- These p... | No |
Example 3.2.2. An example of such a representation is shown in the following figure and table. |  (a) Image\n\n<table><thead><tr><th>No.</th><th>Object name</th><th>Color</th><th>Min. row</th><th>Min. col.</th><th>Inside</th></tr></thead><tr><td>1</td><td>sun</td><td>white</td><td>5</td><td>40</td><td>2</td></tr><... | Yes |
Quadtrees are modifications of T-pyramids. | - Every node of the tree except the leaves has four children (NW: north-western, NE: north-eastern, SW: south-western, SE: south-eastern).\n- the image is divided into four quadrants at each hierarchical level, however it is not necessary to keep nodes at all levels.\n- If a parent node has four children of the same (e... | Yes |
Quiz 4.0.1. Do you remember the example of filtering impulse noise? (2.3.12) | - If pre-processing aims to correct some degradation in the image, the nature of a priori information is important and is used to different extent:\n- no knowledge about the nature of the degradation is used; only very general properties of the degradation are assumed.\n- using knowledge about the properties of the ima... | No |
Brightness transform is a monotonic function: | \[ q = T\left( p\right) \] | No |
Example 4.1.2. Window and Level example. The Khoros workspace for this example is here Window Level Example.\n\nFirst find the minimum and maximum value, decide the start value, bin-width and number of bins for computing the histogram. From the histogram, chose the lower and upper cutoff value. | ## Histogram stretching\n\n- Histogram stretching can be seen as a Window and Level contrast enhancement technique where the window ranges from the minimum to the maximum pixel values of the image.\n\n- This normalization or histogram stretching operation is automatically performed in many display operators.\n\n![73714... | No |
The aim is to produce an image with equally distributed brightness levels over the whole brightness scale. | - Let \( H\left( p\right) \) be the input histogram and that the input gray-scale is \( \left\lbrack {{p}_{0},{p}_{k}}\right\rbrack \) .\n- The intention is to find a monotonic pixel brightness \( q = T\left( p\right) \) such that the output histogram \( G\left( q\right) \) is uniform over the whole output brightness s... | No |
Example 4.2.3. Geometric transform example. | The Khoros workspace for this example is here Geometric transform Example | No |
Linear operations calculate the resulting value in the output image pixel \( g\left( {i, j}\right) \) as a linear combination of brightnesses in a local neighborhood of the pixel \( f\left( {i, j}\right) \) in the input image. | The contribution of the pixels in the neighborhood is weighted by coefficients \( h \)\n\n\[ f\left( {i, j}\right) = \mathop{\sum }\limits_{{\left( {m, n}\right) \in \mathcal{O}}}h\left( {i - m, j - n}\right) g\left( {m, n}\right) \]\n\n\( \left( {4.50}\right) \)\n\n- The above equation is equivalent to discrete convol... | Yes |
Considering the point \( \left( {m, n}\right) \) in the image, the convolution mask is calculated in the neighborhood \( \mathcal{O} \) from the nonlinear formula\n\n\[ h\left( {i, j}\right) = \left\{ \begin{array}{ll} 1, & \text{ for }g\left( {m + i, n + j}\right) \in \left\lbrack {\min ,\max }\right\rbrack \\ 0, & \t... | Note that in the equation above, the interval [min, max] represents valid data. | No |
Example 4.3.3. Rotating mask example. The matlab script from visionbook is is here Rotating mask Example. |  | No |
Given \( {x}_{1} \leq {x}_{2} \leq \cdots \leq {x}_{N} \), then\n\n1. \( \arg \mathop{\min }\limits_{a}\mathop{\sum }\limits_{{i = 1}}^{N}{\left| {x}_{i} - a\right| }^{2} \) is the arithmetic mean of \( {x}_{1},{x}_{2},\cdots ,{x}_{N} \) ; | -Proof. (1) Let\n\n\[ g\left( a\right) = \mathop{\sum }\limits_{{i = 1}}^{N}{\left| {x}_{i} - a\right| }^{2}. \]\n\n\( \left( {4.60}\right) \)\n\nThe result follows immediately by calculus. | No |
Example: computing the gradient and edge magnitude and direction by finite difference. | The Khoros workspace for this example is here Finite Difference Example. | No |
The crucial question is how to compute the the 2nd derivative robustly. | One possibility is to smooth an image first (to reduce noise) and then compute second derivatives. | No |
Consider a bi-cubic facet model\n\[ g\left( {x, y}\right) = {c}_{1} + {c}_{2}x + {c}_{3}y + {c}_{4}{x}^{2} + {c}_{5}{xy} + {c}_{6}{y}^{2} + {c}_{7}{x}^{3} + {c}_{8}{x}^{2}y + {c}_{9}x{y}^{2} + {c}_{10}{y}^{3} \] | - The parameter of which are estimated from a pixel neighborhood(the co-ordinate of the central pixel is \( \left( {0,0}\right) ) \).\n- To determine the model parameters, a least-squares method with singular-value decomposition may be used.\n- Once the facet model parameters are available for each image pixel, edges c... | No |
Let \( \left( {i, j}\right) \) represent the seed pixel, and \( \left( {k, l}\right) \) represent pixels 8-connected to the seed pixel. The adaptive neighborhood of the pixel \( \left( {i, j}\right) \) is defined as a set of pixels \( \left( {k, l}\right) 8 \) -connected to the seed pixel and either satisfying the addi... | \[ \left| {f\left( {k, l}\right) - f\left( {i, j}\right) }\right| \leq {T}_{1} \] (4.116) or satisfying a multiplicative property \[ \frac{\left| f\left( k, l\right) - f\left( i, j\right) \right| }{f\left( {i, j}\right) } \leq {T}_{2} \] (4.117) where \( {T}_{1} \) and \( {T}_{2} \) are parameters of the adaptive neigh... | Yes |
A complete segmentation of an image \( R \) is a finite set of regions \( {R}_{1},\cdots ,{R}_{S} \) , | \[ R = \mathop{\bigcup }\limits_{{i = 1}}^{S}{R}_{i},\;{R}_{i}\bigcap {R}_{j} = \varnothing \] | Yes |
Algorithm 5.2.3. Inner boundary tracing\n\n1. Search the image from top left until a pixel of a new region is found; This pixel \( {P}_{0} \) then has the minimum column value of all pixels of that region having the minimum row value. Pixel \( {P}_{0} \) is a starting pixel of the region border.\n\nDefine a variable di... | \n\nInner boundary tracing: (a) Direction notation, 4-connectivity, (b) 8-connectivity, (c) pixel neighborhood search sequence in 4-connectivity, (d),(e) search sequence in 8-connectivity, (f) boundary tracing in 8- ... | Yes |
Consider the following simple boundary-tracing problem,\n\n  (b)\n\n- The aim is to find the best path (minimum ... | - The main idea of the principle of optimality is: Whatever the path to the node \( E \) was, there exists an optimal path between \( E \) and the end point. In other words, if the optimal path [start point-endpoint] goes through \( E \), then both its parts [start point-E] and [E-end point] are also optimal, respectiv... | Yes |
Theorem 6.4.1. If an image multi-scale analysis \( {T}_{t} \) is causal \( \mu \) and regular then \( I\left( {t, x}\right) = {T}_{t}{\left( I\right) }_{\left( x\right) } \) is a viscosity solution of\n\n\[ \frac{\partial I}{\partial t} = F\left( {{\nabla }^{2}I,\nabla I, I, x, t}\right) \]\n\n(6.6)\n\nwhere the functi... | -Proof We give a simplified proof by assuming that \( I\left( {t, x}\right) \) is \( {C}^{2} \) . A completed and rigorous proof could be found in Alvarez et al.,1993. In a neighborhood of \( \left( {t, x}\right) \), we have\n\n\[ I\left( {t, y}\right) = I\left( {t, x}\right) + < \nabla I\left( x\right), y - x > + \fra... | No |
Theorem 6.4.3. Let \( N = 2 \) . If a multiscale analysis is causal, regular, translation invariant, Euclidean invariant, morphological invariant, \( I\left( {t, x}\right) = {T}_{t}\left( {I}_{0}\right) \left( x\right) \) is the solution of the heat equation | \[ \left\{ \begin{array}{ll} \frac{\partial I}{\partial t} & = \left| {\nabla I}\right| G\left( {\operatorname{div}\left( \frac{\nabla I}{\left| \nabla I\right| }\right) }\right) , \\ {\left. I\right| }_{t = 0} & = {I}_{0}. \end{array}\right. \] (6.9) where \( G \) is a continuous function on \( \mathbf{R} \times \left... | Yes |
Theorem 1.2.2. ([Stein and Weiss,1971, Theorem I.1.3]) If \( f \in {L}^{p}\left( {\mathbf{R}}^{n}\right) ,1 \leq p \leq \infty \), and \( g \in {L}^{1}\left( {\mathbf{R}}^{n}\right) \), then \( h = f * g \) is well-defined and belongs to \( {L}^{p}\left( {\mathbf{R}}^{n}\right) \) . Moreover, | \[ \parallel h{\parallel }_{p} \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{1} \] | Yes |
Theorem 1.3.5. ([Stein and Weiss,1971, Corollary 1.18]) If both \( f \) and \( \widehat{f} \) are integrable, then\n\n\[ f\left( x\right) = {\int }_{{\mathbf{R}}^{n}}\widehat{f}\left( \xi \right) {\mathbf{e}}^{{2\pi i\xi } \cdot x}{d\xi } \]\n\n(1.31)\n\nfor almost every \( x \) . | It follows easily from the above theorem: | No |
Example 1.6.2. The \( \delta \) -function is homogeneous of degree -n in the following sense:\n\n\[ \n{D}_{a}\delta = \frac{1}{{a}^{n}}\delta \n\]\n\n(1.56) | because\n\n\[ \n\left( {{D}_{a}\delta }\right) \left( \varphi \right) = \frac{1}{{a}^{n}}\delta \left( {{D}_{\frac{1}{a}}\varphi }\right) = \frac{1}{{a}^{n}}\delta \left( \varphi \right) .\n\]\n\n(1.57) | Yes |
The partial derivative \( {\partial }_{{x}_{i}}\delta \) is homogeneous of degree \( - n - 1 \) in the following sense: | \[ \left( {{D}_{a}{\partial }_{{x}_{i}}\delta }\right) \left( \varphi \right) = \frac{1}{{a}^{n}}\left( {{\partial }_{{x}_{i}}\delta }\right) \left( {{D}_{\frac{1}{a}}\varphi }\right) = - \frac{1}{{a}^{n}}\delta \left( {{\partial }_{{x}_{i}}{D}_{\frac{1}{a}}\varphi }\right) \] \[ = - \frac{1}{{a}^{n}}\delta \left( {\fr... | Yes |
The Fourier transformation of the Dirac \( \delta \) -function in Example 1.5.4 can be computed as follows. | \[ \widehat{\delta }\left( \varphi \right) = \delta \left( \widehat{\varphi }\right) \] \[ = \widehat{\varphi }\left( 0\right) \] \[ = {\int }_{{\mathbf{R}}^{n}}\varphi \left( x\right) {dx} \] Let \( \mathbf{1} \) be the function of constant value 1. Then, \[ \widehat{\delta } = \mathbf{1}\text{.} \] Because the \( \de... | Yes |
Theorem 1.7.1. [Stein and Weiss, 1971, Theorem IV.3.3] If \( f \) is a radial function in \( {L}^{1}\left( {\mathbf{R}}^{n}\right), n \geq 2 \), i.e., \( f\left( x\right) = {f}_{0}\left( {\parallel x\parallel }\right) \). Then the Fourier transform \( \widehat{f} \) is also radial and has the form \( \widehat{f}\left( ... | \[ {F}_{0}\left( r\right) = \frac{2\pi }{{r}^{\frac{n - 2}{2}}}{\int }_{0}^{\infty }{f}_{0}\left( s\right) {J}_{\frac{n - 2}{2}}\left( {2\pi rs}\right) {s}^{\frac{n}{2}}{ds}. \] | Yes |
Theorem 1.7.3. ([Natterer,2001, p. 195]) For a function \( h \) on \( \left\lbrack {-1, + 1}\right\rbrack \), we have\n\n\[{\int }_{{S}^{n - 1}}h\left( {\theta \cdot \omega }\right) {Y}_{l}\left( \omega \right) {d\omega } = c\left( {n, l}\right) {Y}_{l}\left( \theta \right)\]\n\n(1.93)\n\nwhere\n\n\[c\left( {n, l}\righ... | For a proof of this formula, please refer to [Muller, 1998, § 1.4]. | No |
Theorem 1.7.5. ([Stein and Weiss,1971, Theorem IV.3.10]) Suppose \( n \geq 2 \) and \( f \in \) \( {L}^{2}\left( {\mathbf{R}}^{n}\right) \cap {L}^{1}\left( {\mathbf{R}}^{n}\right) \) has the form \( f\left( x\right) = {f}_{0}\left( {\parallel x\parallel }\right) P\left( x\right) \), where \( P\left( x\right) \) is a so... | \[ {F}_{0}\left( r\right) = \frac{2\pi }{{i}^{l}{r}^{\frac{n + {2l} - 2}{2}}}{\int }_{0}^{\infty }{f}_{0}\left( s\right) {J}_{\frac{n + {2l} - 2}{2}}\left( {2\pi rs}\right) {s}^{\frac{n + {2l}}{2}}{ds}. \] | Yes |
Theorem 2.1.1. ([Natterer,2001, Theorem II.1.1]) Let \( f \in \mathcal{S} \) . For \( \theta \in {S}^{n - 1},\sigma \in \mathbf{R} \) ,\n\n\[ \n\widehat{\left( Rf\right) }\left( {\theta ,\sigma }\right) = \widehat{f}\left( {\sigma \theta }\right) \n\] | Proof.\n\n\[ \n\widehat{\left( Rf\right) }\left( {\theta ,\sigma }\right) = {\int }_{\mathbf{R}}\left( {Rf}\right) \left( {\theta, s}\right) {\mathbf{e}}^{-{2\pi is\sigma }}{ds} \n\]\n\n(2.11)\n\n\[ \n= {\int }_{\mathbf{R}}{\mathbf{e}}^{-{2\pi is\sigma }}{ds}{\int }_{{\theta }^{ \bot }}f\left( {{s\theta } + y}\right) {... | Yes |
Theorem 2.1.2. ([Natterer and Wübbeling,2001, Theorem 2.2]) Let \( f, g \in \mathbb{S} \) . Then\n\n\[ \left( {Rf}\right) * \left( {Rg}\right) = R\left( {f * g}\right) \] | Here the convolution on the left-hand side is in \( {\mathcal{C}}^{n} \) , while it is in \( {\mathbf{R}}^{n} \) on the right-hand side. This theorem can be proved by direct computation. We provide a proof by Theorem 1.2.3, Proof.\n\n\[ {\left( \left( Rf\right) * \left( Rg\right) \right) }^{ \land }\left( {\theta ,\sig... | Yes |
Theorem 2.1.3. ([Natterer,2001, p. 11]) Let \( f \in \mathcal{S} \) . Then\n\n\[ \left( {R{\partial }^{\alpha }f}\right) \left( {\theta, s}\right) = {\theta }^{\alpha }\frac{{d}^{\left| \alpha \right| }}{d{s}^{\left| \alpha \right| }}\left( {Rf}\right) \left( {\theta, s}\right) \] | Proof.\n\n\[ {\left( R\left\lbrack {\partial }^{\alpha }f\right\rbrack \right) }^{ \land }\left( {\theta ,\sigma }\right) = \widehat{{\partial }^{\alpha }f}\left( {\sigma \theta }\right) \;\text{ (by Theorem 2.1.1) } \]\n\n\[ = {\left( 2\pi i\sigma \theta \right) }^{\alpha }\widehat{f}\left( {\sigma \theta }\right) \;\... | Yes |
Theorem 2.1.4. ([Natterer and Wübbeling,2001, p. 10]) \( {R}^{\# } \) is the adjoint to \( R \), i.e., for \( g \in \mathcal{S}\left( {\mathcal{C}}^{n}\right) \), and \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) ,\n\n\[ \n{\int }_{{S}^{n - 1}}{\int }_{\mathbf{R}}g\left( {\theta, s}\right) \left( {Rf}\right) \l... | Proof. For \( g \in \mathcal{S}\left( {\mathcal{C}}^{n}\right) \), and \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \), we have, as in the proof of Theorem 2.1.1, for fixed \( \theta \in {S}^{n - 1} \), with \( x = {s\theta } + y \), where \( y \in {\theta }^{ \bot } \), \n\n\[ \n{\int }_{\mathbf{R}}g\left( {\the... | Yes |
Theorem 2.1.5. ([Natterer,2001, Theorem II.1.3]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) and \( g \in \mathcal{S}\left( {\mathcal{C}}^{n}\right) \) . Then\n\n\[ \left( {{R}^{\# }g}\right) * f = {R}^{\# }\left( {g * \left( {Rf}\right) }\right) . \] | Proof.\n\n\[ \left( {\left( {{R}^{\# }g}\right) * f}\right) \left( x\right) = {\int }_{{\mathbf{R}}^{n}}\left( {{R}^{\# }g}\right) \left( {x - y}\right) f\left( y\right) {dy} \]\n\n\[ = {\int }_{{\mathbf{R}}^{n}}{\int }_{{S}^{n - 1}}g\left( {\theta ,\left( {x - y}\right) \cdot \theta }\right) f\left( y\right) {d\theta ... | Yes |
Theorem 2.1.6. ([Natterer,2001, Theorem II.1.4]) For \( g \in \mathcal{S}\left( {\mathcal{C}}^{n}\right) \), we have\n\n\[ \n{\left( {R}^{\# }g\right) }^{ \land }\left( \xi \right) = \frac{1}{\parallel \xi {\parallel }^{n - 1}}\left\lbrack {\widehat{g}\left( {\frac{\xi }{\parallel \xi \parallel },\parallel \xi \paralle... | Proof. For \( w \in \mathrm{S} \) ,\n\n\[ \n{\int }_{{\mathbf{R}}^{n}}\left( {{R}^{\# }g}\right) \left( x\right) \widehat{w}\left( x\right) {dx} = {\int }_{{S}^{n - 1}}{\int }_{\mathbf{R}}g\left( {\theta, s}\right) \left( {R\widehat{w}}\right) \left( {\theta, s}\right) {dsd\theta }\;\text{ (by Theorem 2.1.4) }\n\]\n\n\... | Yes |
Theorem 2.1.7. ([Natterer,2001, Theorem II.2.1]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right), g = {Rf} \). Then for \( \alpha < n \)\n\n\[ f = \frac{1}{2}{\left( 2\pi \right) }^{1 - n}\left\lbrack {{I}_{-\alpha }{R}^{\# }{I}_{\alpha - n + 1}}\right\rbrack \left( g\right) . \] | Proof. We start out from the Fourier inversion formula in \( {\mathbf{R}}^{n} \) and Theorem 1.8.3,\n\n\[ \left( {{I}_{\alpha }f}\right) \left( x\right) = {\int }_{{\mathbf{R}}^{n}}\frac{1}{{\left( 2\pi \parallel \xi \parallel \right) }^{\alpha }}\widehat{f}\left( \xi \right) {\mathbf{e}}^{{2\pi ix} \cdot \xi }{d\xi }.... | Yes |
Theorem 2.1.12. ([Natterer, 2001, Theorem II.2.2 and II.2.3],[Natterer and Wübbeling, 2001, Theorem 2.7]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right), g = {Rf} \). Then, for \( s, r > 0 \), \[ {g}_{l, k}\left( s\right) = \left| {S}^{n - 2}\right| {\int }_{s}^{\infty }{C}_{l}^{\frac{n - 2}{2}}\left( \frac{s}{... | \[ {f}_{l, k}\left( r\right) = {c}_{n}{r}^{n - 2}{\int }_{r}^{\infty }{\left( {s}^{2} - {r}^{2}\right) }^{\frac{n - 3}{2}}{C}_{l}^{\frac{n - 2}{2}}\left( \frac{s}{r}\right) {g}_{l, k}^{\left( n - 1\right) }\left( s\right) {ds} \] where \[ {c}_{n} = \left\{ \begin{array}{ll} {\left( -1\right) }^{n - 1}\frac{1}{2{\pi }^{... | Yes |
Lemma 2.1.13. For \( s > 0 \) and \( \theta \in {S}^{n - 1} \) , \[ {\int }_{x \cdot \theta = s}f\left( x\right) {dx} = {\int }_{\left\{ \omega \in {S}^{n - 1} : \omega \cdot \theta > 0\right\} }f\left( {\frac{s}{\theta \cdot \omega }\omega }\right) \frac{{s}^{n - 1}}{{\left( \theta \cdot \omega \right) }^{n}}{d\omega ... | Proof. When \( \theta = {e}_{n} = \left( {0,\cdots ,0,1}\right) \), this formula was proved in [Natterer,2001, p. 188]. For general \( \theta \in {S}^{n - 1} \), let \( A \) be an orthogonal transform that maps \( {e}_{n} \) to \( \theta \), i.e., \( \theta = A{e}_{n} \) . After changing the variable by \( x = {Ay} \),... | Yes |
Theorem 2.2.1. ([Natterer and Wübbeling,2001, Theorem 2.11]) Let \( f \in \mathrm{S} \) . Then for \( \theta \in {S}^{n - 1},\xi \in {\theta }^{ \bot }, \)\n\n\[ \n{\left( Pf\right) }^{ \land }\left( {\theta ,\xi }\right) = \widehat{f}\left( \xi \right) \n\] | Proof.\n\n\[ \n{\left( Pf\right) }^{ \land }\left( {\theta ,\xi }\right) = {\int }_{{\theta }^{ \bot }}\left( {Pf}\right) \left( {\theta, x}\right) {\mathbf{e}}^{-{2\pi i\xi } \cdot x}{dx} \n\]\n\n\[ \n= {\int }_{{\theta }^{ \bot }}{\int }_{\mathbf{R}}f\left( {x + {t\theta }}\right) {\mathbf{e}}^{-{2\pi i\xi } \cdot x}... | Yes |
Theorem 2.2.2. ([Natterer and Wübbeling,2001, Theorem 2.12]) Let \( f, g \in \mathbb{S} \) . Then\n\n\[ \left( {Pf}\right) * \left( {Pg}\right) = P\left( {f * g}\right) . \] | Proof. By Theorem 1.2.3 and Theorem 2.2.1\n\n\[ {\left( \left( Pf\right) * \left( Pg\right) \right) }^{ \land }\left( {\theta ,\xi }\right) = \widehat{\left( Pf\right) }\left( {\theta ,\xi }\right) \widehat{\left( Pg\right) }\left( {\theta ,\xi }\right) \]\n\n(2.161)\n\n\[ = \widehat{f}\left( \xi \right) \widehat{g}\le... | Yes |
Theorem 2.2.3. ([Natterer and Wübbeling,2001, p. 18]) For \( g \in \mathrm{S}\left( {T}^{n}\right) \), and \( f \in \mathrm{S}\left( {\mathbf{R}}^{n}\right) \) ,\n\n\[{\int }_{{S}^{n - 1}}{\int }_{{\theta }^{ \bot }}g\left( {\theta, x}\right) \left( {Pf}\right) \left( {\theta, x}\right) {dxd\theta } = {\int }_{{\mathbf... | Proof. For \( g \in \mathcal{S}\left( {T}^{n}\right) \), and \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \), as in the proof of Theorem 2.2.1, for fixed \( \theta \in {S}^{n - 1} \) ,\n\n\[{\int }_{{\theta }^{ \bot }}g\left( {\theta, x}\right) \left( {Pf}\right) \left( {\theta, x}\right) {dx} = {\int }_{{\theta ... | Yes |
Theorem 2.2.4. ([Natterer, 2001, Theorem II.1.3], [Natterer and Wübbeling, 2001, Theorem 2.13]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) and \( g \in \mathcal{S}\left( {T}^{n}\right) \) . Then\n\n\[ \left( {{P}^{ * }g}\right) * f = {P}^{ * }\left( {g * \left( {Pf}\right) }\right) . \] | Proof.\n\n\[ \left( {\left( {{P}^{ * }g}\right) * f}\right) \left( x\right) = {\int }_{{\mathbf{R}}^{n}}\left( {{P}^{ * }g}\right) \left( {x - y}\right) f\left( y\right) {dy} \]\n\n(2.175)\n\n\[ = {\int }_{{\mathbf{R}}^{n}}{\int }_{{S}^{n - 1}}g\left( {\theta ,{E}_{\theta }\left( {x - y}\right) }\right) f\left( y\right... | Yes |
Theorem 2.2.5. ([Natterer,2001, Theorem II.2.1]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right), g = {Pf} \) . Then for \( \alpha < n \) | To prove this theorem, we need the following lemma. | No |
Lemma 2.2.6. ([Natterer, 2001, Eq. (VII.2.8)])\n\n\[ \n{\int }_{{\mathbf{R}}^{n}}h\left( y\right) {dy} = \frac{1}{\left| {S}^{n - 2}\right| }{\int }_{{S}^{n - 1}}{\int }_{{\theta }^{ \bot }}\parallel y\parallel h\left( y\right) {dyd\theta }. \n\] | Proof of Theorem 2.2.5 We start with the Fourier inversion formula\n\n\[ \n\left( {{I}_{\alpha }f}\right) \left( x\right) = {\int }_{{\mathbf{R}}^{n}}\frac{1}{{\left( 2\pi \parallel \xi \parallel \right) }^{\alpha }}\widehat{f}\left( \xi \right) {\mathbf{e}}^{{2\pi ix} \cdot \xi }{d\xi }. \n\]\n\nBy the above lemma,\n\... | Yes |
Theorem 2.2.8. ([Natterer and Wübbeling,2001, p. 18]) Let \( {S}_{0}^{2} \subset {S}^{2} \) meet every equatorial circle of \( {S}^{2} \) (the condition of Orlov [Orlov,1976]). Then \( \left( {Pf}\right) \left( {\theta, x}\right) ,\theta \in {S}_{0}^{2} \) and \( x \in {\theta }^{ \bot } \) , determines \( f \) uniquel... | Indeed, let \( \xi \in {\mathbf{R}}^{3} \) be arbitrary. Then if \( {S}_{0}^{2} \) satisfies Orlov’s uniqueness condition, we can find \( \theta \in {S}_{0}^{2} \) such that \( \theta \bot \xi \), see Figure 2.3. Hence, \( \widehat{f}\left( \xi \right) \) is determined by \( {\left( Pf\right) }^{ \land }\left( {\theta ... | Yes |
Theorem 2.2.11. For \( g \in \mathcal{S}\left( {T}^{n}\right) \), and \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) ,\n\n\[ \n{\int }_{{S}_{0}^{n - 1}}{\int }_{{\theta }^{ \bot }}g\left( {\theta, x}\right) \left( {Pf}\right) \left( {\theta, x}\right) {dxd\theta } = {\int }_{{\mathbf{R}}^{n}}\left( {{P}_{0}^{ * ... | Proof. For \( g \in \mathcal{S}\left( {T}^{n}\right) \), and \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \), we have, as in the proof of Theorem 2.2.3, for fixed \( \theta \in {S}_{0}^{n - 1} \)\n\n\[ \n{\int }_{{\theta }^{ \bot }}g\left( {\theta, x}\right) \left( {Pf}\right) \left( {\theta, x}\right) {dx} = {\i... | Yes |
Theorem 2.2.12. ([Natterer,2001, Theorem II.1.3]) Let \( f \in \mathcal{S}\left( {\mathbf{R}}^{n}\right) \) and \( g \in \mathcal{S}\left( {T}^{n}\right) \) . Then\n\n\[ \left( {{P}_{0}^{ * }g}\right) * f = {P}_{0}^{ * }\left( {g * \left( {Pf}\right) }\right) . | Proof.\n\n\[ \left( {\left( {{P}_{0}^{ * }g}\right) * f}\right) \left( x\right) = {\int }_{{\mathbf{R}}^{n}}\left( {{P}_{0}^{ * }g}\right) \left( {x - y}\right) f\left( y\right) {dy} \]\n\n\( \left( {2.207}\right) \)\n\n\[ = {\int }_{{\mathbf{R}}^{n}}{\int }_{{S}_{0}^{n - 1}}g\left( {\theta ,{E}_{\theta }\left( {x - y}... | Yes |
Theorem 2.2.13. ([Natterer and Wübbeling,2001, Theorem 2.17]) For \( h \in \mathcal{S}\left( {T}_{0}^{n}\right) \) , \[ \widehat{H}\left( \xi \right) = \frac{1}{\parallel \xi \parallel }{\int }_{{S}_{0}^{n - 1} \cap {\xi }^{ \bot }}\widehat{h}\left( {\theta ,\xi }\right) {d\theta }. \] | Proof. \[ {\left( {P}_{0}^{ * }h\right) }^{ \land }\left( \xi \right) = {\int }_{{\mathbf{R}}^{n}}\left( {{P}_{0}^{ * }h}\right) \left( x\right) {\mathbf{e}}^{-{2\pi i\xi } \cdot x}{dx} \] \[ = {\int }_{{S}_{0}^{n - 1}}{\int }_{{\mathbf{R}}^{n}}h\left( {\theta ,{E}_{\theta }x}\right) {\mathbf{e}}^{-{2\pi i\xi } \cdot x... | Yes |
Theorem 2.3.1. Let \( h \) be a function on \( \mathbf{R} \), homogeneous of degree \( 1 - n \) . Then \[ {\int }_{{S}^{n - 1}}\left( {Df}\right) \left( {a,\omega }\right) h\left( {\theta \cdot \omega }\right) {d\omega } = {\int }_{\mathbf{R}}\left( {Rf}\right) \left( {\theta, s}\right) h\left( {s - a \cdot \theta }\ri... | Proof. \[ {\int }_{{S}^{n - 1}}\left( {Df}\right) \left( {a,\omega }\right) h\left( {\theta \cdot \omega }\right) {d\omega } \] \[ = {\int }_{{S}^{n - 1}}{\int }_{0}^{\infty }f\left( {a + {t\omega }}\right) h\left( {\theta \cdot \omega }\right) {dtd\omega } \] \[ = {\int }_{{S}^{n - 1}}{\int }_{0}^{\infty }f\left( {a +... | Yes |
Theorem 2.3.3. (Tuy’s formula, [Tuy, 1983]) Suppose that the source curve A satisfies Tuy's condition. Then,\n\n\[ f\left( x\right) = \frac{1}{2\pi i}{\int }_{{S}^{2}}\frac{1}{{a}^{\prime }\left( \lambda \right) \cdot \theta }\frac{\partial }{\partial \lambda }\left\lbrack {{\left( Df\right) }^{ \land }\left( {a\left( ... | Proof. For \( \theta \in {S}^{2} \) and \( \lambda \in I \) ,\n\n\[ {\left( Df\right) }^{ \land }\left( {a\left( \lambda \right) ,\theta }\right) = {\int }_{{\mathbf{R}}^{3}}\left( {Df}\right) \left( {a\left( \lambda \right), y}\right) {\mathbf{e}}^{-{2\pi i\theta } \cdot y}{dy} \]\n\n(2.304)\n\n\[ = {\int }_{{\mathbf{... | Yes |
Theorem 2.3.4. (Weighted Tuy’s formula, [Zhao et al., 2005b]) Suppose that the weight function satisfies\n\n\[ \mathop{\sum }\limits_{{\lambda \in \Lambda \left( {x,\theta }\right) }}{\omega }_{x}\left( {\lambda ,\theta }\right) = 1,\;\text{ for a.e. }\theta \in {S}^{2}. \]\n\n(2.314)\n\nThen\n\n\[ f\left( x\right) = \... | Proof. By (2.311), for \( x \in \operatorname{supp}\left( f\right) \) ,\n\n\[ \frac{1}{2\pi i}\mathop{\sum }\limits_{{\lambda \in \Lambda \left( {x,\theta }\right) }}\frac{{\omega }_{x}\left( {\lambda ,\theta }\right) }{{a}^{\prime }\left( \lambda \right) \cdot \theta }\frac{\partial }{\partial \lambda }\left\lbrack {{... | Yes |
Theorem 3.2.3. (Kullback’s Theorem) If the maximum entropy distribution density \( \widehat{\pi } \) of \( \theta \) subject to the constraints (3.19) exists, then\n\n\[ \widehat{\pi }\left( \theta \right) = \frac{{\pi }_{0}\left( \theta \right) {\mathbf{e}}^{\mathop{\sum }\limits_{{k = 1}}^{m}{\lambda }_{k}{g}_{k}\lef... | Proof. Consider\n\n\[ G\left( \pi \right) = - {\int }_{\Theta }\Pr \left( \theta \right) \log \frac{\Pr \left( \theta \right) }{{\pi }_{0}\left( \theta \right) }{d\theta } + \mathop{\sum }\limits_{{k = 1}}^{m}{\lambda }_{k}\left\lbrack {E\left\lbrack {{g}_{k}\left( \theta \right) }\right\rbrack - {\mu }_{k}}\right\rbra... | Yes |
Theorem 4.1.2. The distribution \( \Pi \) of \( X = \left( {X}_{s}\right) \) is determined by its local characteristics. | Proof. We will verify that for any \( x = \left( {x}_{i}\right) \) and \( y = \left( {y}_{i}\right) \) ,\n\n\[ \frac{\Pi \left( x\right) }{\Pi \left( y\right) } = \mathop{\prod }\limits_{{i = 1}}^{N}\frac{\Pi \left( {{x}_{i} \mid {x}_{1},\cdots ,{x}_{i - 1},{y}_{i + 1},\cdots ,{y}_{N}}\right) }{\Pi \left( {{y}_{i} \mid... | Yes |
Example 4.3.2. Let \( \left\{ {{X}_{n},0 \leq n \leq N}\right\} \) be a Markov process with state space \( \Lambda, P\left( {{X}_{0} = }\right. \) \( \lambda ) = \nu \left( \lambda \right) > 0 \), and transitions \( {P}_{n}\left( {\lambda ,\delta }\right) = \Pr \left( \left( {{X}_{n + 1} = \delta \mid {X}_{n} = \lambda... | Proof. By definition, \( \left( {X}_{n}\right) \) is Markov if \( \forall m \geq 0 \)\n\n\[ \n\Pr \left( {{X}_{m + 1} = {x}_{m + 1} \mid {X}_{j} = {x}_{j},0 \leq j \leq m}\right) = \Pr \left( {{X}_{m + 1} = {x}_{m + 1} \mid {X}_{m} = {x}_{m}}\right) .\n\]\n\n(4.15)\n\nWe have, e.g.,\n\n\[ \n\Pr \left( {{X}_{0} = {x}_{0... | Yes |
Theorem 4.5.1. Let \( \mathcal{G} \) be a neighborhood system on \( S \) . Then \( \Pi \) is a Gibbsian random field w.r.t \( \mathcal{G} \) if and only if \( \Pi \) is a Markov random field w.r.t \( \mathcal{G} \), in which case \( \left\{ {V}_{A}\right\} \) in (4.24) is a Gibbsian potential. | Proof. \ | No |
Theorem 6.2.1. The EM algorithm stated above induces a sequence of estimates with increasing log-likelihood of the observed data:\n\n\[ L\left( {y \mid {\theta }^{\text{old }}}\right) \leq L\left( {y \mid {\theta }^{\text{new }}}\right) \]\n\n(6.15)\n\nwhere equality holds if and only if\n\n\[ Q\left( {{\theta }^{\text... | Proof. By Bayes' rule,\n\n\[ \Pr \left( {x \mid y,\theta }\right) = \frac{\Pr \left( {y \mid x,\theta }\right) \Pr \left( {x \mid \theta }\right) }{\Pr \left( {y \mid \theta }\right) } \]\n\nThen\n\n\[ L\left( {y \mid \theta }\right) = \log \Pr \left( {y \mid \theta }\right) \]\n\n(6.18)\n\n\[ = L\left( {x \mid \theta ... | Yes |
Lemma 6.2.2. Let \( {a}_{i} \) and \( {b}_{i}, i = 1,\cdots, m \), be non-negative numbers. Then the inequality\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{m}{a}_{i}\log \frac{{a}_{i}}{{b}_{i}} \geq a\log \frac{a}{b} \]\n\n(6.26)\n\nholds, where \( a = \mathop{\sum }\limits_{{i = 1}}^{m}{a}_{i} \) and \( b = \mathop{\sum }\... | It is easy to find that Eq. (6.24) follows from a continuous form of the entropy inequality.\n\nMore details can be found in [Ihara, 1993, p. 23, p. 29]. | No |
Theorem 6.2.3. The EM algorithm stated above induces a sequence of estimates with increasing a posteriori or penalized likelihood:\n\n\[ \Phi \left( {\theta }^{\text{old }}\right) \leq \Phi \left( {\theta }^{\text{new }}\right) \]\n\n(6.31)\n\nwhere equality holds if and only if\n\n\[ \Phi \left( {{\theta }^{\text{new ... | Proof. As in the proof of Theorem 6.2.1, we arrive at the same inequality Eq. (6.25). Adding \( P\left( {\theta }^{\prime }\right) - P\left( \theta \right) \) to both sides of Eq. (6.25), we have\n\n\[ \Phi \left( \theta \right) - \Phi \left( {\theta }^{\prime }\right) \geq \phi \left( {\theta \mid {\theta }^{\prime }}... | Yes |
We assume a circular trajectory with a third generation configuration as in \( §{7.3} \). This is called the standard fan beam scanning in [Natterer and Wübbeling, 2001]. | \[ \left( {{V}_{\Omega } * f}\right) \left( x\right) = {R}^{\# }\left( {v * g}\right) \left( x\right) = r{\int }_{0}^{2\pi }{d\beta }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{v}_{\Omega }\left( {x \cdot \theta - r\sin \alpha }\right) g\left( {\beta ,\alpha }\right) \cos {\alpha d\alpha }, \] where \( \theta = \theta \... | Yes |
Theorem 8.3.1. (Katsevich's FBP formula, [Katsevich, 2002, Katsevich, 2003, Katsevich, 2004, Zhao et al.,2005b] Let \( \Gamma \) be a regular curve in \( \mathbf{R}3 \) parameterized by \( a\left( t\right), t \in \mathbf{R} \) . For each \( x \in \Omega \) on a chord, the set \( \left\{ {\lambda \in {I}_{x} : x \cdot \... | Proof. We start with the inverse Fourier transform of the cone-beam data\n\n\[ \left( {Df}\right) \left( {a\left( t\right) ,\theta }\right) = {\int }_{{\mathbf{R}}^{3}}\widehat{Df}\left( {a\left( t\right), y}\right) {\mathbf{e}}^{{2\pi i\theta } \cdot y}{dy} \]\n\n\[ = {\int }_{{S}^{2}}{d\sigma }{\int }_{0}^{\infty }\w... | Yes |
Theorem 9.2.1. \( {A}^{ + } : D\left( {A}^{ + }\right) \rightarrow \mathcal{X} \) is a closed densely defined linear operator which is bounded if and only if \( R\left( A\right) \) is closed. | Proof. To see this, note first that \( D\left( {A}^{ + }\right) = R\left( A\right) + R{\left( A\right) }^{ \bot } \) is evidently dense in \( \mathcal{Y} \) . The linearity of \( {A}^{ + } \) follows easily from Eq. (9.6). To see that \( {A}^{ + } \) is closed, note that if\n\n\[ \left\{ {b}_{n}\right\} \subset D\left(... | Yes |
Theorem 9.3.1. ([Natterer, 2001, Theorem II.1.6])\n\n(a) For each \( \theta \in {S}^{n - 1} \), the operator\n\n\[ \n{R}_{\theta }\left( f\right) \left( s\right) = \left( {Rf}\right) \left( {\theta, s}\right) \n\]\n\nis a linear bounded operator from \( {L}^{2}\left( {B}^{n}\right) \) to \( {L}_{2}\left( {\left\lbrack ... | Proof. For \( f \) with support in \( {B}^{n} \), we have from the Cauchy-Schwartz inequality, by Eq. (2.6),\n\n\[ \n{\left| {R}_{\theta }\left( f\right) \left( s\right) \right| }^{2} = {\left| {\int }_{{\theta }^{ \bot }}f\left( s\theta + y\right) dy\right| }^{2} \n\]\n\n\[ \n= {\left| {\int }_{y \in {\theta }^{ \bot ... | Yes |
Theorem 9.4.3. (Paley-Wiener Theorem, [Stein and Weiss, 1971, Theorem III.4.9]) Suppose that \( F \in {L}^{2}\left( {\mathbf{R}}^{n}\right) \) . Then \( F \) is the Fourier transform of a function vanishing outside a symmetric body \( K \) if and only if \( F \) is the restriction to \( {\mathbf{R}}^{n} \) of an entire... | Proof. If \( F \) is the Fourier transform of a function \( f \), vanishing outside \( K \), then it is easy to check that\n\n\[ F\left( z\right) = {\int }_{{\mathbf{R}}^{n}}f\left( t\right) {\mathbf{e}}^{-{2\pi i}\langle z, t\rangle }{dt} = {\int }_{K}{\mathbf{e}}^{{2\pi }\langle y, t\rangle }f\left( t\right) {\mathbf... | No |
Theorem 9.4.4. The Fourier transform of \( \downarrow \downarrow \downarrow \) is itself: | Proof. By definition, for \( \varphi \in \mathcal{S} \) ,\n\n\[ !!{!}_{\left( h\right) }^{ \land }\left( \varphi \right) = !!{!}_{\left( h\right) }\left( \widehat{\varphi }\right) \]\n\n\[ = \mathop{\sum }\limits_{{\mathbf{k} \in {\mathbf{Z}}^{n}}}\delta \left( {\cdot - h\mathbf{k}}\right) \left( \widehat{\varphi }\rig... | Yes |
Theorem 9.4.5. ([Natterer,2001, Theorem III.1.1]) Let \( f \) be b-band-limited, and let \( h \leq \frac{1}{2b} \). Then \( f \) is uniquely determined by the values \( f\left( {h\mathbf{k}}\right) ,\mathbf{k} \in {\mathbf{Z}}^{n} \), and in \( {L}^{2}\left( {\mathbf{R}}^{n}\right) \), \[ f\left( x\right) = \mathop{\su... | Proof. Since \( \widehat{f}\left( \xi \right) \) vanishes outside \( {\left\lbrack -\frac{1}{2h},\frac{1}{2h}\right\rbrack }^{n} \), it can be extended to a periodic function \( {\widehat{f}}_{ * }\left( \xi \right) \) on \( {\mathbf{R}}^{n} \) of period \( \frac{1}{h} \). Then it has the following Fourier series expan... | Yes |
Theorem 9.4.6. ([Natterer, 2001, Theorem III.1.2]) Let \( f \) be b-band-limited, and let \( h < \frac{1}{2b} \) . Let \( \gamma \in {C}^{\infty }\left( {\mathbf{R}}^{n}\right) \) vanish for \( \parallel \xi \parallel > 1 \) and\n\n\[{\int }_{{\mathbf{R}}^{n}}\gamma \left( \xi \right) {d\xi } = 1\]\n\nThen\n\n\[f\left(... | Proof. Again we start from Eq. (9.78), which holds in \( {\left\lbrack -\frac{1}{2h},\frac{1}{2h}\right\rbrack }^{n} \) . Since supp \( \left( \widehat{f}\right) \subset {\left\lbrack -b, b\right\rbrack }^{n} \) ,\n\nEq. (9.78) holds in \( {\left\lbrack -a, a\right\rbrack }^{n} \) where \( a = \frac{1}{2h} + d \) and \... | Yes |
Theorem 9.4.7. ([Natterer,2001, Theorem III.1.3]) Let \( f \in \mathcal{S} \). Then, there is a \( {L}^{\infty } \) function \( {\chi }_{x} \) with \( \left| {\chi }_{x}\right| \leq 1 \) such that (cf. Eq. (9.77))\n\n\[ \left( {{S}_{h}f - f}}\right) \left( x\right) = 2{\int }_{{\mathbf{R}}^{n} \smallsetminus {\left\lbr... | Proof. Eq. (9.87) is just Eq. (9.74).\n\nTo prove Eq. (9.88), we compute the Fourier transform of \( f\overset{h}{ * }g \) by its definition as follows\n\n\[ {\left( f\overset{h}{ * }g\right) }^{ \land }\left( \xi \right) = {h}^{n}\mathop{\sum }\limits_{\mathbf{l}}{\left( f\left( \cdot - h\mathbf{l}\right) \right) }^{ ... | No |
Theorem 9.5.2. ([Natterer,2001, Theorem III.2.1]) Let \( A \) be \( m \) -resolving, and \( f \in {C}_{0}^{\infty }\left( {B}^{n}\right) \) , and \( \lambda > \frac{1}{2} \) . If \( {Rf}\left( {\theta , \cdot }\right) \) vanishes for \( \theta \in A \), then\n\n\[ \left( {Rf}\right) \left( {\theta, s}\right) = {\left( ... | Proof. According to Theorem 2.1.17, we have the expansion\n\n\[ \left( {Rf}\right) \left( {\theta, s}\right) = {\left( 1 - {s}^{2}\right) }^{\lambda - \frac{1}{2}}\mathop{\sum }\limits_{{l = 0}}^{\infty }{C}_{l}^{\lambda }\left( s\right) {h}_{l}\left( \theta \right) ,\]\n\nwith \( {h}_{l} \in {\mathcal{H}}_{l}^{\prime ... | Yes |
Theorem 9.5.4. ([Natterer,2001, Theorem III.2.2]) Let \( A \) be \( m \) -resolving and \( f \in {C}_{0}^{\infty }\left( {B}^{n}\right) \) . If \( {Rf}\left( {\theta , \cdot }\right) \) vanishes for \( \theta \in A \), then,\n\n\[ \n{\int }_{\left| \sigma \right| \leq {\vartheta m}}\left| {\widehat{Rf}\left( {\theta ,\... | Proof. We use Theorem 9.5.2 with \( \lambda = 0 \), obtaining\n\n\[ \n\left( {Rf}\right) \left( {\theta, s}\right) = {\left( 1 - {s}^{2}\right) }^{-\frac{1}{2}}\mathop{\sum }\limits_{{l > m}}^{\infty }{T}_{l}\left( s\right) {h}_{l}\left( \theta \right) ,\n\]\n\n(9.117)\n\nwhere \( {T}_{l} \) is the Chebychev polynomial... | Yes |
Theorem 9.5.5. ([Natterer,2001, Theorem III.2.3]) Let \( A \) be \( m \) -resolving and \( f \in {C}_{0}^{\infty }\left( {B}^{n}\right) \) . If \( {Rf}\left( {\theta , \cdot }\right) \) vanishes for \( \theta \in A \), then, | Proof. From Theorem 2.1.1 and Theorem 9.5.4,\n\n\[ \n{\int }_{\left| \xi \right| \leq {\vartheta m}}\left| {\widehat{f}\left( \xi \right) {d\xi } = {\int }_{{S}^{n - 1}}{\int }_{0}^{\vartheta m}\left| {\widehat{f}\left( {\sigma \theta }\right) }\right| {d\sigma d\theta }}\right| \n\]\n\n\[ \n= {\int }_{{S}^{n - 1}}{\in... | Yes |
Theorem 9.5.6. ([Natterer, 2001, Theorem III.2.3.4]) Let \( A \) be m-resolving and \( f \in {C}_{0}^{\infty }\left( {B}^{n}\right) \) . If \( {Rf}\left( {\theta , \cdot }\right) \) vanishes for \( \theta \in A \), then,\n\n\[ \parallel f{\parallel }_{{L}^{\infty }\left( {B}^{n}\right) } \leq \frac{1}{1 - \eta \left( {... | Proof. We have\n\n\[ \left| {f\left( x\right) }\right| = {\int }_{{\mathbf{R}}^{n}}\widehat{f}\left( \xi \right) {\mathbf{e}}^{{2\pi i\xi } \cdot x}{d\xi } \]\n\n\[ \leq {\int }_{\left| \xi \right| \leq {\vartheta m}}\left| {\widehat{f}\left( \xi \right) }\right| {d\xi } + {\int }_{\left| \xi \right| \geq {\vartheta m}... | Yes |
Theorem 9.5.8. ([Natterer,2001, Theorem III.3.1]) Let \( f \in {C}_{0}^{\infty }\left( {B}^{2}\right) \), and let\n\n\[ g\left( {\varphi, s}\right) = \left( {Rf}\right) \left( {\theta, s}\right) ,\;\theta = \left( \begin{matrix} \cos \varphi \\ \sin \varphi \end{matrix}\right) . \]\n\nFor \( 0 < \vartheta < 1 \) and \(... | Proof. See [Natterer,2001, pp. 71 - 73]. The set \( K \) comes in through the analysis based on the Debye's Debye's asymptotic formula for Bessel function, Eq. (1.80). | No |
Proposition 12.5.1.\n\n\[ \mathcal{R}{\left\lbrack {T}_{{\Gamma }_{P}}\right\rbrack }^{ \bot } = L\left\lbrack {{H}_{0,{\Gamma }_{P}}^{2}\left( \Omega \right) }\right\rbrack \] | Proof. If \( q \in L\left\lbrack {{H}_{0,{\Gamma }_{P}}^{2}\left( \Omega \right) }\right\rbrack \) with \( q = L\left\lbrack p\right\rbrack \) for some \( p \in {H}_{0,{\Gamma }_{P}}^{2}\left( \Omega \right) \), then for \( v = {T}_{{\Gamma }_{P}}\left\lbrack \psi \right\rbrack \in \mathcal{R}\left\lbrack {T}_{{\Gamma ... | Yes |
Theorem 12.5.2. Assume that the \( \left( {{BLT}\left( P\right) }\right) \) problem is solvable. For any couple \( \left( {{g}^{ - }, g}\right) \) such that\n\n\[ \n{N}_{{\Gamma }_{P}}\left\lbrack {{g}^{ - } + {2g}}\right\rbrack + g \in {H}^{\frac{1}{2}}\left( {\Gamma }_{P}\right) \n\]\n\n(12.77)\n\nthere is one specia... | Because the minimal norm source solution \( {q}_{H} \bot L\left\lbrack {{H}_{0,{\Gamma }_{P}}^{2}\left( \Omega \right) }\right\rbrack \), we have, by Green’s formula (12.67), for any \( \left. {v \in {H}_{0}^{2}\left( \Omega \right) \subset {H}_{0,{\Gamma }_{P}}^{2}\left( \Omega \right) }\right\rbrack \) ,\n\n\[ \n{\in... | No |
Theorem 12.5.3. The minimal norm solution \( {q}_{H} \) cannot possess a compact support within \( \Omega \) unless it is zero. | Proof. If \( {q}_{H} \) is of compact support within \( \Omega \), it satisfies the partial differential equation (12.79) and the following boundary conditions: \( {\left. {\gamma }_{0}\left\lbrack {q}_{H}\right\rbrack \right| }_{{\Gamma }_{P}} = 0,{\left. {\gamma }_{1}\left\lbrack {q}_{H}\right\rbrack \right| }_{{\Gam... | Yes |
Theorem 12.5.4. If \( {q}_{0} \in {L}^{2}\left( \Omega \right) ,{q}_{0} \neq 0 \), with compact support inside \( \Omega \) is a solution to the BLT problem, then \( {q}_{0} \) can not be of minimal norm and must have a non-radiating, i.e., un-observable part, as contribution from \( {H}_{0,{\Gamma }_{P}}^{2}\left( \Om... | Proof. If \( {q}_{0} \) does not have a non-radiating part, then \( q = {q}_{H} \) is the minimal norm source and satisfies (12.79). Hence, it follows that \( {q}_{0} = 0 \) as in the proof of the above theorem. | No |
Theorem 13.2.3. Assume the conditions \( {C1} - {C3}, C{4}^{ * },{C5} \) and \( {C6} \) hold. For \( \lambda \in \Lambda \), if \( \left. {{q}_{1}\left( {\lambda, y}\right) = \mathop{\sum }\limits_{{s = 1}}^{S}{g}_{s}\left( {\lambda ,\left| \right| y - {y}_{s}\left| \right| }\right) {\chi }_{{B}_{{r}_{0}^{s},{r}_{1}^{s... | \[ {\int }_{{r}_{0}^{s}}^{{r}_{1}^{s}}{r}^{N - 1}{\varphi }_{C\left( s\right) }\left( {\lambda, r}\right) {g}_{s}\left( {\lambda, r}\right) {dr} = {\int }_{{R}_{0}^{\tau \left( s\right) }}^{{R}_{1}^{\tau \left( s\right) }}{r}^{N - 1}{\varphi }_{C\left( s\right) }\left( {\lambda, r}\right) {G}_{\tau \left( s\right) }\le... | Yes |
Proposition 13.4.1. Let \( {q}_{0}\left( \lambda \right) \) be a RBF source distribution in (13.27). Assume that the condition D1 and D2 hold for \( {q}_{0} \) . Then it follows that\n\n\[ \n{I}_{s}\left( {\lambda }_{2}\right) = \omega {I}_{s}\left( {\lambda }_{1}\right) ,\;s = 1,\cdots, S.\n\]\n\n(13.39) | Proof. We use the mathematical induction on \( S \) . The conclusion is obvious when \( S = 1 \) . Now assume that the conclusion holds for \( S = K \) . We are to prove that it holds for \( S = K + 1 \) . Consider all the \( K + 1 \) balls \( B\left( {{y}_{s},{R}_{s}}\right), s = 1,\cdots, K + 1 \) . There is one that... | Yes |
Lemma 13.4.3. For \( t > 0 \) , \n\n\[ \n\left\lbrack {{\beta }^{\prime }\left( t\right) + t{\beta }^{\prime \prime }\left( t\right) }\right\rbrack \frac{\beta \left( t\right) }{t} > {\beta }^{\prime }{\left( t\right) }^{2} \n\] | Proof. Let \( x\left( t\right) = {\beta }^{\prime }\left( t\right) + t{\beta }^{\prime \prime }\left( t\right) \) . Then we have \n\n\[ \nx\left( t\right) = \left( {1 + \frac{1}{{t}^{2}}}\right) \sinh t - \frac{\cosh t}{t} \n\] \n\n\[ \n{t}^{2}x\left( t\right) = \left( {1 + {t}^{2}}\right) \sinh t - t\cosh t \n\] \n\n\... | Yes |
Theorem 15.4.1. If an image multi-scale analysis \( {T}_{t} \) is causal and regular then \( I\left( {t, x}\right) = \) \( {T}_{t}{\left( I\right) }_{\left( x\right) } \) is a viscosity solution of\n\n\[ \frac{\partial I}{\partial t} = F\left( {{\nabla }^{2}I,\nabla I, I, x, t}\right) \]\n\n\( \left( {15.12}\right) \)\... | Proof We give a simplified proof by assuming that \( I\left( {t, x}\right) \) is \( {C}^{2} \) . A completed and rigorous proof could be found in Alvarez et al.,1993. In a neighborhood of \( \left( {t, x}\right) \), we have\n\n\[ I\left( {t, y}\right) = I\left( {t, x}\right) + < \nabla I\left( x\right), y - x > + \frac... | No |
Property 1. Positivity: By maximum principle, the diffusion solution \( I\left( {x, y, t}\right) \) is everywhere nonnegative, because of the non-negativity of the initial value and the adiabatic boundary condition. | However, we should remark that this property is not mathematically proved, especially for the interested choices of diffusion coefficient \( c \), see \( § \) 15.4.3, | No |
Property 2. Conservative Flux: By Gaussian theorem, | \[ \frac{\partial }{\partial t}{\int }_{D}{Idxdy} = {\int }_{D}\frac{\partial I}{\partial t}{dxdy} = {\int }_{D}\operatorname{div}\left( {g\left( \left| {\nabla I\left( {x, y, t}\right) }\right| \right) \nabla I}\right) {dxdy} \] \[ = {\int }_{\partial D}g\left( \left| {\nabla I\left( {x, y, t}\right) }\right| \right) ... | Yes |
Theorem 15.4.6. Let \( u \) satisfy the uniformly parabolic differential inequality (15.40) with bounded coefficients in a domain \( \Omega \) in \( \left( {\mathbf{x}, t}\right) \) -space \( {\mathbf{R}}^{n} \times \mathbf{R} \) and suppose that the maximum of \( u \) in \( {\Omega M} \) is attained at a point \( P = ... | \[ \frac{\partial u}{\partial \nu } > 0 \] | Yes |
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