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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 \( K \) : ![00eeb6ce-d106-4d6c-bb86-c4abb4702764_83_0.jpg](images/00eeb6ce-d106-4d6c-bb86-c4abb4702764_83_0.jpg) 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 ![00eeb6ce-d106-4d6c-bb86-c4abb4702764_83_1.jpg](images/00eeb6ce-d106-4d6c-bb86-c4abb4702764_83_1.jpg)\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![73714111-2e23-48d8-954e-774c7bcef84a_23_1.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_23_1.jpg)\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.
![73714111-2e23-48d8-954e-774c7bcef84a_43_0.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_43_0.jpg) (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.
![73714111-2e23-48d8-954e-774c7bcef84a_70_0.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_70_0.jpg)
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...
![73714111-2e23-48d8-954e-774c7bcef84a_127_0.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_127_0.jpg)\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![73714111-2e23-48d8-954e-774c7bcef84a_142_1.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_142_1.jpg) ![73714111-2e23-48d8-954e-774c7bcef84a_142_2.jpg](images/73714111-2e23-48d8-954e-774c7bcef84a_142_2.jpg) (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