... ǫ/4]}; (27) that is, these cosets are all the places where fW is not “too close” to being an indicator function 3.1 Construction of the Function g To construct the function g with the properties claimed ... instead of sets S, we have a function f : Fn → [0, 1] We have not bothered to optimize the p conclusion of the theorem (to the same extent as we did Theorem 1) given the method of proof, though much ... ∆−2 Let V be the subspace of Fn generated by the elements of A, and then let W = V ⊥ p In general, we will not have that Fn = V ⊕ W , but that does not matter for the argument p that follows...
... Convex sets and convex functions taking the infinity value Illustration tvnguyen (University of Science) Convex Optimization 11 / 108 Chapter Convex sets and convex functions taking the infinity value ... from sets to functions A common device to construct convex functions on IRn is to construct a convex set F in IRn+1 and then take the function whose graph is the lower boundary of F in the following ... Chapter Convex sets and convex functions taking the infinity value Infimal convolution We introduce the functional operation which corresponds to the addition of epigraphs as sets in IRn+1 Let f1 , f2...
... subset K ⊂ X then the homotopy {ft }t [0,1] can be chosen holomorphic and uniformly close to f0 on K Note that the Oka property of Y is just the Oka principle for sections of the trivial (product) ... (corresponding to the endpoints of P ) are holomorphic on Cn , the above construction can be performed so that these two maps remain unchanged, thereby showing that the basic CAP implies the one-parametric ... FRANC FORSTNERIC The hypothesis in Theorem 0.1 will be referred to as the convex approximation property (CAP) of the manifold Y The converse implication is obvious and hence the two properties are...
... (X; R) The nontrivial part in the proof of the theorem is to show that the sequence (7.1) is exact or in other words that Poincar´’s lemma holds true e for Whitney functions The essential tool ... compact set on infinitely many jets of the argument F The following theorem will not be needed in the rest of this work but appears to be of independent interest Since the proof goes along the ... obviously subanalytic and always regularly situated), the statement of the thefor subanalytic sets Proof Since the sheaf M is finitely generated projective we can reduce ∞ the claim to the case M =...
... this article, HL participated in some study of this article All authors read and approved the final manuscript Competing interests The authors declare that they have no competing interests Received: ... 3Department of Applied Mathematics, The Hong Kong Polytechnic University, Kowloon, Hong Kong, China Authors’ contributions CL completed the main part of this article, XY presented the ideas of this ... prequasiinvex with respect to the same h on Γ Proof Suppose, on the contrary, f is not a semistrictly prequasiinvex function with ¯ respect to h on Γ, then there exist points x, y Î Γ with f (x) ≠...
... following the approach of [3], we establish the following result Theorem 3.3 Let all above-stated hypotheses hold and let conditions of Theorem 2.7 be satisfied Then, the W-solution u of Dirichlet problem ... the sequel of paper, G will be a “regular set.” In order to obtain our regularity result on G, we need the following further hypotheses ◦ Hypothesis◦ 3.1 There exists a constant c > such that ... function: nq ∈ Lt (Ω) with t > , ν q −n nt ν ∈ Lt (Ω) with t > qt − n (2.5) We put q = nqt/(n(1 + t) − qt) We can easily prove that a constant c0 > exists such ◦ that if u ∈ W 1,q (ν,Ω), the following...
... Pc0 that intersect ω(0) are called the nth step pieces of the Markov partition Note that for n ≥ the collection of all the nth step pieces is a Markov partition; we call it a refinement of the ... Since the Fatou set is totally invariant, we can replace x with a finite iteration f i (x) at any stage of the argument For each x ∈ F (f ), under iterating f i (x) converge to an attracting (super- ... is R > such that j(Bσ (c0 ) × a∈A Ua ) ⊂ BR (0) Proof of the Theorems In this section we prove the theorems; the proof is divided into parts Let f , fn be rational functions We have the following...
... decreasing its adhesion and thus also the gluing strength The greater the temperature difference, the more important the effect Adhesive reactivity is the rate of resole conversion to three-dimensional ... increase The temperature of the central zone of the set at larger formats continuously increases to the temperature of pressing plates It has been found that in construction 3×, a temperature of ... coefficient of compressibility etc The extent of basic tests was determined statistically and the results were also evaluated statistically The paper results in the proposal of pressing parameters...
... 1 t t2 H(x (t) ) t and max 1 t t1 x≥x (t) H(x) = µ t (8) Indeed, if ≤ t ≤ t2 then x ≥ x (t) ≥ and H(x) /t ≤ H(x (t) ) /t ≤ µ If t2 ≤ t ≤ t1 then H(x) /t ≤ H(2) /t2 = H(x (t2 )) /t2 ≤ µ Now we are ready to ... result of Lev [7] Using an affine transformation of variables his theorem can be stated as follows Theorem Let A be a finite set of integers and let h be a positive integer satisfying h > 2κ − Then there ... Acknowledgments I would like to thank referees for many valuable comments Due to their suggestions we were able to prove the main result of the note in its present sharp form the electronic journal...
... denotes the transition matrix, Ω denotes the state space, and π denotes the unique stationary distribution Recall that the variation distance between two distributions µ, ν on Ω is defined as dT ... not just independent sets It will soon be clear that the results hold for the identical chain only defined on the set of independent sets Given a configuration t ⊂ V , the transitions t → t+ 1 ... space is the set of independent sets Redefining the state space to the set of all subsets is a crucial element in extending the original proof to graphs with triangles The technique, first used by...
... want to prove that A intersects at most 6k −7 other members of M Let B ∈ M intersect A Choose a coordinate system on the plane so that the center of A is the origin and the center of B lies on the ... assume that the center of A is the origin O = (0, 0) Let z be the rightmost point on the X-axis that belongs to A If z = (0, 0), then A is an interval with the center O and G is an interval graph ... of the intersection graph Theorem Let G be the intersection graph of translations of a fixed compact convex set in the plane with ω(G) = k, k ≥ Then the maximum degree of G is at most 6k − This...
... R(7, 5), mentioned in the Introduction As is mentioned there the condition s ≥ is not needed if the set has more than points, but without it the set of the six intersection points of three circles, ... A set has property C(r) if there is a round set containing all its points with the exception of at most r Trivial examples of sets with R (t, s) are sets with C (t − s) So we can state the theorems ... Round SetsThe theorems mentioned in the Introduction could suggest that a set with R (t, s), if sufficiently large, is trivial in the sense that it has R∗ (t, s) This is of course true for the pairs...
... turns out to be negative, then we can improve the bound of Theorem 2: Proposition Let p = 2n + be a prime, and let T denote the set of perfect dominating n−1 sets of the form (1) Every (S, T ) where ... 0), the unit vector with at the i-th coordinate Let p = 2n + be a prime and S be a perfect dominating set in G, and let f be the characteristic function of S, i.e f (x) = if x ∈ S and f (x) = otherwise ... 1}, Note that V = V∅ To prove Theorem we start from D = ∅ At every step, if D does not extend uniquely to S, then there exists a vertex v ∈ S such that dim VD∪{v} < dim VD ; we add v to D Since...
... boundary of Da at the point ta with the r x-coordinate equal to − (the angle a ta a is right) It implies that ta lies outside the region L and that the line xa crosses Sb (the left part of the boundary ... intersection of S \ {T } is non-empty Then arbitrary two points x1 ∈ (S \ {T }) and x2 ∈ T pierce S The triangle T intersects some other triangle Let U ∈ S be the triangle such that rU is the ... 2.3.2 There are two distinct triangles L, R ∈ Z such that x3 is on the left line of R and on the right line of L Then the statement easily follows from the (4, 3)-property for the triangles S, T...
... Consider the n translates C − q1 , , C − qn of C Count the number of incidences between these curves and the elements of P Notice that if the point p + q belongs to S, then p is incident to C ... point-curve incidences which occur in such a configuration Notice that C + ti and C + tj intersect in at most two points for i = j Furthermore, for any two distinct points pµ and pν , there exist ... n Notice that the set of all midpoints of the connecting segments of an n-element set P can be expressed as (P ⊕ P ), so that M (n, n) is an upper bound on the quantity studied by Halman et al...
... in a tree In this section, we turn to the number of matchings in a graph This is also known as the Hosoya index, or the Z-index in mathematical chemistry For a rooted tree T , let Z (T ) be the ... paper is to prove Conjecture In fact, we prove a stronger result For a rooted tree (T, r), let i0 (T, r) denote the number of independent sets not covering the root Let D(m) = {(i0 (T, r), i (T )) ... interesting paper of Wagner [5] looks at the number of independent sets modulo m Wagner showed that the proportion of trees on n vertices with the number of independent sets divisible by m tends...
... the currently known best constant o coefficients) Sort the lines in L by the increasing order of their slopes (break ties arbitrarily) Denote by Pi the set of points in P that are incident to the ... chain Set the length of each line segment to be at least the diameter of the point set P The chain C has n + vertices including two endpoints Now we can the electronic journal of combinatorics ... points of P are properly between and Then, the gradient of the ith parabola is at x = and + 2ε at x = Let ε be so small that the intervals [ai , + 2ε] are all disjoint: Namely, the gradient of the...
... that there is a point Q, not lying on a secant line to B Project B from the point Q onto a hyperplane through P and not through Q It is clear that the number of (p0 +1)-secants through P to the ... intersection of an Fp -subline with an Fp -linear set; all possibilities for the size of the intersection that are obtained in this statement, can occur (see [7]) The bound on the characteristic ... 1)-secant The intersection B ′ ∩ L is by the induction hypothesis an Fpe -linear set Since B ∩ L = B ′ ∩ L, the statement follows Finally, we extend Theorem (ii) to a theorem on k-blocking sets in...
... calculation in the CBCT data sets of the thorax and pelvis patient HU-DS10F0 was applied to the CBCT data sets of the head patients - Tables were generated separately for the three different patient ... thorax patients the lung was delineated in the CBCT image sets Results Phantom Study The image quality and the CT values of the CBCT data set were different to the planning CT This is illustrated ... Consequently, the treatment isocentre was not in the centre of the patient but in the centroid of the peripheral lesion Incomplete patient models were acquired due to the FOV size of 41 cm in diameter...