DSpace at VNU: Global Analysis: Functional Analysis Examples c-1 tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án...
Global Analysis Functional Analysis Examples c-1 Leif Mejlbro Download free books at Leif Mejlbro Global Analysis Download free eBooks at bookboon.com Global Analysis © 2009 Leif Mejlbro & Ventus Publishing ApS ISBN 978-87-7681-533-2 Disclaimer: The texts of the advertisements are the sole responsibility of Ventus Publishing, no endorsement of them by the author is either stated or implied Download free eBooks at bookboon.com Contents Global Analysis Contents Introduction Metric Spaces Topology 13 Continuous mappings 21 Topology 28 Sequences 34 Semi-continuity 41 Connected sets, differentiation a.o 48 Addition and multiplication by scalars in normed vector spaces 53 Normed vector spaces and integral operators 57 10 Differentiable mappings 63 11 Complete metric spaces 67 12 Local Existence and Uniqueness Theorem for Autonomous Ordinary Differential Equations 71 13 Euler-Lagrange’s equations 75 Download free eBooks at bookboon.com Introduction Global Analysis Introduction This is the first book containing examples from Functional Analysis We shall here deal with the subject Global Analysis The contents of the following books are Functional Analysis, Examples c-2 Topological and Metric Spaces, Banach Spaces and Bounded Operators Topological and Metric Spaces (a) Weierstraß’s approximation theorem (b) Topological and Metric Spaces (c) Contractions (d) Simple Integral Equations Banach Spaces (a) Simple vector spaces (b) Normed Spaces (c) Banach Spaces (d) The Lebesgue integral Bounded operators Functional Analysis, Examples c-3 Hilbert Spaces and Operators on Hilbert Spaces Hilbert Spaces (a) Inner product spaces (b) Hilbert spaces (c) Fourier series (d) Construction of Hilbert spaces (e) Orthogonal projections and complement (f) Weak convergency Operators on Hilbert Spaces (a) Operators on Hilbert spaces, general (b) Closed operators Functional Analysis, Examples c-4 Spectral theory Spectrum and resolvent The adjoint of a bounded operator Self-adjoint operators Isometric operators Unitary and normal operators Positive operators and projections Compact operators Download free eBooks at bookboon.com Introduction Global Analysis Functional Analysis, Examples c-5 Integral operators Hilbert-Schmidt operators Other types of integral operators www.sylvania.com We not reinvent the wheel we reinvent light Fascinating lighting offers an ininite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges An environment in which your expertise is in high demand Enjoy the supportive working atmosphere within our global group and beneit from international career paths Implement sustainable ideas in close cooperation with other specialists and contribute to inluencing our future Come and join us in reinventing light every day Light is OSRAM Click on the ad to read more Download free eBooks at bookboon.com Metric Spaces Global Analysis Metric Spaces Example 1.1 Let (X, dX ) and (Y, dY ) be metric spaces Define dX×Y : (X × Y ) × (X × Y ) → R+ by dX×Y ((x1 , y1 ) , (x2 , y2 )) = max (dX (x1 , x2 ), dY (y1 , y2 )) Show that dX×Y is a metric on X × Y Show that the projections pX : X × Y → X, pX (x, y) = x, pY : X × Y → Y, pY (x, y) = y, are continuous mappings The geometric interpretation is that dX×Y compares the distances of the coordinates and then chooses the largest of them 3.5 2.5 1.5 0.5 Figure 1: The points (x1 , y1 ) and (x2 , y2 ), and their projections onto the two coordinate axes MET We have assumed that dX and dY are metrics, hence dX×Y ((x1 , y1 ), (x2 , y2 )) = max (dX (x1 , x2 ), dY (y1 , y2 )) ≥ max(0, 0) = If dX×Y ((x1 , y1 ), (x2 , y2 )) = max (dX (x1 , x2 ), dY (y1 , y2 )) = 0, then dX (x1 , y1 ) = and dY (y1 , y2 ) = Using that dX and dY are metrics, this implies by MET for dX and dY that x1 = x2 and y1 = y2 , thus (x1 , y1 ) = (x2 , y2 ), and MET is proved for dX×Y MET From dX and dY being symmetric it follows that dX×Y ((x1 , y1 ), (x2 , y2 )) = max (dX (x1 , x2 ), dY (y1 , y2 )) = max (dX (x2 , x1 ), dY (y2 , y1 )) = dX×Y ((x2 , y2 ), (x1 , y1 )) , and we have proved MET for dX×Y Download free eBooks at bookboon.com Metric Spaces Global Analysis MET The triangle inequality If we put in (x, y), we get dX (x1 , x2 ) ≤ dX (x1 , x) + dX (x, x2 ) ≤ dX×Y ((x1 , y1 ), (x, y)) + dX×Y ((x, y), (x2 , y2 )) , and analogously, dY (y1 , y2 ) ≤ dX×Y ((x1 , y1 ), (x, y)) + dX×Y ((x, y), (x2 , y2 )) Hence the largest of the numbers dX (x1 , x2 ) and dY (y1 , y2 ) must be smaller than or equal to the common right hand side, thus dX×Y ((x1 , y1 ), (x2 , y2 )) = max (dX (x1 , x2 ), dY (y1 , y2 )) ≤ dX×Y ((x1 , y1 ), (x, y)) + dX×Y ((x, y), (x2 , y2 )) , and MET is proved Summing up, we have proved that dX×Y is a metric on X × Y Since pX : X × Y → X fulfils dX (pX ((x, y)), pX ((x0 , y0 ))) = dX (x, x0 ) ≤ dX×Y ((x, y), (x0 , y0 )) , we can to every ε > choose δ = ε Then it follows from dX×Y ((x, y), (x0 , y0 )) < ε that dX (pX ((x, y)), pX ((x0 , y0 ))) ≤ dX×Y ((x, y), (x0 , y0 )) < ε, and we have proved that pX is continuous The proof of pY : X × Y → Y also being continuous, is analogous Example 1.2 Let (S, d) be a metric space For every pair of points x, y ∈ S, we set d(x, y) = d(x, y) + d(x, y) Show that d is a metric on S with the property ≤ d(x, y) < for all x, y ∈ S + Hint: You may in suitable way use that the function ϕ : R+ → R0 defined by ϕ(t) = t , 1+t t ∈ R+ 0, is increasing MET Obviously, d(x, y) = d(x, y) ≥ 0, + d(x, y) and if d(x, y) = 0, then d(x, y) = 0, hence x = y MET From d(x, y) = d(y, x) follows that d(x, y) = d(y, x) d(x, y) = = d(y, x) + d(x, y) + d(y, x) Download free eBooks at bookboon.com Metric Spaces Global Analysis 1.4 1.2 0.8 0.6 0.4 0.2 Figure 2: The graph of ϕ(t) and its horizontal asymptote MET We shall now turn to the triangle inequality, d(x, y) ≤ d(x, z) + d(z, x) Now, d(x, y) ≤ d(x, z) + d(z, y), and ϕ(t) = t =1− ∈ [0, 1[ 1+t 1+t for t ≥ 0, is increasing Since a positive fraction is increased, if its positive denominator is decreased (though still positive), it follows that d(x, y) = d(x, y) = ϕ(d(x, y)) + d(x, y) d(x, z) + d(z, y) + d(x, z) + d(z, y) d(z, y) d(x, z) + + d(x, z) + d(z, y) + d(x, z) + d(z, y) d(z, y) d(x, z) + + d(x, z) + d(z, y) ≤ ϕ(d(x, z) + d(z, x)) = = = = d(x, z) + d(z, y), and we have proved that d is a metric Now, ϕ(t) ∈ [0, 1] for t ∈ R+ , thus d(x, y) = ϕ(d(x, y)) ∈ [0, 1[ for all x, y ∈ S, hence ≤ d(x, y) < for all x, y ∈ S + Remark 1.1 Let ϕ : R+ → R0 satisfy the following three conditions: ϕ(0) = 0, and ϕ(t) > for t > 0, ϕ is increasing ≤ ϕ(t + u) ≤ ϕ(t) + ϕ(u) for all t, u ∈ R+ If d is a metric on S, then ϕ ◦ d is also a metric on S The proof which follows the above, is left to the reader ♦ Download free eBooks at bookboon.com Metric Spaces Global Analysis Example 1.3 Let K be an arbitrary set, and let (S, d) be a metric space, in which ≤ d(x, y) ≤ for all x, y ∈ S Let F (K, S) denote the set of mappings f : K → S Define D : F (K, S) × F (K, S) → R+ by D(f, g) = sup d(f (t), g(t)) t∈K Show that D is a metric on F (K, S) Let t0 ∈ K be a fixed point in K and define Evt0 : F (K, S) → S Evt0 (f ) = f (t0 ) by Show that Evt0 is continuous (Evt0 is called an evolution map.) 0.5 –0.5 –1 Figure 3: The metric D measures the largest point-wise distance d between the graphs of two functions over each point in the domain t ∈ K First notice that since ≤ d(x, y) ≤ 1, we have D(f, g) = sup d(f (t), g(t)) ≤ for all f, g ∈ F (K, S) t∈K Without a condition of boundedness the supremum could give us +∞, and D would not be defined on all of F (K, S) × F (K, S) MET Clearly, D(f, g) ≥ Assume now that D(f, g) = sup d(f (t), g(t)) = t∈K Then d(f (t), g(t)) = for all t ∈ K, thus f (t) = g(t) for all t ∈ K This means that f = g, and MET is proved MET is obvious, because D(f, g) = sup d(f (t), g(t)) = sup d(g(t), f (t)) = D(g, f ) t∈K t∈K MET It follows from d(f (t), g(t)) ≤ d(f (t), h(t)) + d(h(t), g(t)) for all t ∈ K, that D(f, g) = sup d(f (t), g(t)) ≤ sup {d(f (t), h(t)) + d(h(t), g(t))} t∈K t∈K 10 Download free eBooks at bookboon.com Metric Spaces Global Analysis The maximum/supremum of a sum is of course at most equal to the sum of each of the maxima/suprema, so we continue the estimate by D(f, g) ≤ sup d(f (t), h(t)) + sup d(h(t), g(t)) = D(f, h) + D(g, h), t∈K t∈K and MET is proved Summing up, we have proved that D is a metric on F (K, S) Since d (Evt0 (f ), Evt0 (g)) = d (f (t0 ), g(t0 )) ≤ sup d(f (t), g(t)) = D(f, g), t∈K we can to every ε > choose δ = ε, such that if D(f, g) < δ = ε, then d (Evt0 (f ), Evt0 (g)) ≤ D(f, g) < ε, and the map Evt0 : F (K, S) → D is continuous 360° thinking Discover the truth at www.deloitte.ca/careers © Deloitte & Touche LLP and affiliated entities 11 Click on the ad to read more Download free eBooks at bookboon.com Metric Spaces Global Analysis Example 1.4 Example 1.1 (2) and Example 1.3 (2) are both special cases of a general result Try to formulate such a general result + Let (X, dX ) and (T, dY ) be two metric spaces, and let ϕ : R+ → R0 be a continuous and strictly increasing map (at least in a non-empty interval of the form [0, a]) with ϕ(0) = Then the inverse map ϕ−1 : [0, ϕ(a)] → [0, a] exists, and is continuous and strictly increasing with ϕ −1 (0) = Theorem 1.1 Let f : X → Y be a map If dY (f (x), f (y)) ≤ ϕ (dX (x, y)) for all x, y ∈ X, then f is continuous Proof We may without loss of generality assume that < ε < a Choose δ = ϕ −1 (ε) If x, y ∈ X satisfy dX (x, y) < δ = ϕ−1 (ε), then we have for the image points that dY (f (x), f (y)) ≤ ϕ (dX (x, y)) < ϕ ϕ−1 (ε) = ε, and it follows that f is continuous Examples In the previous two examples, ϕ(t) = t, t ∈ R+ Clearly, ϕ is continuous and strictly increasing, and ϕ(0) = Another example is given by ϕ(t) = c · t, t ∈ R+ , where c > is a constant Of more sophisticated examples we choose √ ϕ(t) = exp(t) − 1, ϕ(t) = ln(t + 1), ϕ(1) = t, ϕ(t) = sinh(t), ϕ(t) = t, ϕ(t) = Arctan t, etc etc 12 Download free eBooks at bookboon.com Topology Global Analysis Topology Example 2.1 Let (S, d) be a metric space For x ∈ S and r ∈ R+ let Br (x) denote the open ball in S with centre x and radius r Show that the system of open balls in S has the following properties: If y ∈ Br (x) then x ∈ Br (y) If y ∈ Br (x) and < s ≤ r − d(x, y), then Bs (y) Br (x) If d(x, y) ≥ r + s, where x, y ∈ S, and r, s ∈ R+ , then Br (x) and Bs (y) are mutually disjoint We define as usual Br (x) = {y ∈ S | d(x, y) < r} Figure 4: The two balls Br (x) and Br (y) and the line between the centres x and y Notice that this line lies in both balls If y ∈ Br (x), then it follows from the above that d(x, y) < r Then also d(y, x) < r, which we interpret as x ∈ Br (y) Figure 5: The larger ball Br (x) contains the smaller ball Bs (y), if only < s ≤ r − d(x, y) If z ∈ Bs (y), then it follows from the triangle inequality that d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + s ≤ d(x, y) + {r − d(x, y)} = r, which shows that z ∈ Br (x) This is true for every z ∈ Bs (y), hence Bs (y) Br (x) 13 Download free eBooks at bookboon.com ... operators Isometric operators Unitary and normal operators Positive operators and projections Compact operators Download free eBooks at bookboon.com Introduction Global Analysis Functional Analysis, ... Differential Equations 71 13 Euler-Lagrange’s equations 75 Download free eBooks at bookboon.com Introduction Global Analysis Introduction This is the first book containing examples from Functional Analysis. .. Operators on Hilbert Spaces (a) Operators on Hilbert spaces, general (b) Closed operators Functional Analysis, Examples c-4 Spectral theory Spectrum and resolvent The adjoint of a bounded operator