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Lecture10.Pdf Bootcamp Christoph Thiele Summer 2012 Metric spaces Definition 1 A metric space is a pair (M; d) consisting of a set M with a function + d : M × M ! R0 with the following properties that hold for all x; y; z 2 M 1. Zero diestance for identical element: d(x; x) = 0. 2. Non-zero distance for non-identical elements: If x 6= y, then d(x; y) > 0 3. Symmetry: d(x; y) = d(y; x) 4. Triangle inequality: d(x; z) ≤ d(x; y) + d(y; z) We list some examples: q 1. D with distance d(x; y) = jx − yj = (x − y)2 q 2. R with distance d(x; y) = jx − yj = (x − y)2 n 3. R , i.e. the set of tuples x = (x1; : : : ; xn) with xi 2 R with distance d(x; y) = q Pn 2 i=1(xi − yi) n 4. R with distance d(x; y) = sup1≤i≤n jxi − yij q 2 P1 2 5. l (N), the set of sequecnes x : N ! R with distance d(x; y) = i=1(xi − yi) 1 6. l (N), the set of bounded sequences, with distance d(x; y) = supi2N jxi − yij 1 P1 7. l (N), the set of absolutely summable sequences, with d(x; y) = i=1 jxi − yij 8. The set BV (D) of functions of bounded variation on D, i.e. the set of pairs (f; g) with f; g right continuous (say) monotone increasing bounded functions modulo the equivalence relation (f; g) ∼ (f 0; g0) if f + g0 = f 0 + g with distance d((f; g); (h; k)) = inf[j sup(f 0+k0)(x))−(f 0+k0)(0)j+j sup(g0+h0)(x)−(g0+h0)(0)j] x x where the infimum is taken over all representatives of ther respective the equiv- alence classes. 1 9. The space C([0; 1]) of continuous functions f : [0; 1] ! R with distance sup jf(x) − g(x)j x2[0;1] + + 10. The space C(R0 ) of continuous functions f : R0 ! R with distance sup jf(x) − g(x)j + x2R0 + + 11. The space C0(R0 ) ⊂ C(R0 ) of functions satisfying lim sup jf(x)j = 0 x!1 with the same distance. 2 We show the triangle inequality in the case of l (N). Let xi and yi two sequecnes P 2 P 2 with i xi < 1 and i yi < 1. s X X 2 X 2 jxiyij ≤ ( xi )( yi ) i i i Note that this inequality is homgeneous, if we multiply xi or yi by a positive real numfber, the inequality remains true. hence it suffices to prove this inequality under the assumpiton X 2 X 2 xi = yi = 1 i i But then the inequality becomes X X 2 X 2 X 2 2 2 jxiyij ≤ ( xi ) + ( yi ) = (x1 + yi ) i i i i But then the inequality follwos by the comparison principle since for each i 2 2 ±2xiyi ≤ xi + yi since 2 0 ≤ (xi ± yi) We conclude the Cauchy Schwarz inequality s X X 2 X 2 xiyi ≤ ( xi )( yi ) i i i But then we have X 2 X 2 X 2 X (xi + yi) = xi + yi + 2 xiyi i i i i 2 s s s X 2 X 2 X 2 X 2 X 2 X 2 2 ≤ xi + yi + 2 ( xi )( yi ) = ( xi + yi ) i i i i i i Upon setting xi = ai − bi and yi = bi − ci this becomes the triangle inequality for the sequences ai, bi, ci We also show there tirangle inequality for the space BV (D). For simplicity of notation we restrict attention to the space of functions therein which vanish at 0, which can be expressed using pairs of functions vanishing at 0. Then the distance is expressed as d((f; g); (h; k)) = inf[sup(f 0 + k0)(x) + sup(g0 + h0)(x)] x x where again the infimum is taken over all choices of representatives of the given class. (exercise: do the reducdtion of the general caes to this caes.) Let (fg; ); (h; i); (j; k) be three pairs of functions. Assume that the first two are chosen appropriate representatives so that d((f; g); (h; i)) + ≥ sup(f + i)(x) + sup(g + h)(x)] x x Assume that (j; k) is chosen so that for some equivalent pair h0; i0 to h; i d((j; k); (h; i)) + ≥ sup(j + i0)(x) + sup(k + h0)(x)] x x Then we have h + i0 = h0 + i and (f; g) is equivalent to (f + i + h0; g + h + i0). Hence d((f; g); (j; k)) ≤ sup(f + i + h0 + k)(x) + sup(g + h + i0 + j)(x) x x ≤ sup(f + i)(x) + sup(g + h)(x) + sup(h0 + k)(x) + sup(i0 + j)(x) x x x x d((f; g); (h; i)) + + d((j; k); (h; i)) + Since was arbitrary, this proves the triangle inequality. Furthermore, if (M; d) is a metric space, then for any subset M 0 of M we have the metric subspace (M 0; d0) with d0 the restriction of d to M 0. A space satisfying only properties 1; 3; 4 of a metric space can be transformed into a metric space by passing to equivalence classes. Lemma 1 Assume a space M and a distance function d on M satisfies properties 1; 3; 4 of a metric space. Then the relation defined by x ∼ y if d(x; y) = 0 is an equiv- alence relation on M The distance function d(x; y) depends only on the equivalence classes of x and y. the space M~ of equivalence classes with the distance function d~ induced by d is a metric space. Proof: We first show that ∼ is an equivalence relation. Reflexivity: For all x 2 M we have d(x; x) = 0 by property 1 and thus x ∼ x. Symmetry: follows from symmetry of d. Transitivity: If x ∼ y and y ∼ z then d(x; z) ≤ d(x; y) + d(y; z) = 0 + 0 = 0. hence x ∼ z. 3 Next we show that the distance function depends only on the equivalence class of x; y. let x ∼ x0 and y ∼ y0. Then d(x; y) ≤ d(x; x0) + d(x0; y0) + d(y0; y) = d(x0; y0) and similarly we have the reverse ineuqlity. thus d(x; y) = d(x0; y0). Finally we show that M~ is a metric space with metric d~. properties 1; 3; 4 follow immediately form the corresponding properties of M. Property 2 not that if x and y are not equivalent, then d(x; y) 6= 0 and hence d~(x; y) 6= 0. 2 The above reasoning suggests that it makes sense to talk about almost equality of points in a metric space when they have small distance, where smallness may be determined by some parameter that may depend on the context. If two elements have distance at most , then each is in the ball around the other. Definition 2 (Open ball, cone) The open r ball about a point x in a metric space (M; d), denoted by Br(x), is the set of all points y such that d(x; y) < The cone B(x) associated with a point x in a metric space (M; d) is the set of all pairs (y; r) in M × D which satisfy y 2 Br(x). The following lemma shows that a cone has aproperty that can be stated without reference to the point x. Lemma 2 If B(x) is the cone associated with x, then for all points (y; r) in the cone B(x) there exists > 0 such that inf sup d(z; y) + ≤ r δ z:(z,δ)2B(x) and for all points (y; r) not in the cone we have inf sup d(z; y) ≥ r δ z:(z,δ)2B(x) Proof: First let (y; r) 2 B(x). Pick such that d(x; y) + = r. Then for all (z; δ) 2 B(x) we have d(z; y) + ≤ d(z; x) + d(x; y) + = d(z; x) + r ≤ δ + r Hence sup d(z; y) + ≤ δ + r z:(z,δ)2B(x) Taking infimum over δ gives inf sup d(z; y) + ≤ r δ z:(z,)2B(x) Now let (y; r) 2= B(x), then sup d(z; y) ≥ d(x; y) ≥ r z:(z,δ)2B(x) 4 and taking infimum over δ proves the desired estimate. 2 A cone is the natural pendant for metric spaces of a Dedekind cut for ordered spaces. This suggest we will want to do extensions of metric spaces like D to obtain + metric spaces like R0 . We begin with a few definitions. Definition 3 A subset M 0 of a metric space M is called dense if for every (x; r) 2 0 M × D we have y 2 M such that y 2 Br(x). Definition 4 A metric space is called separable, if it has a countable dense subspace. Exercise 1 Prove any number of the following: 1. l2(N) is separable. 2. l1(N) is not separable. 3. BV (D) is not separable. 4. C([0; 1]) is separable. Definition 5 Let (M; d) be a metric space and M 0 be a dense subspace. A set B in M 0 × D is called a Dedekind cone, if for all r there is a point (y; r) 2 B and for all (y; r) 2 B there exists such that inf sup d(y; z) + ≤ r δ z:(z,δ)2B and for all (y; r) 62 D inf sup d(y; z) ≥ r δ z:(z,δ)2B Given a cone B we can define for r 2 D the set (\ball") Br of all points y with (y; r) 2 B.
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