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Metrics a Metric Is a Way of Measuring the Distance Between Objects in a Set

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Metrics a Metric Is a Way of Measuring the Distance Between Objects in a Set Metrics A metric is a way of measuring the distance between objects in a set. Given a set S, a metric on S is a function d from S S to R such that for all x; y; z S, × 2 (i) d(x; x) = 0; (ii) d(x; y) > 0; x = y; 6 (iii) d(x; y) = d(y; x); (iv) d(x; z) d(x; y) + d(y; z) ≤ For most metrics, the first three properties are usually obviously satisfied. The fourth property, known as the Triangle Inequality, commonly requires a bit more effort to verify. Examples: The following functions are metrics on the stated sets: 1. S = R; d(x; y) = x y : j − j (i) d(x; x) = x x = 0 = 0 j − j j j (ii) x y 0; and x y = 0 if and only if x y = 0; that is x = y. j − j ≥ j − j − (iii) d(y; x) = y x = x y = d(x; y). j − j j − j (iv) For any real number, x x . ≤ j j Therefore x + y 2 = (x + y)2 j j = x2 + 2xy + y2 x 2 + 2 x y + y 2 ≤ j j j jj j j j =( x + y )2 j j j j Since x + y 0, we have j j j j ≥ x + y x + y j j ≤ j j j j and hence x z = (x y) + (y z) x y + y z : j − j j − − j ≤ j − j j − j 1 2 2. S = Rn; n 1=2 2 d(x; y) = (x1 yi) ! X − i=1 1=2 = (x y):(x y) − − = x y jj − jj This is usually referred to as the Euclidean metric. To prove (iv), consider (x + ty):(x + ty) = x 2 + 2t(x:y) + t2 y 2 jj jj jj jj where t is a real variable. Since this quadratic expression in t is always 0, the discriminant ≥ 4 x 2 y 2 4(x:y)2 jj jj jj jj − is non-negative. Therefore x:y x y ≤ jj jj jj jj x + y 2 = x 2 + 2x:y + y 2 jj jj jj jj jj jj 2 x 2 + 2 x y + y 2 = x + y ≤ jj jj jj jj jj jj jj jj jj jj jj jj x + y x + y jj jj ≤ jj jj jj jj and d(x; z) = x z jj − jj = x y + y z jj − − jj x y + y z ≤ jj − jj jj − jj d(x; y) + d(y; z) ≤ 3. S = C; d(z1; z2) = z1 z2 : j − j Since 2 2 1=2 z1 z2 = (x1 x2) + (y1 y2) j − j − − this is covered by the previous example. 4. S = Rn; n d(x; y) = xi yi X j − j i=1 This is sometimes known as the taxi cab metric. 3 Since xi zi = xi yi + yi zi xi yi + yi zi j − j j − − j ≤ j − j j − j d(x; z) d(x; y) + d(y; z) : ≤ 5. S = Rn; d(x; y) = max xi yi i j − j This is known as the sup metric. d(x; z) = xj zj for some j j − j xj yj + yj zj ≤ j − j j − j d(x; y) + d(y; z) ≤ 6. S arbitrary; d(x; y) = 1; x = y 6 d(x; x) = 0 : (This is known as the discrete metric.) To prove the triangle inequality, we note that if z = x, d(x; z) = 0 d(x; y) + d(y; z) ≤ for any choice of y, while if z = x then either z = y or x = y (at least) so that 6 6 6 d(x; y) + d(y; z) 1 = d(x; z) ≥ 7. S is the set of all real continuous functions on [a; b]. 1 b 2 2 d(f; g) = Z (f(x) g(x)) dx! : a − This is the continuous equivalent of the Euclidean metric in Rn. The proof of the triangle inequality follows the same form as in that case. 8. S as in 7. d(f; g) = max f(x) g(x) : a x b j − j ≤ ≤ This is the continuous equivalent of the sup metric. The proof of the triangle inequality is virtually identical. 4 9. S arbitrary; d(x; y) = D(x; y)=(1 + D(x; y)), where D(x; y) is another metric on S. d(x; y) + d(y; z) d(x; z) − D(x; y)(1 + D(y; z) + D(x; z) + D(y; z)D(x; z)) = (1 + D(x; y))(1 + D(y; z))(1 + D(x; z)) D(y; z)(1 + D(x; y) + D(x; z) + D(x; y)D(x; z)) + (1 + D(x; y))(1 + D(y; z))(1 + D(x; z)) D(x; z)(1 + D(y; z) + D(x; y) + D(y; z)D(x; y)) − (1 + D(x; y))(1 + D(y; z))(1 + D(x; z)) D(x; y) + D(y; z) D(x; z) = − (1 + D(x; y))(1 + D(y; z))(1 + D(x; z)) 2D(x; y)D(y; z) + D(x; y)D(y; z)D(x; z) + (1 + D(x; y))(1 + D(y; z))(1 + D(x; z)) 0 ≥ since the denominator is positive, and the numerator is non-negative. 10. Let f be a twice differentiable real function defined for x 0, and such that ≥ f(0) = 0(a) f 0(x) > 0 x 0(b) ≥ f 00(x) 0 x 0(c) ≤ ≥ Then if d(x; y) is a metric on the set S, so is f(d(x; y)). Since f 0(x) > 0, we have from the Mean Value Theorem f(x) f(y) = f 0(θ)(x y) − − f(x) > f(y) ; x > y : In particular, since f(0) = 0, f(x) > 0 ; x > 0 : If a 0 and s 0, then by the Mean Value Theorem we also have ≥ ≥ f 0(a + s) f 0(s) = f 00(θ)s 0 − ≤ f 0(a + s) f 0(s) ≤ and if b 0 also ≥ b b Z f 0(a + s) ds Z f 0(s) ds 0 ≤ 0 f(a + b) f(a) f(b) f(0) = f(b) − ≤ − f(a + b) f(a) + f(b) ≤ Since d(x; z) d(x; y) + d(y; z), ≤ f(d(x; z)) f(d(x; y) + d(y; z)) ≤ f(d(x; y)) + f(d(y; z) ≤ 5 Three important inequalities Definitions: If p and q are both > 1, we say that they are dual indices if 1 1 + = 1 : p q For x Rn, we define 2 n 1=p p x p = xi ! jj jj X j j i=1 Young's inequality. For x and y positive real numbers, and dual indices p and q, 1 1 xy xp + yq : ≤ p q Proof Fix y > 0, and consider the function 1 f(x) = xy xp − p for x > 0. p 1 p 2 f 0(x) = y x − ; f 00(x) = (p 1)x − < 0 − − − p 1 1=(p 1) Therefore f has a global maximum when y = x − , x = y − , and 1 p 1+1=(p 1) 1 p=(p 1) 1 q xy x y − y − = y − p ≤ − p q H¨older'sinequality For x and y in Rn, n xiyi x p y q X j j ≤ jj jj jj jj i=1 where p and q are dual indices. Proof If either x = 0 or y = 0 the result is trivially true. Otherwise, for each i xi yi j j 0; j j 0 x p ≥ y q ≥ jj jj jj jj and therefore by Young's inequality p q xi yi 1 xi 1 yi j jj j j j p + j j q x p y q ≤ p x p q y q jj jj jj jj jj jj jj jj Summing from i = 1 to n, we have 1 n 1 1 xiyi + = 1 : x p y q X j j ≤ p q jj jj jj jj i=1 6 Minkowski's inequality For x and y in Rn, and p > 1, x + y p x p + y p : jj jj ≤ jj jj jj jj n p p 1 x + y = xi + yi xi + yi − jj jjp X j j j j i=1 n n p 1 p 1 xi xi + yi − + yi xi + yi − ≤ X j jj j X j jj j i=1 i=1 j n 1=q (p 1)q x p xi + yi − ! ≤ jj jj X j j i 1 − n 1=q (p 1)q + y p xi + yi − ! jj jj X j j i 1 − p=q n 1=p 2 p 3 = x p + y p xi + yi ! jj jj jj jj X j j 4 i=1 5 p 1 = x p + y p x + y − jj jj jj jj jj jjp Minkowski's inequality shows that d(x; y) = x y p jj − jj is a metric on Rn. If we consider the limits p 1 and p , we obtain the taxicab and sup metrics which we have already! seen. ! 1 The Euclidean metric corresponds to p = 2. These metrics can be extended to infinite real sequences xi 1 : f gi=1 p If l is the set of all sequences xi for which f g 1 p xi ; p 1 X j j ≥ i=1 converges, then the norm 1=p 1 p x p = xi ! jj jj X j j i=1 induces a metric on lp in the same way as the finite form does on Rn.
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