a. Define a function called an inner product on pairs of points x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) in R n by

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1 Real Analysis Homework 1 Solutions 1. Show that R n with the usual euclidean distance is a metric space. Items a-c will guide you through the proof. a. Define a function called an inner product on pairs of points x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) in R n by x, y = x 1 y 1 + x 2 y x n y n. Show that for any x R n, x, x 0 and is equal to 0 if and only if x = 0 (i.e. the inner product is positive definite). Also show that for any x, y R n, x, y = y, x (i.e. the inner product is symmetric). Third, show that the inner product is linear in each component, that is for α, β R and x, y, z R n show αx + βy, z = α x, z + β y, z to establish linearity in the first coordinate, then use the symmetry of the inner product to give linearity in the second coordinate. Proof. Since x, x = x x x 2 n and each coordinate is real, this sum is always nonnegative. If x = 0, then this sum is = 0. If this sum is zero, then every single coordinate must also be zero, thus we have a function so that x, x 0 and is equal to zero if and only if x = 0. Therefore, is positive definite. [Note that when we say x = 0, this means that x is the origin in R n, not that x is the real number 0.] From the commutativity of multiplication, x, y = x 1 y x n y n = y 1 x 1 + y n x n = y, x. With z = (z 1, z 2,..., z n ), then Therefore αx + βy = α(x 1,..., x n ) + β(y 1,..., y n ) = (αx 1,..., αx n ) + (βy 1,..., βy n ) = (αx 1 + βy 1,..., αx n + βy n ). αx + βy, z = (αx 1 + βy 1 )z (αx n + βy n )z n = α(x 1 z x n z n ) + β(y 1 z 1 + y n z n ) = α x, z + β y, z, 1

2 which establishes linearity in the first variable. To get linearity in the second, let α, β R and x, y, z R n. Then using the symmetry of, and that it s linear in the first variable, x, αy + βz = αy + βz, x = α y, x + β z, x = α x, y + β x, z. Therefore it is linear in the second variable, too. b. For x R n define the norm of x by x = x, x 1 2. Note that this gives the standard euclidean distance in R n of x from the origin. We now establish the Cauchy-Schwarz-Bunyakovsky inequality: x, y x y. This is a significant inequality, and is essential in the study of inner product spaces. Let λ R and x, y R n. Consider the inequality from positive definiteness λx + y, λx + y 0. Use linearity and symmetry to rewrite this as x, x λ x, y λ + y, y 0. Think of this as a quadratic in the variable λ, and since the left side is always nonnegative, the discriminant in the quadratic formula is nonpositive, so that From this deduce that Proof. For λ R and x, y R n, 4 x, y 2 4 x, x y, y 0. x, y x y. 0 λx + y, λx + y = λ x, λx + y + y, λx + y = λ 2 x, x + λ x, y + λ y, x + y, y = λ 2 x, x + 2λ x, y + y, y. 2

3 By considering this as a quadratic in the variable λ and that it is always nonnegative, it either has a repeated real root or no real roots at all. Therefore the discriminant in the quadratic formula must be at most 0, i.e. b 2 4ac 0. This translates to This leads to (2 x, y ) 2 4 x, x y, y 0. 4 x, y 2 4 x, x y, y, which by taking square roots and substituting in where appropriate, x, y x y. c. Define the function d(x, y) = x y, x y 1 2. We show that d is a metric on R n. i. Show that d is positive definite. ii. Show that d is symmetric. iii. Consider three points x, y, z R n. Since R n is a linear space and that in the function d we are considering differences of points, we may assume y = 0 for simplicity, so x y = x and y z = z. With this simplification, we have d(x, y) = x and d(y, z) = z. Show that d(x, z) 2 = x 2 2 x, z + z 2. Using the Cauchy-Schwarz inequality, show that d(x, z) 2 x x z + z 2. Taking square roots, conclude that d(x, z) x + z = d(x, y) + d(y, z). Hence, R n is a metric space using the metric derived from the inner product. Proof. We prove the positive definiteness of d since w, w 0 and is equal to zero if and only if w = 0. The form w, w 1 2 also has this property. By replacing w with x y establishes that d is positive definite. Symmetry is attained by using linearity in both components and observing d(y, x) 2 = y x, y x = (x y), (x y) = x y, x y = x y, x y = d(x, y) 2 3

4 For the triangle inequality, let x, y, z R n. If we define x = x y, y = 0, z = z y, by definition, d(x, y ) = d(x, y), d(y, z ) = d(y, z), and d(x, z ) = d(x, z), then we are able to use x, y, z in place of x, y, z. Therefore we may assume without loss of generality that y = 0. Then d(x, y) = x and d(y, z) = z. Consider the following calculation: d(x, z) 2 = x z, x z = x, x x, z z, x + z, z = x 2 2 x, z + z 2 x x, z + z 2 x x z + z 2 = ( x + z ) 2. = (d(x, y) + d(y, z)) 2 So we arrive at d(x, z) 2 (d(x, y) + d(y, z)) 2, and by taking square roots we get d(x, z) d(x, y) + d(y, z). d. Given any real vector space V, if there is a function, which satisfies i. x, x 0, and x, x = 0 if and only if x = 0 (positive definiteness), ii. x, y = y, x (symmetry), and iii. αx + βy, z = α x, z + β y, z for all α, β R and x, y, z V, then we say V is a real inner product space. The Cauchy-Schwarz inequality follows from these axioms, so V becomes a metric space under the metric d(x, y) = x y, x y 1 2. If we want to consider complex vector spaces we have to change things a slight bit. First, we have to make sure the function respects complex linearity, so we must consider α, β C instead of the reals. Second, the symmetric property must account for the complex variables as well. So we replace ii. with ii. x, y = y, x. so V equipped with such a function is a complex inner product space. e. There are a lot of inner product spaces out there. For example C[0, 1] is a real (complex) inner product space with inner product f, g = 1 0 ( f(x)g(x) dx = 1 0 ) f(x)g(x) dx. 4

5 f. There are a lot of spaces which are not inner product spaces. For example, the metrics d 1 and d on R 2 do not come from inner products. Similarly, the metrics d 1 and d on C[0, 1] don t come from inner products. 2. Given a metric space (X, d) define a function d : X X R by d (x, y) = d(x, y) 1 + d(x, y). a. Show that (X, d ) is a metric space. Hint: When showing the triangle inequality, at some point you will have something like d(x, y)+d(y, z) in the denominator, and you replace this with d(x, z) which will make the fraction larger (but that s ok since you re showing an inequality). Proof. The function d is positive definite since the function f(t) = t 1+t is nonnegative on the nonnegative real numbers and is equal to zero exactly when t = 0. The function d is symmetric since d (x, y) = d(x, y) d(y, x) = 1 + d(x, y) 1 + d(y, x) = d (y, x). We show the triangle inequality for d by starting with the triangle inequality for d. For simplicity, let a = d(x, z), b = d(x, y) and c = d(y, z). Then a b + c a + (ab + ac + abc) b + c + (ab + ac + abc) a + (ab + ac + abc) b + c + (ab + ac + abc) + (bc + abc + bc) a(1 + b)(1 + c) b(1 + a)(1 + c) + c(1 + a)(1 + b) a 1 + a b 1 + b + c 1 + c, where the second line arises from that adding equal amounts to each side of an inequality preserves the inequality, the third line arises from that bc + abc + bc 0, the fourth line is algebra, and the fifth line results from division by the positive number (1 + a)(1 + b)(1 + c). 3. Show that an arbitrary union of open sets is open. Proof. Let U α be open for every α A, where A is an index set. Let x U = α A U α. Since U is a union, there is some β A so that x U β. Thus there is some r > 0 so that B r (x) U β. Since U β U, this implies B r (x) U. Since x was an arbitrary element of U, this shows that U is open. 5

6 4. Let X be a metric space and S X. Prove that ext(s) = (X\S). Proof. We show ext(s) = (X\S) by double containment. ( ). Let x ext(s). Then there some r > 0 so that B r (x) S =. Using the same r, x is also a point in X\S so that B r (x) X\S. Therefore x (X\S). ( ). Let x (X\S). Then there is some r > 0 so that B r (x) X\S. Thus B r (x) S =. But this shows that x ext(s). 5. Show that R\Z is open in R. Proof. Let x / Z be a real number. Then x < x x + 1, where is the greatest integer function. Let r = min{x x, x + 1 x}. Then for y R, if x y < r, then y > x and y < x + 1, so y / Z. Hence B r (x) R\Z, so R\Z is open. Implicit in this argument is that the integers x and x + 1 are the two closest integers to x. The archimedean principle will show this to be true. 6. Prove that in R, Q = R. Hint: between any two real numbers, there is a rational. Proof. We want to show that for any x R and r > 0, then B r (x) contains a point in Q and a point in R\Q. Case 1. Let x R\Q and r > 0. Then there is a rational q between x and x + r. Thus B r (x) contains a rational. This ball automatically contains x R\Q. Therefore x Q. Case 2. Let x Q and r > 0. There is some rational q so that 0 < q < r. Then 0 < q π < r as well, and the number q π is irrational. Thus the irrational number x + q π is contained in B r(x). Thus any neighborhood of x contains a rational (namely x) and an irrational. Therefore x Q. Thus Q = R in R. This proof also contains the idea that between any two reals there is an irrational as well. This and the hint are consequences of the archimedean principle. K&F 6.1. Give an example of a metric space R and two open spheres B r1 (x) and B r2 (y) in R such that B r1 (x) B r2 (y) although r 1 > r 2. Proof. Your example may be different. Let R be a nonempty discrete metric space, x = y R, and 0 < r 2 < r 1 < 1. Then B r1 (x) = B r2 (x), and set containment follows from equality. For an example of strict containment, consider R = {(x, y) R 2 x 0} {( 1, 0)}. Then B 2 (( 1, 0)) B 11 ((0, 0)). Note that you can t get 10 strict containment if x = y. 6