Solutions to assignment 5. x y = x z + z y x z + z y < = 5. x y x z = x (y z) x y z x y + ( z) x ( y + z ) < 2 (3 + 4) = 14,

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1 Solutions to assignment 5 8..: a) x y = x z + z y x z + z y < = 5 b) x y x z = x (y z) x y z x y + ( z) x ( y + z ) < 2 (3 + 4) = 4, where I used the Cauchy-Schwartz inequality in the first step, and then the triangle inequality. c) x (y z) y (x z) = x y x z y x + y z = z (y x) z y x < 6. d) x y 2 x x = (x y) (x y) x x = x x 2x y + y y x x = y (y 2x) y y 2x < 2 = 2. e) x z y z = (x y) z = x y z sin θ x y z < 6, where θ is the angle between x y and z. Note: a b a b is always true for the same reason as above. f) Note that the book has a misprint: they shouldn t be taking the norm of x (y z), since that s a scalar. Instead, they should have the absolute value, which I have below. Anyway, x (y z) x y z x y z < 6,

2 where for the first step I used the Cauchy-Schwartz inequality, and for the second step I used the inquality which I observed in part e. 8..3: If x = 0 or y = 0, one clearly has equality in the Cauchy-Schwartz inequality. If x is parallel to y, then there is a scalar t so that x = ty. In this case, x y = t x 2 = t x x = x y. Thus, if x = 0 or y = 0 or x is parallel to y, then there is equality in the Cauchy-Schwartz inequality. For the other direction, suppose that x y = x y. From the proof of the Cauchy-Schwartz inequality, if y 0, the only time one can have equality is if x ty = 0 for t = x y 2. But this means that y there must exist a scalar t so that x ty = 0, i.e., x = ty. Thus either x = 0, or y = 0, or they are parallel, by the definition on page : Take the cube to have unit sides, so that we want to find the angle between (,,,, ) and (, 0, 0,, 0). This will satisfy cos θ = = (,,,, ) (, 0, 0,, 0) (,,,, ) (, 0, 0,, 0), n ( so that the angle they want is arccos n ). Plugging into a calculator gives the approximation they ask for when n = : For any N, we have N a k b k N a 2 k N b 2 k, by applying the Cauchy-Schwartz inequality to the vectors ( a,, a N ) and ( b,, b N ). But since a2 k and b2 k are positive term series, the sum of each series is larger than or equal to each partial sum, thus N a k b k b 2 k <. Since we have a uniform bound for the sequence of partial sums of a kb k, it follows (since this is a non-negative term series) that a kb k converges, hence a kb k converges absolutely. 8..0: First the l norm, then the l norm. i) x = n i= x i is obviously non-negative, and if the sum is zero, then all of the x i s are zero (we can t get back to zero if one of the x i s is non-zero). a 2 k 2

3 ii) αx = n i= αx i = n i= α x i = α x. iii) x + y = n x i + y i i= n ( x i + y i ) = i= i= i= n n x i + y i = x + y, using the triangle inequality for absolute value. The reverse triangle inequality x y x y either follows similarly from above, using the reverse triangle inequality for absolute value, or once we have the triangle inequality, we can say x = x y + y x y + y, then subtract y from both sides. i) x = max ( x,, x n ) is clearly non-negative, and if the maximum is zero, all the x i s are zero. ii) αx = max ( αx,, αx n ) = max ( α x,, α x n ) = α max ( x,, x n ), since it s pretty clear you can factor out a non-negative factor from a maximum of a set of non-negative numbers. iii) Suppose k is such that x k + y k = max ( x + y,, x n + y n ). Then x + y = x k + y k x k + y k x + y. The reverse triangle inequality follows from this as above. 8.3.: I m not going to try to sketch these electronically, since they are pretty obvious and it s sort of a hassle. Also, I told you not to worry about connectivity, since I hadn t gotten to that in class yet. a) Open. b) Closed. c) Neither open nor closed. d) Open. e) Closed. 3

4 8.3.2: It s probably easiest to first prove that closed balls are closed sets. Then V is open because it s the intersection of an open ball with the complement of a closed ball, i.e., the intersection of two open sets, and E is closed because it s the intersection of a closed ball with the complement of an open ball, i.e., the intersection of two closed sets. However, if you aren t quite this sneaky, here s how to do this directly : To show that V is open, take x V, and set ε = min (r x a, x a s). Then if y B ε (x), then and y a = y x + x a y x + x a < ε + x a r, y a = (x a) (x y) so B ε (x) V, and V is open. x a x y > x a ε x a ( x a s), To show that E is closed, take x / E. Then either x a < s or x a > r. If x a < s, then (since we ve shown that open balls are open sets), there is an ε > 0 so that B ε (x) B s (a). If x a > r, set ε = x a r. Then y B ε (x) will imply that y a > r as in the second half of the proof of openness of the set V. In either case, a point exterior to E will have an open neighborhood which is disjoint from E. Thus E is closed. a) First, suppose that V is open. For each x V there is ε (x) > 0 so that B ε(x) (x) V. I claim that V = x V B ε(x) (x). Indeed, if x V then x B ε(x) (x), hence an element of the union. Conversely, since B ε(x) (x) V for all x, their union is contained in V. Thus the sets are equal. The other way is easier: if V is a union of open balls, then it s a union of open sets, which we ve seen by theorem is open. b) The question posed by the book is a little vague, so in class I told them to look at the claim that E is closed iff there is a collection of closed balls whose union is E, and to either prove or give a counterexample to each direction of the claim. If E is closed, it s certainly a union of closed balls, since (taking r = 0 in the book s definition), a single point is a closed ball. Thus every set is a union of closed balls. This gives a counterexample to the other direction: take E to be any set which is not closed. Then E = x E B 0 (x), a union of closed balls. (Slight abuse of notation: B 0 (x) is the closed ball of radius 0 whose center is x, i.e., the point x itself.) 4

5 8.3.9: Take a / E. Since E c is open, there is an ε > 0 so that B ε (a) E c, i.e., if x a < ε, then x E c. Thus inf x E x a ε > 0. 5