Math 131C Homework 3 Solutions From Rudin Chapter 9 Problem 3 Suppose Ax Ay Then A x y 0 x y Problem 4 These results follow immediately from the definition of a vector space and linearity You don t have to verify all the vector space axioms since the range and kernel are subsets of vector spaces it suffices to show they are subspaces i e that they are closed under addition and scalar multiplication Problem 5 Given A L Rn R let yA Ae1 Ae2 Aen where e1 en is the standard basis for Rn By linearity Ax x yA for all x Rn The uniqueness of yA is immediate Now by Cauchy Schwarz we have kAk sup x yA sup x yA yA x 1 But if x yA yA x 1 we have x 1 and Ax yA so kAk yA So kAk yA Problem 6 For x y 6 0 0 we can compute Dj f x y via the usual differentiation formulas just as in Math 32A or the like At 0 0 we consider the definition of partial derivatives D1 f 0 0 lim h 0 f 0 h 0 f 0 0 0 0 lim 0 h 0 h h Similarly D2 f 0 0 0 So the partial derivatives of f exist everywhere in R2 However f is not continuous at 0 0 since f a a 12 for all a R So there is no neighborhood of the origin on which f x y f 0 0 f x y 12 Problem 7 Fix x E and 0 Since the partial derivatives of f are bounded in E say by M the Mean Value Theorem implies that f a hej f a hM for all a E and all h R such that a hej E 1 j n Now choose nM s t B x E we can choose such a since E is open Pn Pk Now pick an arbitrary y in B x Write y x 1 hj ej v0 0 and vk 1 hj ej Note vn y x It s pretty easy to see that hj for all j Now we have 1 f y f x f x y x f x n X f x vj f x vj 1 j 1 n X j 1 n X j 1 n X f x vj f x vj 1 hj M M j 1 where we ve used the Mean Value Theorem result from the first paragraph So y x f y f x and f is continuous at x Problem 8 Let x x1 xn and let Ej be the j cross section of E through x i e Ej x R x1 xj 1 x xj 1 xn E Now define fj Ej R by fj x f x1 xj 1 x xj 1 xn Since f has a local maximum at x fj has a local maximum at xj Ej for all j By single variable calculus this implies fj0 xj 0 for all j But since fj0 xj Dj f x VERIFY this implies f 0 x 0 by Theorem 9 17 in Rudin Problem 9 To show f is constant fix some x0 E and consider the set A x E f x f x0 We want to show that A E but since E is connected and A is nonempty it suffices to show that A is both open and closed in the relative topology on E Since E is open the relative topology coincides with the usual topology on Rn First we show A is open Take any x A Since E is open there is an open ball Br x E Now Br x is convex so Theorem 9 19 or its corollary in Rudin implies that f is constant on Br x So since x A this implies Br x A So A is open To show A is closed use an exactly analogous argument to show that Ac is open So A is both closed and open which implies that A E Problem 10 This whole problem is easier to think about if you draw a picture in two dimensions Fix any 0 0 0 0 x0 x1 xn E We want to show that for all x x1 x2 xn E f x f x0 As 0 0 0 0 in Problem 8 define f1 x f x x2 xn Note that f10 x D1 f x x2 xn so f10 x 0 0 0 for all x at which f1 is defined Now take any x1 s t x1 x2 xn E By convexity we have 0 0 0 x x2 xn E for all x1 x x1 so f1 is defined for all such x Since f10 x 0 for all these 2 0 0 0 x the Mean Value Theorem shows that f1 x1 f1 x1 So f x0 f x1 x2 xn as desired Note that the proof just given only required that E be convex in the first variable i e we can relax the convexity condition to a x2 xn E and b x2 xn E x x2 xn E a x b But as noted the proof doesn t work for arbitrary connected regions Draw the union of the following three regions in R2 A x y 2 x 2 0 y 1 B x y 2 x 1 2 y 0 C x y 1 x 2 2 y 0 Take f x y 0 y y x y A x y B x y C Then clearly f x 0 on the union of these regions but f does not depend only on y The problem is that we can t apply the Mean Value Theorem because of the gap between B and C 3