CAAM 402/502 Spring 2013 Homework 2 Solutions
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CAAM 402/502 Spring 2013 Homework 2 Solutions 1. Problem XVII.1.1 in Lang. Let E be a vector space and let v1; :::; vn 2 E be a basis for E. Show that any linear map λ : E ! F into the normed vector space F is continuous. Proof. Let us denote the norm of F by k·kF . We also equip the finite dimensional vector space E with Pn the norm k · k1 defined in terms of the basis fv1; :::; vng as follows. Given x 2 E, let x = i=1 αivi where the scalars αi are uniquely determined by x. Then kxk1 = maxi jαij. We brake the proof into 2 claims. Claim 1: The linear map λ :(E; k · k1) ! (F; k · kF ) is continuous. Proof: In virtue of the result found in Lang page 455, we may equivalently prove that λ is bounded. Let x 2 E be arbitrary. There Pn exist scalars fα1; :::; αng uniquely determined by x such that x = i=1 αivi. Then n n n X X X kλ(x)kF = αiλ(vi) ≤ jαijkλ(vi)kF ≤ max jαij kλ(vi)kF = Ckxk1 i i=1 F i=1 i=1 Pn where C = i=1 kλ(vi)kF is clearly independent of x. Hence λ is bounded. Claim 2: Let k · kE be any norm on E. Then the linear map λ :(E; k · kE) ! (F; k · kF ) is continuous. Proof: Again we show boundedness. We use Theorem 4.3 (Lang page 145) on the equivalence of norms for finite-dimensional spaces. Then there exists a constant K > 0 such that kxk1 ≤ KkxkE; for all x 2 E Then, from Claim 1, we have that, kλ(x)kF ≤ Ckxk1 ≤ CKkxkE; for all x 2 E which makes λ a bounded map. 2. Let f :[−1; 1] ! R de defined as 3x2; if x 2 \ [−1; 1]; f(x) = Q 0; otherwise. Show that f is not differentiable at any point in [−1; 1] except for x = 0. Proof. Since differentiability implies continuity, then f cannot be differentiable at points where it is not continuous. Now we show that f is differentiable at x = 0 and that f 0(0) = 0. Consider h 6= 0 and 2 f(h) − f(0) f(h) 3h 0 ≤ = ≤ = 3jhj: h jhj jhj So in the limit as h ! 0, we obtain that f 0(0) = 0. 1 3. Problem XVII.1.4 in Lang. Let E, F and G be normed vector spaces. A map λ : E × F ! G is said to be bilinear if it satisfies the conditions, λ(v; w1 + w2) = λ(v; w1) + λ(v; w2) λ(v1 + v2; w) = λ(v1; w) + λ(v2; w) λ(cv; w) = cλ(v; w) = λ(v; cw) for all v, vi 2 E, w, wi 2 F and c 2 R (a) Show that a bilinear map λ is continuous if and only if there exists (C > 0) such that for all (v; w) 2 E × F we have jλ(v; w)j ≤ Cjvjjwj Proof. \)", λ is continuous ) λ is continuous at (0; 0). This means that there exist δ1; δ2 > 0 such that jvij ≤ δ1; jwij ≤ δ2 ) jλ(vi; wi) − λ(0; 0)j ≤ 1 We have λ(0; 0) = 0 from third property of bilinear map. Hence jvij ≤ δ1; jwij ≤ δ2 ) jλ(vi; wi)j ≤ 1 v w Consider for any v 2 E, w 2 F , v = δ and w = δ . Clearly jv j = δ and jw j = δ . i jvj 1 i jwj 2 i 1 i 2 Therefore we have, v w δ1 δ2 1 λ δ1; δ2 ≤ 1 ) jλ(v; w)j ≤ 1 ) jλ(v; w)j ≤ jvjjwj ) jλ(v; w)j ≤ Cjvjjwj jvj jwj jvj jwj δ1δ2 1 with C = . δ1δ2 \(" Assume that there exists C > 0 such that jλ(v; w)j ≤ Cjvjjwj for all (v; w) 2 E × F . For any > 0 and (vi; wi) 2 E × F , pick δ = min 1; C(1 + jvij + jwij) jλ(v; w) − λ(vi; wi)j = jλ(v − vi; wi) + λ(v − vi; w − wi) + λ(vi; w − wi)j ≤ jλ(v − vi; wi)j + jλ(v − vi; w − wi)j + jλ(vi; w − wi)j ≤ Cjv − vijjwij + Cjv − vijjw − wij + Cjvijjw − wij 2 < Cδjwij + Cδ + Cδjvij for jv − vij < δ; jw − wij < δ < Hence λ is continuous at every (vi; wi) 2 E × F (b) Let v 2 E be fixed. Show that if λ is continuous, then the map λv : F ! G given by w ! λ(v; w) is a continous linear map. Proof. Let α; β 2 R and w1; w2 2 F . λv(αw1+βw2) = λ(v; αw1+βw2) = λ(v; αw1)+λ(v; βw2) = αλ(v; w1)+βλ(v; w2) = αλv(w1)+βλv(w2) Hence λv is linear. jλv(w)j = jλ(v; w)j ≤ Cjvjjwj = Cvjwj ) jλv(w)j ≤ Cvjwj Cv = Cjvj Hence λv is bounded and linear. Hence λv is a continuous linear map. 2 4. Let S be an open and connected set in Rn, and consider the differentiable function f : S ! Rm with derivative f 0(c) = 0 (the linear map zero), for all c 2 S. Prove that f is constant. Hint: Use the result of Problem 1 below, which says that S is polygonally connected. Proof. Fix a 2 S and let x 2 S. Since S is open and connected, it is polygonally connected (Problem 1 for 502). Thus, there exist points x0 = a; x1; ··· ; xk = x such that L(xj−1; xj) (line segment between two points) is in S for all j = 1; ··· ; k. By the fundamental theorem of calculus we have that Z 1 0 f(xj) − f(xj−1) = f (txj + (1 − t)xj−1)dt (xj − xj−1) = 0 0 for j = 1; ··· ; k. Summing over j we get k X 0 = (f(xj) − f(xj−1)) = f(x) − f(a): j=1 Hence, f(x) = f(a) for all x 2 S, that is, f is constant. 3 ADDITIONAL PROBLEMS FOR CAAM 502 1. Let S be an open and connected set in Rn. Prove that S is polygonally connected, ie., any two points x and y in S can be joined by a polygonal line. Proof. Since S is connected, then the only set two sets in S that are both open and closed (with respect to S) are the whole set S and the empty set . Assume that S is not empty and let a 2 S. Let C ⊂ S denote the set of all the points in S that are polygonally connected to a. This set C is clearly nonempty since a is polygonally to itself. Now we claim that C is both open and closed with respect to S. We proceed to prove openness first. Let x 2 C ⊂ S. Since S is open in Rn, then there exists r > 0 such that Br(x) ⊂ S. Let y 2 Br(x) be arbitrary. Notice that the line segment from x to y stays within Br(x) ⊂ S (by convexity). Now since x 2 C there is a polygonal line in S from a to x. We simply join this line with the line segment from x to y to obtain the desired polygonal line from a to y. Hence we have that Br(x) ⊂ C which implies that C is open in S. Finally we proceed to show that C is closed with respect to S or equivalent that the complement S n C is open. Take x 2 S n C arbitrary (no polygonall line from a to x). Since S is open in Rn, then there exists r > 0 such that Br(x) 2 S. No point y in the ball Br(x) can belong to C, otherwise (like before) we could join a line segment to connected to the center of the ball which is x, and this cannot happen since x2 = C. Hence Br(x) ⊂ S n C which makes S n C open in S. Since a was chosen arbitrarily in S, then we conclude that S is polygonally connected. 2. Let E and F be normed vector spaces and denote by L2(E; F ) the space of continuous bilinear maps of E × E ! F . Let λ 2 L2(E; F ) and define kλk to be the greatest lower bound of all numbers C > 0 such that kλ(v1; v2)kF ≤ Ckv1kEkv2kE 8v1; v2 2 E Show that kλk is a norm in λ 2 L2(E; F ) and verify that its definition is equivalent to kλ(v ; v )k kλk = sup 1 2 F v1;v26=0 kv1kEkv2kE Proof. First we verify the equivalence of definitions. This follows from the fact that the greatest lower bound of all C's that satisfy the above inequality coincides with the least upper bound of all C's that do not satisfy the same inequality. Now we check the axioms of a norm. N1. kλk ≥ 0 and kλk = 0 iff λ = 0 kλk is the greatest lower bound of all C's that satisfy the above inequality. Hence kλ(v ; v )k 1 2 F ≤ kλk kv1kEkv2kE k·kF is a norm, hence kλ(v1; v2)kF ≥ 0 ) kλk ≥ 0. If kλk = 0 ) kλ(v1; v2)kF = 0, 8v1; v2 2 E ) λ(v1; v2) = 0 , 8v1; v2 2 E. Hence λ = 0 N2. For any α 2 R kαλk = jαjkλk For any C > 0 satisfying the given inequality, we have kαλ(v1; v2)kF = jαjkλ(v1; v2)kF ≤ jαjCkv1kEkv2kE If kλk is the greatest lower bound of all possible numbers C, jαjkλk is the greatest lower bound of all possible numbers jαjC.