midterm1-practice-solutions

midterm1-practice-solutions - MATH 131B 1ST PRACTICE...

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Unformatted text preview: MATH 131B 1ST PRACTICE MIDTERM SOLUTIONS Problem 1. State the book’s definition of: (a) Uniform convergence of a sequence of function (b) A metric (c) Pointwise convergence of a sequence of functions (d) A continuously differentiable function of two variables (e) A continuous function of two variables (f) The supremum norm f ∞ Solution. See the following pages: (a), p. 167; (b), 196; (c), 164; (d), 155; (e) 152; (f) 175. Problem 2. Prove that a uniform limit of a sequence of continuous functions is also continuous. Solution. See proof of Theorem 5.2.1 on pages 169-170. Problem 3. Let fn be a sequence of functions that converges uniformly on a b to a function f . Show that £¢¡ ¤ f x dx § ¦¥ a b ¦¥ fn x dx a ¨ ¤ b Give an example which shows that the conclusion does not hold if we only assume pointwise convergence. Solution. For the first part, see proof of Theorem 5.2.2 on p. 170. For the second part, see Example 4 on page 166. Problem 4. Let C 1 0 1 be the space of functions on 0 1 which have a continuous derivative. Let  ¨  © £¢¡ £¢¡ f f f ∞ ∞ (a) Show that f is a norm (b) Show that C 1 0 1 is complete in this norm.    © © £¢¡ ©  Solution. (a) We must prove: (i) f 0 iff f 0; (ii) α f α f and (iii) f g g. If f 0 then for sure f ∞ 0. Thus f 0 since f ∞ is a norm. Conversely, if f 0, then f 0. Since f supx f x supy f y , we have that α f α f. Similarly, ©   ¨  ¦ ¥ ¦¥ ¦¥        ¦ ¥ ¦ ¥ ¦ ¥  ¦ ¥     ¦ ¥  sup f x gx sup f y gy y sup g x x sup f y y g 1  x sup g y y ¦ ¥ sup f x  x f  "! © ©  © ¦¥   g ©  © f  ©! © ! ©!  f MATH 131B 1ST PRACTICE MIDTERM SOLUTIONS 2 (b) We must prove that if f n is a Cauchy sequence, then f n converges in this norm to a continuously differentiable function. Note that f f ∞ and f f ∞ . Thus if fn is a Cauchy sequence with respect to the norm , then fn is a Cauchy sequence for the uniform norm ∞ . Since the space of continuous functions is complete in the uniform norm, it follows that the sequence f n converges in uniform norm to a function g C 0 1 . Moreover, since f n form a Cauchy sequence in the uniform norm, then fn 0 form a Cauchy sequence and thus converge. We now apply Theorem 5.2.3 with x0 0. The conclusion is that the sequence f n converges uniformly to a function f , which is continuously differentiable and whose derivative is g. Note that we thus have that f C 1 0 1 . Since fn f uniformly, it follows that f f n ∞ 0, by Proposition 5.3.2. Similarly, since fn g f uniformly, f fn ∞ 0. Thus f fn f fn ∞ f fn ∞ 0, so that f fn in the norm . ¥ £ ¢  "¡  § © © §   © § § © © ¦¥ £ §  §  ¨ £¢¡ ¦ § § ¦ £¢¡ © ¤ £   § © § § Problem 5. Endow with the following metric ρ: ρ x y 1 if x y and ρ x y 0 if x y. (a) Show that ρ is a metric. (b) Show that a sequence xn is Cauchy for this metric if and only if it is eventually constant; i.e., for some N , xn xm whenever n m N . (c) Show that is complete with respect to this metric.  ¦¢¥ © ¦¢¥ © ¦ ¢ © © © © ¦¢ © ©  © ¢¥ ¢ ¦¢ ©© © ¦ ¢ ¥ ¦ ¢ ¥ ¦¢¥ ¦ © Solution. (a) It is trivial to see that ρ x y ρ y x and that ρ x y 0 iff x y. For the triangle ρ x y ρ y z . The left hand side is 1 unless x z. The inequality, we must show that ρ x z right hand size is either 0, 1 or 2, and so it is at least as big as the left hand in the last two cases. If the right hand side is 0, then ρ x y ρyz 0 so x y z and the tringle inequality still holds. (b) Let ε 1 2. Then there is an N so that for all n m N , ρ xn xm 1 2. But this means that xn xm for all n m N . (c) Let xn be a Cauchy sequence. From part (b) we know that there is an N so that xn xm for all n m N . Fix m N and let x xm . Then xn x. Indeed, for any ε 0, ρ xm xn 0, since xn xm . we have that for all n N , ρ x xn   ¦ ¦¢¥  § ¢ ¥ ¥ ¦¢¥ © ¦¢¥ ¢ © ¥ ¦ ¢¥ n2 x 1 n4 x 2 © converges pointwise to 0 on 0 1 , but does not converge uni- £¢¡  © ¢  © © ©  Problem 6. Show that fn formly. Solution. For each fixed x, ∞ does not converge  § 1 2, so that fn  ∞  0  ©  Thus the sequence converges pointwise. 1 1 On the other hand, since f n n2 2 , it follows that f n to zero. Thus fn do not converge to zero uniformly. x2 ∞1 n4 § n x n2 © n lim ¨ n2 x ∞ 1 n 4 x2 lim © ¦¥ Problem 7. Let f : 0 1 0 1 be a continuously differentiable function. Suppose that f x 1 for all x 0 1 . Show that f is a “uniform contraction”: there is a constant 0 C 1 so that for fy Cx y. all x y 0 1 , f x  ¦ ¥       ¦ ¥  ¦ ¥  ©  £¢¡ £¢¡¦ ¦ ¥ § ¦ ¥ ¦ ¥ £ ¢ ¡ ¦ ¢ §   £¢ ¡ ¦ £ ¢ ¡ §£ ¢ ¡ Solution. Since f x is continuous, so is f x . Hence f x attains its maximum at some point x0 0 1 . It follows that if we set C f x0 , then C 1 and f z f x0 C for all z in 0 1 . © ¦ ¥   ¦ ¥  ¦ ¥  1ST PRACTICE MIDTERM SOLUTIONS By applying the Mean Value Theorem, we obtain that for all x y so that f x fy f z x y . We conclude that £¢¡¦ ¢ § Cx y ¢ §   ¦ ¥  © ¦¥ § ¦ ¥  § ¥ ¦ ¥ © ¦¥ § ¦ ¥ where C as above is a constant strictly less than 1. y  fzx  fy  ¦ fx 0 1 there is a point z 3 x y, £¢¡¦ MATH 131B ...
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