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Contents 1 Sequence and Series of Functions 1.1 Pointwise and Uniform Convergence . . . . . . . . . . . . . . 1.1.1 Pointwise Convergence . . . . . . . . . . . . . . . . . . 1.1.2 Uniform Convergence . . . . . . . . . . . . . . . . . . 1.1.3 Uniform Convergence and Continuity . . . . . . . . . . 1.2 Uniform Convergence and Differentiability . . . . . . . . . . . 1.3 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Convergence of Series of Functions . . . . . . . . . . . 1.3.2 Uniform Convergence, Continuity and Differentiability 1.4 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Domain of Pointwise Convergence of Power Series . . 1.4.2 Uniform Convergence of Power Series . . . . . . . . . 1.5 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Convergence of Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 4 5 7 9 12 12 15 17 18 20 23 27 2 The Riemann Integral 2.1 The Definition of Riemann Integral . . . . 2.1.1 Integrability . . . . . . . . . . . . . 2.1.2 Criterion of Integrability . . . . . . 2.2 Integrating Functions with Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 33 36 38 40 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Sequence and Series of Functions In this chapter we follow the path of investigation of the sequence of real numbers, and extend the scope to the universe of functions. To make a direct comparison of the difficulties, the cardinality of all functions over R is 2 to the power of cardinality of R. A real number r can be regarded as a function taking constant value r, which in turn is a trivial special case. Therefore we cannot hope to understand the sequence of general functions to the same depth and clarity as sequence of real numbers, and this chapter may be considerably harder to learn than the previous one. However from the application perspective, it is very important to study a sequence of functions, sometimes due to the imperfectness of human understanding of unknown relations in the physical world, and other times a deliberation of keeping formulae simple to calculate. For example, the Taylor series of a function f is used to calculate an approximate value of f at a certain point, without knowing the formula of f explicitly. The key point of making the sequence of functions a fruitful object to study is to define various notions of ”convergence”. Surely more complicated than the limit of real numbers, when we say a sequence of functions (fn ) has a limit f , we need to specify what kind of limit. We can see that the ✏-N language is as powerful as it was in defining the limit of real numbers. 1.1 Pointwise and Uniform Convergence Let us look at an example of sequence of functions. To denote a sequence of functions, we use subscript notations fn : A ! R, meaning that for each n = 1, 2, 3, · · · , fn is a real-valued function on A. We may evaluate the whole sequence {fn } at one point x 2 A, and denote the resulting sequence of real numbers by (fn (x)) (indexed by n); or we can create another sequence of real numbers xn 2 A, and pair the function fn with xn , thus we get a sequence of real numbers (fn (xn )) (still indexed by n). These notations are self-explanatory once we accept their meanings, and are ubiquitous in this chapter. 3 Math 142B Lecture Notes (v. 2021/08/24) 1.1.1 Yucheng Tu 4 Pointwise Convergence Definition 1.1.1. For each n 2 N, let fn be a function defined on a set A ⇢ R. The sequence (fn ) of functions converges pointwise on A to a function f if for all x 2 A, the sequence of real numbers fn (x) converges to f (x). Example 1.1.2. Let fn : R ! R be defined as fn (x) = (x2 + nx)/n. Let f (x) = x be the identity function, then for any x 2 R: ⇣ x2 ⌘ x2 + nx = lim + x = x = f (x) n!1 n!1 n n lim fn (x) = lim n!1 hence the sequence (fn ) converges point-wise to f (on R, usually omitted when (fn ) and f have the same domain. Sometimes a sequence of continuous functions (even differentiable, smooth ones) may point-wise converge to a discontinuous function. Example 1.1.3. Let fn : [0, 1] ! R be defined as fn (x) = xn . From the behavior of power functions, when |x| < 1, limn!1 xn = 0, when x = 1, limn!1 xn = 1. Therefore the pointwise limit of fn must be a discontinuous function: ( 0 when 0 x < 1 f (x) = 1 when x = 1 Remark. Example 1.1.3 often serves as a counterexample to reject the naive guess that when (fn ) are continuous, their point-wise limit is also continuous. A more formal way to state this fact is the following non-interchangeability of limits: lim lim fn (x) 6= lim lim fn (x) n!1 x!x0 x!x0 n!1 in general. (1.1) Exercise 1.1.4. By computing the double limits (taking inner limits first, and then the outer limit), argue that the inequality (1.1) holds when we choose fn as in Example 1.1.3 and x0 = 1. With Example 1.1.3 in mind, we see that pointwise limit f of sequence of functions (fn ) does not preserve the continuity property of fn . In a certain sense pointwise convergence is too weak, namely it treats the domain of fn as individual points, so at different points x, y 2 A, the convergence of (fn (x)) to f (x) and (fn (y)) to f (y) can be two independent limits. If we look closer into Example 1.1.3 by choosing x = 0.9 and y = 0.999, here are what we get: It is clear that even though 0.9n ! 0 and 0.999n ! 0 as n ! 1, the latter evidently converges slower than the former. Intuitively, this causes the limit function f to be not continuous at x = 1, since the closer x is to 1 (x < 1), the slower fn (x) converges to 0. (version 2021/08/24) n 0.9n 0.999n 1 0.9 0.999 2 0.81 0.998 5 5 0.59 0.995 10 0.349 0.99 20 0.122 0.98 50 0.005 0.951 100 2.65E 0.905 5 200 7.05E 0.818 10 500 1.32E 0.606 23 Exercise 1.1.5. Let fn : R ! R be defined as follows: fn (x) = |x|n 1 |x|n + 1 find the pointwise limit of fn . (Hint: consider three cases: |x| < 1, |x| = 1 and |x| > 1. 1.1.2 Uniform Convergence Now we introduce a stronger notion of convergence: uniform convergence. Definition 1.1.6 (Uniform Convergence). Let (fn ) be a sequence of functions defined on A ⇢ R. Let f : A ! R be another function on A. We say fn converges uniformly on A to f if: for every ✏ > 0, there is N 2 N such that |fn (x) f (x)| < ✏ for every n N and x 2 A. This definition appears no-where near the simple pointwise convergence (Definition 1.1.1). However, the pointwise convergence can be written in a similar language, just by switching the phrase ”for all x 2 A” to the head: Definition 1.1.7 (Rephrasing 1.1.1). Let (fn ) be a sequence of functions defined on A ⇢ R. Let f : A ! R be another function on A. We say fn converges uniformly on A to f if: for all x 2 A and all ✏ > 0, there is N 2 N such that |fn (x) f (x)| < ✏ for every n N. Remark. In the definition of uniform convergence, the domain point x 2 A is referenced at the end, meaning that N 2 N must be chosen only depending on a specific ✏ > 0, without specifying x; in point-wise convergence, we begin with ”for all x 2 A”, hence N is chosen to be depending on the specific x 2 A and ✏ > 0. So it requires more effort to verify uniform convergence than point-wise convergence. Exercise 1.1.8. By manipulating Definition 1.1.6 and 1.1.7, show that if (fn ) uniformly converges to f on A, then (fn ) also pointwise converges to f on A. Example 1.1.9. Let fn : R ! R be defined as fn (x) = x2 1 +n n2N then the pointwise limit of fn is f (x) := lim n!1 x2 1 =0 +n Math 142B Lecture Notes (v. 2021/08/24) Yucheng Tu 6 since the denominator x2 + n ! 1 when x is fixed and n ! 1. Let us check if the convergence is actually uniform in the sense of Definition 1.1.6. The key observation is that since 8x 2 R, x2 0, if one choose N 2 N such that 1/N < ✏ (we can!), then |fn (x) f (x)| = 1 1 1 0, 9N 2 N such that |xn 0| < ✏, 8n N and 8x 2 [0, 1). Recall the different convergence rate of xn ! 0 when x = 0.9 and x = 0.999. Can N be big enough so that at least for ✏ = 1/2 and n = N , we have xn < ✏ for all x 2 [0, 1)? That is already q a very moderate test case. However, we can not pass the case, because 1 x = ✏N = N 1 2 < 1 would be contradiction with xN < ✏. In other words: 9✏(= 12 ) > 0, such that 8N 2 N, 9n(= N ) 1 N and 9x(= ✏ N ) 2 [0, 1) and |xn 0| = ✏ ⌅ ✏. This is a direct negation of the previous statement, a disproof of fn uniformly converges to f on [0, 1). In conclusion, on [0, 1], fn does not converge uniformly to f . Generally, proving that fn converges uniformly to f on A requires us to follow the definition 1.1.6. On the contrary, if fn does not converge uniformly to f , we need to prove the negation of definition 1.1.6. Uniformly Cauchy Sequence Recall that Cauchy sequence plays important role in the proof of existence of a limit. For a sequence of functions, we can introduce this idea into the study of uniform convergence. It is particularly useful when we study series of functions (infinite sum of functions). Definition 1.1.11 (Uniformly Cauchy). A sequence of functions (fn ) defined on A ⇢ R is called uniformly Cauchy if: 8✏ > 0, 9N 2 N, such that |fn (x) fm (x)| < ✏ for all m, n N and x 2 A. (version 2021/08/24) 7 Theorem 1.1.12. A sequence of functions (fn ) defined on A ⇢ R converges uniformly to f on A if and only if (fn ) is uniformly Cauchy. Since we will not use this theorem until we talk about series of functions, let us omit its proof here. 1.1.3 Uniform Convergence and Continuity The introduction of uniform convergence is to provide a solid criterion that guarantees nice properties of the limit function. We will discuss two theorems on how uniform convergence can achieve that in a satisfactory manner. Theorem 1.1.13. Let {fn } be a sequence of functions defined on A ⇢ R, and each fn is continuous at some c 2 A. If fn converges uniformly to a function f on A, then f is continuous at c. We need to show that for every ✏ > 0, there is a such that, if |x c| < , then |f (x) f (c)| < ✏. The key ideas: (1) fn (x) and fn (c) are close to f (x) and f (c), respectively; (2) fn (x) is close to fn (c) since fn is continuous. Therefore using triangle’s inequality we obtain a very short proof. Proof. Let ✏ > 0 and c 2 A be given. By the uniform convergence of fn , we can find N 2 N such that |fn (x) f (x)| < ✏/3 8x 2 A and n 2 N. In particular c 2 A, hence |fn (c) f (c)| < ✏/3. By continuity of fn , there is a that |x c| < implies |fn (x) fn (c)| < ✏/3. > 0 such Hence |f (x) whenever |x c| < f (c)| = |f (x) fn (x) + fn (x) |f (x) fn (x)| + |fn (x) ✏ ✏ ✏ < + + 3 3 3 =✏ fn (c) + fn (c) fn (c)| + |fn (c) f (c)| f (c)| and x 2 A. Therefore f is continuous at c 2 A. Theorem 1.1.13 can be applied to decide non-uniform convergence in its counter-positive form. The statement is: Corollary 1.1.14 (Counter-positive of Theorem 1.1.13). Let {fn } be a sequence of functions defined on A ⇢ R, and each fn is continuous at some c 2 A. If f is not continuous at c, then fn does not converge uniformly to f on A. Example 1.1.15 (Another proof of Example 1.1.3). Proof. For each n, fn (x) = xn is a continuous function on [0, 1]. However, the limit function f is not continuous at x = 1. By Corollary 1.1.14, we conclude that fn does not converge to f uniformly on [0, 1]. Math 142B Lecture Notes (v. 2021/08/24) Yucheng Tu 8 Unfortunately, we cannot say more about uniform convergence when both the sequence (fn ) and limit function f are continuous on A. We should keep the following example in mind as it appears hand-crafted other than natural: Example 1.1.16 (Shifting Peaks). Let (fn ) (n 8 > : 0 2) be defined as: x 2 [0, 1/n] x 2 (1/n, 2/n] x 2 [2/n, 1] The graph of fn can be sketched by connecting four points (0, 0), (1/n, 1), (2/n, 0) and (1, 0) consecutively by line segements. The peak point on the graph, (1/n, 1) has a constant level y = 1 but is moving left towards 0 (x = 1/n). What is a pointwise limit of (fn )? Does (fn ) convergence uniformly on [0, 1] to the limit function? For pointwise limit we need to fix an arbitrary x 2 [0, 1]. Note that potentially we need to use different formulae for fn (x) based on whether x is in [0, 1/n], (1/n, 2/n] or [2/n, 1]. A quick observation is that as n ! 1, [2/n, 1] can eventually cover every x 2 (0, 1], and once x 2 [2/n, 1], we have x 2 [2/(n + 1), 1], x 2 [2/(n + 2), 1], and so on. Therefore fn (x) is 0 when n > 2/x. Here is a typical example of (fn (x)): ⇥, ⇥, ⇥, · · · , ⇥, 0, 0, 0, 0, 0, 0, 0, · · · here ⇥ represents a nonzero value of fn (x) when n is small. This observation make us believe that lim fn (x) = 0 when x 2 (0, 1] n!1 On the other hand, fn (0) = n · 0 = 0 for each n, hence the pointwise limit of (fn ) is the zero function on [0, 1]. Exercise 1.1.17. Provide a rigorous proof that fn converges pointwise to the zero function on [0, 1]. Now we turn to the uniform convergence problem in Example 1.1.16. The most urgent question is to decide whether to prove or disprove it, because their proofs are so different and we do not want to waste effort on a wrong decision. The following useful theorem provides a more systematic way in doing so: Theorem 1.1.18. Let (fn ) be a sequence of functions on A ⇢ R and f : A ! R. (fn ) converges to f uniformly on A if and only if the following holds: lim n!1 sup |fn (x) x2A f (x)| = 0 (version 2021/08/24) 9 Before we prove the theorem, let us apply it to Example 1.1.16. We consider the following quantity for each n: sup |fn (x) x2[0,1] f (x)| = sup |fn (x)| = |fn (1/n)| = 1 x2[0,1] since f is the zero function and fn has it maximum 1 at x = 1/n. Then we can see lim sup |fn (x) f (x)| = lim 1 = 1 6= 0 n!1 n!1 x2[0,1] So Theorem 1.1.18 immediately asserts that (fn ) does not converge uniformly to f on [0, 1]. We can see it is very easy to apply this theorem once we can determine the expression supx2A |fn (x) f (x)|. Now let us see why Theorem 1.1.18 is true. Proof of Theorem 1.1.18. ()) Assume that (fn ) converges uniformly to f , hence by definition, for every ✏, there is N 2 N such that |fn (x) f (x)| < ✏ whenever n N and x 2 A. From here we just rephrase the part ”|fn (x) f (x)| < ✏ for all x 2 A” as supx2A |fn (x) f (x)| ✏. Now the definition above becomes: for every ✏, there is N 2 N such that supx2A |fn (x) f (x)| ✏ whenever n N. This is almost the same as the definition of limit, and indeed implies the definition of limit: lim sup |fn (x) f (x)| = 0. n!1 x2A (() Note that the arguments in ()) are equivalent deductions, hence assuming the last limit equation, we can reverse the deductions to get that (fn ) converges to f uniformly on A. (Please redo this part in detail to make sure you understand the logic.) Remark. Theorem 1.1.18 provides a natural criterion to measure how different two functions can be. The quantity supx2A |f (x) g(x)| is 0 if and only if f and g are the same function on A. Uniform convergence thus give rise to a nice way to define the distance between different functions. 1.2 Uniform Convergence and Differentiability In this section we study the relation between uniform convergence and differentiability of the limit function. We might guess that similar to Theorem 1.1.13, if a sequence of differentiable functions (fn ) converges to f uniformly, then f is also differentiable. However, it turns out that uniform convergence does not guarantee differentiability of the limit function, partly because derivative of function is less well-behaved than the function itself. Let us start with an example. Example 1.2.1. Let (fn ) be defined as: fn (x) = r x2 + 1 n x2R Math 142B Lecture Notes (v. 2021/08/24) Yucheng Tu 10 The pointwise limit of (fn ) is f (x) = lim fn (x) = lim n!1 n!1 r x2 + p 1 = x2 n which is actually the absolute function |x|. It requires some algebraic to show p p technique b that (fn ) converges uniformly to f on R. (Hint: try to use the formula a b = paa+p to b manipulate |fn (x) f (x)|) On the other hand, each fn is a differentiable function at x = 0. To see this we compute q q x2 + n1 02 + n1 x2 + n1 02 + n1 x q q lim = lim ⇣q ⌘ = lim q x!0 x!0 x!0 x 0 1 1 2 2 2 x x + + 0 + x +1+ 1 n n n n p As x ! 0, the numerator goes to 0, but the denominator is greater than 2 1/n, a positive constant. Hence we can see the limit is 0, i.e. fn0 (0) = 0 for each n. However, f (x) = |x| is simply not differentiable at x = 0. From this example we can see that differentiability can get lost even when the convergence is uniform. Exercise 1.2.2. Complete the proof that fn uniformly converges to f on R. Did we miss any essential assumption here? It is a basic understanding that derivative controls how the function changes, so maybe we should start from the uniform convergence of the derivative fn0 instead of uniform convergence of fn . Here is a nice theorem that verifies our idea. Theorem 1.2.3. Let (fn ) be a sequence of differentiable functions on an open interval (a, b). Suppose fn0 converges uniformly to g on (a, b), and fn has a pointwise limit f , then f 0 = g. Remark. Since the definition of derivative is not written in ✏- language in the textbook, here we assert the equivalence: f 0 (c) = L () lim x!c f (x) x f (c) =L c () 8✏ > 0, 9 > 0, such that f (x) x f (c) c L < ✏ whenever |x c| < . The second equivalence follows from the equivalent definition of functional limit, see Definition 4.2.1B on page 116 of the textbook. Proof. Let us fix an arbitrary c 2 (a, b) and try to prove that f 0 (c) = g(c). By the definition of derivative, we need to show: 8✏ > 0, 9 > 0 such that f (x) f (c) x c g(c) < ✏ whenever |x c| < . (version 2021/08/24) 11 Like the proof of Theorem 1.1.13, we need to use fn as a bridge to use the triangleinequality trick. Unlike the previous proof, as derivative fn0 is involved, we need another ingredient: the Mean Value Theorem for derivatives. First we split f (x)x fc (c) g(c) into three parts, for some fixed N to be chosen later: h f (x) i ⇥ ⇤ f (c) fN (x) fN (c) i h fN (x) fN (c) 0 0 + fN (c) + fN (c) g(c) x c x c x c f (x) f (c) fN (x) fN (c) fN (x) fN (c) 0 0 + fN (c) + fN (c) g(c) x c x c x c (1.2) (1.3) (1) Let us deal with the first term. By pointwise convergence of (fn ) to f , we have f (x) x f (c) c fN (x) x fN (c) fn (x) fn (c) fN (x) fN (c) = lim n!1 c x c x c [fn (x) fN (x)] [fn (c) fN (c)] = lim n!1 x c (1.4) (1.5) Now we can apply the MVT for derivatives to the function fn fN : there is y in between c and x such that [fn (x) fN (x)] [fn (c) fN (c)] = (fn fN )0 (y) x c The uniform convergence of (fn0 ) implies that (fn0 ) is a uniformly Cauchy sequence, therefore 0 given ✏ > 0, one can choose N1 such that |fn0 (y) fN (y)| < ✏/3 for any y 2 (a, b) and 1 n N1 . After taking the limit n ! 1 in equation 1.4, we have f (x) x f (c) c fN1 (x) x fN1 (c

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