MATH 401 - NOTES Sequences of functions Pointwise and Uniform Convergence Fall 2005 Previously, we have studied sequences of real numbers. Now we discuss the topic of sequences of real valued functions. A sequence of functions {fn} is a list of functions (f1, f2,...) such that each fn maps a given subset D of R into R. I. Pointwise convergence Definition. Let D be a subset of R and let {fn} be a sequence of functions defined on D. We say that {fn} converges pointwise on D if lim fn(x) exists for each point x in D. n→∞ This means that lim fn(x) is a real number that depends only on x. n→∞ If {fn} is pointwise convergent then the function defined by f(x) = lim fn(x), n→∞ for every x in D, is called the pointwise limit of the sequence {fn}. Example 1. Let {fn} be the sequence of functions on R defined by fn(x) = nx. This sequence does not converge pointwise on because lim fn(x) = ∞ R n→∞ for any x > 0. Example 2. Let {fn} be the sequence of functions on R defined by fn(x) = x/n. This sequence converges pointwise to the zero function on R. Example 3. Consider the sequence {fn} of functions defined by nx + x2 f (x) = for all x in . n n2 R Show that {fn} converges pointwise. Solution: For every real number x, we have: 2 x x 1 2 1 lim fn(x) = lim + = x lim + x lim = 0 + 0 = 0 n→∞ n→∞ n n2 n→∞ n n→∞ n2 1 Thus, {fn} converges pointwise to the zero function on R. Example 4. Consider the sequence {fn} of functions defined by sin(nx + 3) fn(x) = √ for all x in R. n + 1 Show that {fn} converges pointwise. Solution: For every x in R, we have −1 sin(nx + 3) 1 √ ≤ √ ≤ √ n + 1 n + 1 n + 1 Moreover, 1 lim √ = 0. n→∞ n + 1 Applying the squeeze theorem for sequences, we obtain that lim fn(x) = 0 for all x in . n→∞ R Therefore, {fn} converges pointwise to the function f ≡ 0 on R. Example 5. Consider the sequence {fn} of functions defined by fn(x) = 2 n n x for 0 ≤ x ≤ 1. Determine whether {fn} is pointwise convergent. Solution: First of all, observe that fn(0) = 0 for every n in N. So the sequence {fn(0)} is constant and converges to zero. Now suppose 0 < x < 1 then n2xn = n2en ln(x). But ln(x) < 0 when 0 < x < 1, it follows that lim fn(x) = 0 for 0 < x < 1 n→∞ 2 Finally, fn(1) = n for all n. So, lim fn(1) = ∞. Therefore, {fn} is not n→∞ pointwise convergent on [0, 1]. n Example 6. Let {fn} be the sequence of functions defined by fn(x) = cos (x) for −π/2 ≤ x ≤ π/2. Discuss the pointwise convergence of the sequence. Solution: For −π/2 ≤ x < 0 and for 0 < x ≤ π/2, we have 0 ≤ cos(x) < 1. 2 It follows that lim (cos(x))n = 0 for x 6= 0. n→∞ Moreover, since fn(0) = 1 for all n in , one gets lim fn(0) = 1. Therefore, N n→∞ {fn} converges pointwise to the function f defined by 0 if − π ≤ x < 0 or 0 < x ≤ π f(x) = 2 2 1 if x = 0 Example 7. Consider the sequence of functions defined by n fn(x) = nx(1 − x) on [0, 1]. Show that {fn} converges pointwise to the zero function. Solution: Note that fn(0) = fn(1) = 0, for all n ∈ N. Now suppose 0 < x < 1, then n ln(1−x) n ln(1−x) lim fn(x) = lim nxe = x lim ne = 0 n→∞ n→∞ n→∞ because ln(1 − x) < 0 when 0 < x < 1. Therefore, the given sequence con- verges pointwise to zero. Example 8. Let {fn} be the sequence of functions on R defined by n3 if 0 < x ≤ 1 f (x) = n n 1 otherwise Show that {fn} converges pointwise to the constant function f = 1 on R. Solution: For any x in R there is a natural number N such that x does not belong to the interval (0, 1/N). The intervals (0, 1/n) get smaller as n → ∞. Therefore, fn(x) = 1 for all n > N. Hence, lim fn(x) = 1 for all x. n→∞ The formal definition of pointwise convergence Let D be a subset of R and let {fn} be a sequence of real valued functions defined on D. Then {fn} converges pointwise to f if given any x in D and given any ε > 0, there exists a natural number N = N(x, ε) such that |fn(x) − f(x)| < ε for every n > N. Note: The notation N = N(x, ε) means that the natural number N depends on the choice of x and ε. 3 I. Uniform convergence Definition. Let D be a subset of R and let {fn} be a sequence of real valued functions defined on D. Then {fn} converges uniformly to f if given any ε > 0, there exists a natural number N = N(ε) such that |fn(x) − f(x)| < ε for every n > N and for every x in D. Note: In the above definition the natural number N depends only on ε. Therefore, uniform convergence implies pointwise convergence. But the con- verse is false as we can see from the following counter-example. Example 9. Let {fn} be the sequence of functions on (0, ∞) defined by nx f (x) = . n 1 + n2x2 This function converges pointwise to zero. Indeed, (1 + n2x2) ∼ n2x2 as n gets larger and larger. So, nx 1 1 lim fn(x) = lim = lim = 0. n→∞ n→∞ n2x2 x n→∞ n But for any ε < 1/2, we have 1 1 1 fn − f = − 0 > ε. n n 2 Hence {fn} is not uniformly convergent. Theorem. Let D be a subset of R and let {fn} be a sequence of continuous functions on D which converges uniformly to f on D. Then f is continuous on D 4 Homework Problem 1. 4 n Let {fn} be the sequence of functions on [0, 1] defined by fn(x) = nx(1−x ) . Show that {fn} converges pointwise. Find its pointwise limit. Problem 2. 1 Is the sequence of functions on [0, 1) defined by fn(x) = (1 − x) n pointwise convergent? Justify your answer. Problem 3. Consider the sequence {fn} of functions defined by n + cos(nx) f (x) = for all x in . n 2n + 1 R Show that {fn} is pointwise convergent. Find its pointwise limit. Problem 4. n Consider the sequence {fn} of functions defined on [0, π] by fn(x) = sin (x). Show that {fn} converges pointwise. Find its pointwise limit. Using the above theorem, show that {fn} is not uniformly convergent. 5.