I. Pointwise convergence


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1 MATH 40  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 {f n } is a list of functions (f, f 2,...) such that each f n maps a given subset D of R into R. I. Pointwise convergence Definition. Let D be a subset of R and let {f n } be a sequence of functions defined on D. We say that {f n } converges pointwise on D if f n(x) exists for each point x in D. This means that f n (x) is a real number that depends only on x. If {f n } is pointwise convergent then the function defined by f(x) = f n (x), for every x in D, is called the pointwise it of the sequence {f n }. Example. Let {f n } be the sequence of functions on R defined by nx. This sequence does not converge pointwise on R because for any x > 0. Example 2. Let {f n } be the sequence of functions on R defined by x/n. This sequence converges pointwise to the zero function on R. Example 3. Consider the sequence {f n } of functions defined by nx + x2 for all x in R. n 2 Show that {f n } converges pointwise. Solution: For every real number x, we have: ( ) ( f x n(x) = n + x2 n = x + x 2 2 n ) = = 0 n 2
2 Thus, {f n } converges pointwise to the zero function on R. Example 4. Consider the sequence {f n } of functions defined by sin(nx + 3) for all x in R. Show that {f n } converges pointwise. Solution: For every x in R, we have sin(nx + 3) Moreover, = 0. Applying the squeeze theorem for sequences, we obtain that f n(x) = 0 for all x in R. Therefore, {f n } converges pointwise to the function f 0 on R. Example 5. Consider the sequence {f n } of functions defined by n 2 x n for 0 x. Determine whether {f n } is pointwise convergent. Solution: First of all, observe that f n (0) = 0 for every n in N. So the sequence {f n (0)} is constant and converges to zero. Now suppose 0 < x < then n 2 x n = n 2 e n ln(x). But ln(x) < 0 when 0 < x <, it follows that f n(x) = 0 for 0 < x < Finally, f n () = n 2 for all n. So, f n () =. Therefore, {f n } is not pointwise convergent on [0, ]. Example 6. Let {f n } be the sequence of functions defined by cos n (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) <. 2
3 It follows that (cos(x))n = 0 for x 0. Moreover, since f n (0) = for all n in N, one gets f n (0) =. Therefore, {f n } converges pointwise to the function f defined by { 0 if π f(x) = x < 0 or 0 < x π 2 2 if x = 0 Example 7. Consider the sequence of functions defined by nx( x) n on [0, ]. Show that {f n } converges pointwise to the zero function. Solution: Note that f n (0) = f n () = 0, for all n N. Now suppose 0 < x <, then f n(x) = nxe n ln( x) = x ne n ln( x) = 0 because ln( x) < 0 when 0 < x <. Therefore, the given sequence converges pointwise to zero. Example 8. Let {f n } be the sequence of functions on R defined by { n 3 if 0 < x n otherwise Show that {f n } converges pointwise to the constant function f = on R. Solution: For any x in R there is a natural number N such that x does not belong to the interval (0, /N). The intervals (0, /n) get smaller as n. Therefore, for all n > N. Hence, f n(x) = for all x. The formal definition of pointwise convergence Let D be a subset of R and let {f n } be a sequence of real valued functions defined on D. Then {f n } 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 f n (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
4 I. Uniform convergence Definition. Let D be a subset of R and let {f n } be a sequence of real valued functions defined on D. Then {f n } converges uniformly to f if given any ε > 0, there exists a natural number N = N(ε) such that f n (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 converse is false as we can see from the following counterexample. Example 9. Let {f n } be the sequence of functions on (0, ) defined by nx + n 2 x 2. This function converges pointwise to zero. Indeed, ( + n 2 x 2 ) n 2 x 2 as n gets larger and larger. So, f nx n(x) = n 2 x = 2 x n = 0. But for any ε < /2, we have f n ( n ) f Hence {f n } is not uniformly convergent. ( ) = n 2 0 > ε. Theorem. Let D be a subset of R and let {f n } be a sequence of continuous functions on D which converges uniformly to f on D. Then f is continuous on D 4
5 Homework Problem. Let {f n } be the sequence of functions on [0, ] defined by nx( x 4 ) n. Show that {f n } converges pointwise. Find its pointwise it. Problem 2. Is the sequence of functions on [0, ) defined by ( x) n convergent? Justify your answer. pointwise Problem 3. Consider the sequence {f n } of functions defined by n + cos(nx) 2n + for all x in R. Show that {f n } is pointwise convergent. Find its pointwise it. Problem 4. Consider the sequence {f n } of functions defined on [0, π] by sin n (x). Show that {f n } converges pointwise. Find its pointwise it. Using the above theorem, show that {f n } is not uniformly convergent. 5
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