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 + 0 = 0 n 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
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
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 counter-example. 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
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