Sequences of Functions

Let

a sequence of real functions defined on a common interval I of ℝ. An example is

and we will write f_n(x) = sin nx. The difference with the sequences that we already studied is that the terms are no longer real numbers but function.

Deinition 8.2.1 (Pointwise convergence). Let I be a nonempty subset of ℝ. A sequence of real-valued functions {f_n}, where for each n ∈ ℕ, f_n: I → ℝ is said to converge pointwise on I to a function f iff for each x₀ ∈ I the sequence {f_n(x₀)} converges to a real number f(x₀). Thus, {f_n} converges pointwise on I to the function f iff given ε > 0 and x₀ ∈ I there is an N ∈ ℕ (depending on ε and the point x₀) such that n > N implies |f_n(x₀) − f(x₀)| < ε. The function f is called the pointwise limit of the sequence {f_n} on I. Clearly, the pointwise limit of a sequence, if it exists, it is unique.  □

Example 8.2.2. For each natural number n, define

Since {f_n(1)} is a constant sequence whose constant values is 1.

On the other hand,

Thus, the sequence of functions {f_n} converges pointwise on [0,1] to the function f defined by

$$f(x)=\{\begin{array}{ll}1,& \text{if }x=1\\ 0,& \text{if }0\le x<1.\end{array}$$

Observe that in this example we have a sequence of continuous functions that converges pointwise to a discontinuous function. ■

Example 8.2.3.Consider f_n(x) = nx, x ∈ ℝ⁺. Since lim_n→∞ f_n(x) = ∞ for any x > 0, the sequence {f_n} does not converge pointwise on ℝ⁺. Consider f_n(x) = cosⁿ x, − π/2 ≤ x ≤ π/2. Clearly, the sequence {f_n} converges pointwise to the function,

$$f(x)=\{\begin{array}{ll}0,& \text{if }x\in [-\pi /2,0)\cup (0,\pi /2]\\ 1,& \text{if }x=0.\end{array}$$

Example 8.2.4. Consider f_n(x) = n/(nx + 1), x ∈ (0, 1). Clearly, the sequence {f_n} converges pointwise to the function f(x) = 1/x, x ∈ (0, 1). Since |f_n(x)| < n, x ∈ (0, 1) each f_n is bounded on (0,1), but the pointwise limiting function f(x) = 1/x is not bounded on (0, 1). Hence, the pointwise limit function of bounded functions is not necessarily bounded.  ■

Example 8.2.5. Consider f_n(x) = xⁿ; clearly f_n converges pointwise on [0,1] to the function

$$f(x)=\{\begin{array}{ll}0,& \text{if }0\le x<1\\ 1,& \text{if }x=1.\end{array}$$

Here each f_n is continuous and differentiable on [0, 1], but f is neither differentiable nor continuous at x = 1. Hence, the pointwise limit function of continuous (differentiable) functions is not necessarily continuous (differentiable). ■

From the preceding examples it is clear that for a given sequence of functions pointwise convergence is only of limited importance, it does not guarantee that f is continuous as f_n and therefore, we need a stronger concept which we introduce now.

Definition 8.2.6. The sequence {f_n} of functions converges uniformly on I to the function f if, for any ε > 0, there is a n₀ such that

we write

Example 8.2.7. Consider f_n(x) = xⁿ and f(x) = 0. We shall show that f_n → f uniformly on [0, a] for any a < 1, and pointwise but not uniformly, on [0, 1). For 0 < a < 1, let ε > 0 and N > [ln ε/ ln a], so that x ∈ [0, a] and n > N implies that

Hence, x_n → 0 uniformly on [0, a], a < 1. Now suppose that x_n → 0 uniformly on [0, 1). Then, for ε = 1/8 there exists N ∈ ℕ such that for all n > N and x ∈ [0, 1), |x_n − 0| < 1/8. But, for x = 1 − 1/n this implies (1 − 1/n) n < 1/8, and hence as n → ∞, e −1 < 1/8, which is a contradiction. ■

Remarks. It is obvious that uniform convergence implies pointwise convergence. Further: the uniform limit, if it exists, has to be equal to the pointwise limit: We note that if {f_n} converges pointwise to f on I, and if {f_n} converges uniformly on I, then {f_n} converges uniformly to f on I. Thus, to test the uniform convergence by the above definition, it becomes necessary to find the pointwise limit function (if it exists).

Example 8.2.8. Consider f_n(x) = (sin nx)/n, x ∈ ℝ. The sequence {f_n} converges pointwise to f(x) = 0, x ∈ ℝ. Now since |f_n(x) − 0| = | sin nx|/n ≤ 1/n < ε for all x ∈ ℝ, n > 1/ε, the convergence is in fact uniform. Note that the sequence {f'_n} of derivatives f'_n(x) = cos nx does not converge pointwise on ℝ. Hence, there exists differentiable functions f_n and f such that f_n → f uniformly, but lim_{n → ∞} f'_n(x) does not exist at least for some x.  ■

From the previous example we see that in general uniform convergence does not guarantee that

$${(\underset{n\to \mathrm{\infty }}{lim}{f}_n(x))}^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{f}_n^{\prime }(x)\text{\hspace{1em} 8.2.1}$$

holds.

Thus the uniform convergnce of continous functions implies the limit function f to be also continous, on the same domain but differentiabily is not guarantee.

Theorem 8.2.9. Let {f_n} be a sequence of bounded functions which converge uniformly on I to the function f. Then, f is bounded on I.

Proof. In the definition of uniform convergence of {f_n} on I let ε = 1, so that there exists an N ∈ ℕ such that |f_n(x) − f(x)| < 1 for all x ∈ I, n ≥ N, which implies that |f(x)| ≤ 1 + |f_n(x)|. Now since each f_n is bounded on I, the result follows.  □

Theorem 8.2.10. Let {f_n} be a sequence of continuous functions on the domain I. Assume that the sequence converges uniformly on the domain I to the function f. Then f is continuous on I.

Proof. Let x₀ ∈ I. We shall show that f is continuous at x₀. Let ε > 0. Uniform convergence implies that there exists an integer N, such that if n > N, then |f_n(x) − f(x)| < ε/3 for all x ∈ I. Then in particular since f_n is continuous at x₀, there exsts a δ >, in which we have |f_n(x) − f(x₀)| < ε/3 for all x ∈ I with |x − x₀| < δ. We have that if x in I and |x − x₀| < δ

Remark 8.2.11. In Theorem 8.2.10 uniform convergence is a sufficient condition but not a necessary condition. For this, consider the sequence of continuous functions with f_n (x) = nx/(1 + n²x²), x ∈ ℝ, which converges pointwise to a continuous function f(x) = 0, x ∈ ℝ (see exercise 1). Since f_n (1/n) = 1/2, n ≥ 1 the convergence is not uniform.  ■

Theorem 8.2.12. Let {f_n} be a sequence of Riemann integrable functions on [a, b], which converge uniformly to the function f. Then, f is Riemann integrable on [a, b].

Proof. Since Riemann integrable functions on [a, b] are bounded and the convergence of the sequence {f_n} on [a, b] is uniform, from Theorem 8.2.9 it follows that the limiting function f is also bounded on [a, b].  □

Theorem 8.2.13. Let {f_n} be a sequence of Riemann integrable functions on [a, b], which converge uniformly to the function f. Then, f is Riemann integrable on [a, b], and

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_a^b{f}_n(x)\text{d}x={\int }_a^b\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)\text{d}x={\int }_a^{b}f(x)\text{d}x$$

Since the sequence {f_n} converges uniformly to f, there exists N ∈ ℕ such that for all x ∈ [a, b],

In view of Theorem 8.2.12, the function f is Riemann integrable on [a, b]. Thus, together with Theorem 7.2.1, the functions f_n, f, and f_n − f are Riemann integrable on [a, b]. Hence, from Threom 8.2.9, for all x ∈ [a, b], it follows that

$$\begin{array}{rl}|{\int }_a^b{f}_n(x)\text{d}x-{\int }_a^{b}f(x)\text{d}x|& =|{\int }_a^b[f_n(x)-f(x)]|\\ & \le {\int }_a^b|[f_n(x)-f(x)]|\text{d}x<{\int }_a^b\frac{\epsilon }{b-a}\text{d}x=\epsilon \end{array}$$

Remark 8.2.14. In Theorem 8.2.10 uniform convergence is a sufficient condition but not a necessary condition. To show this, we consider again (see Remark 8.2.11) the sequence of continuous functions f_n(x) = nx/(1 + n²x²), x ∈ [0, 1], which converges pointwise to a continuous function f(x) = 0, x ∈ [0, 1], but not uniformly to f(x) = 0, x ∈ [0, 1]. For this sequence, we have

The following result gives sufficient conditions for the equality in (Eq. 8.2.1).

Theorem 8.2.15. Let {f_n} be a sequence of differentiable functions on [a, b]. Suppose that {f_n} converges pointwise on [a, b] to f, and {f'_n} converges uniformly on [a, b] to g, then f is continuously differentiable on [a, b], further {f_n} converges uniformly to f and {f'_n} converges uniformly to f' on [a, b], i.e., lim_n→∞ f'_n(x) = f'(x) = g(x), x ∈ [a, b].

Proof. Since {f'n} convergece uniformly to g [a,b]. Theorem 8.2.13 implies that the function g is continuous on [a. b]. Thus, for each x ∈ [a,b], Theorem 8.2.14 implies that

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_a^x{f}_n^{\prime }(t)\text{d}t={\int }_a^{x}g(t)\text{d}t$$

Now, from Corollary 7.3.4, we have

which in view of lim_{n → ∞} f_n(x) = f(x) and lim_n→∞ f_n(a) = f(a) implies that

But, from The Fundamental Theorem of Calculus, we find f′(x) = g(x), x ∈ [a, b]. □

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