7

UNIFORM CONVERGENCE AND APPLICATIONS

7.1POINTWISE AND UNIFORM CONVERGENCE

We have encountered several operations that are performed on functions or sequences of functions, including differentiation, integration, and the taking of limits. Often, two operations are performed in sequence, and it may make a difference in which order the operations are carried out. We illustrate this point with some examples.

7.1.1Examples of Invalid Interchange of Operations

Let

For each fixed x, ƒ_n(x) → ƒ(x) as n → ∞, where ƒ(x) = 0 for x ≤ 0, and f(x) = 1 for x > 0 (see Fig. 7.1.1). We consider the operations (on f_n(x)) of taking the limit as x → 0 and taking the limit as n → ∞.

Figure 7.1.1 Example 7.1.1(a)

We obtain

But

which does not exist, since f(x) → 1 as x → 0⁺, and f(x) → 0 as x → 0⁻. Note also that each ƒ_n is continuous everywhere, and f_n converges to ƒ pointwise; that is, f_n(x) → f(x) for each x, but ƒ is discontinuous at x =0.

(b) Let

In this case, f_n → ƒ pointwise, where f(x) = 0 for all x. We consider the operations of integration and taking the limit as n → ∞. We obtain

but

Thus, although f_n → f pointwise, ; in other words, the limit of the integral of ƒ_n is not the integral of the limit of ƒ_n.

(c) Let f_n(x) = (l/n)sin nx; again ƒ_n → ƒ pointwise, where ƒ ≡ 0. In this case we consider differentiation and the limit as n → ∞. We have

which does not exist unless x is an integer multiple of 2π. But

Thus, f_n → ƒ pointwise, but that is, the limit of the derivative of f_n is not the derivative of the limit.

Pointwise convergence of f_n to ƒ may be visualized geometrically via a vertical line test. For a fixed x, the vertical line through (x, 0) travels a distance |ƒ_n(x) − ƒ(x)| between the graphs of ƒ_n and ƒ at x. Equivalently, we can sketch |f_n − f|; the vertical line will then travel between 0 and |f_n(x) − f(x)|. This distance must approach 0 as n → ∞. See Fig. 7.1.2 for the case f_n(x) = ne^−nx, x > 0, f(x) = 0.

The idea of uniform convergence may be illustrated by a horizontal line test. For each ∊ > 0, we draw the horizontal line through (0, ∊) and ask whether the entire graph of |ƒ_n − ƒ| will lie below the line for all sufficiently large n. If this is possible for all ∊ > 0, then ƒ_n → ƒ uniformly. In the example illustrated in Fig. 7.1.2 we have f_n(x) → n as x → 0, so there is no way to squash the entire graph of |f_n − ƒ| below ∊; see Fig. 7.1.3.

7.1.2Definitions

Let ƒ, f₁, f₂, ... be real-valued functions defined on the arbitrary set E. We say that f_n converges to ƒ pointwise on E if for each x ∈ E, f_n(x) → f(x) as n → ∞. We say that f_n converges to ƒ uniformly on E if

Figure 7.1.2 Vertical Line Test for Pointwise Convergence

In other words, given ∊ > 0 there is a positive integer N (depending only on ∊) such that for all x ∈ E and all n ≥ N, |f_n(x) − f(x)| < ∊. Geometrically, the entire graph of |f_n − ƒ| is squashed below ∊ for all sufficiently large n.

Figure 7.1.3 Horizontal Line Test for Uniform Convergence

7.1.3Example

Let f_n(x) = n²x(1 − x)ⁿ, 0 ≤ x ≤ 1; f_n converges pointwise to the zero function ƒ. To check for uniform convergence, we will compute sup_0≤x≤1 |f_n(x) − ƒ(x)|; note that the sup is actually a maximum by continuity of |f_n − ƒ|. If we take the derivative of n²x(1 − x)ⁿ and set the result equal to 0, we obtain x = 1/(n + 1); hence,

Since n/n + 1 → 1, n²/n + 1 → ∞, and (1 + r/m)^m → e^r as m → ∞ for any r, the above expression approaches infinity as n → ∞. Thus, f_n does not converge uniformly to ƒ.

We have already noted (see Fig. 7.1.2) that in Example 7.1.1(b), ƒ_n does not converge uniformly to ƒ. A similar analysis using a horizontal line test shows that the conclusion is the same in Example 7.1.1(a). In Example 7.1.1(c), f_n does converge uniformly to ƒ (since |f_n(x)| ≤ 1/n → 0), but this is not enough to ensure that df_n/dx → df/dx.

Problems for Section 7.1

    1.In Example 7.1.1(a), cut the domain of the ƒ_n down to (0, ∞), and evaluate lim_x→0 lim_n→∞ ƒ_n(x) and lim_n→∞ lim_x→0 f_n(x).

    2.Let ƒ_n(x) = xⁿ/(2 + 3xⁿ), 0 ≤ x ≤ 1. Show that ƒ_n converges point-wise, and determine whether the sequence converges uniformly on [0, 1].

    3.Let f_n(x) = xⁿ/(n + xⁿ), 0 ≤ x ≤ 1. Show that ƒ_n converges point-wise, and determine whether or not the sequence converges uniformly on [0, 1].

    4.Show that the sequence of Problem 3 converges uniformly on [1+δ, ∞) for any δ > 0, but does not converge uniformly on [1, ∞).

    5.The sequence f_n(x) = xⁿe^−nx converges pointwise to 0 on [0, ∞). Does the sequence converge uniformly? Explain.

7.2UNIFORM CONVERGENCE AND LIMIT OPERATIONS

We now discuss the relation between uniform convergence and interchange of limit operations.

7.2.1THEOREM. A uniform limit of continuous functions is continuous. That is, if E is a subset of a metric space, f₁, f₂,·... are continuous real-valued functions on E, and f_n → ƒ uniformly on E, then f is continuous on E. In particular, if x₀ ∈ E,

    Proof. If x₀ ∈ E, then

Given ∊ > 0, the uniform convergence of ƒ_n to ƒ allows us to find a positive integer N such that |f(x) − f_n(x)| < ∊/3 for all n ≥ N and all x ∈ E; set x = x₀ to obtain |f_n(x₀) − ƒ(x₀)| < ∊/3· Fix n ≥ N; since f_n is continuous at x₀, there is a δ > 0 such that |f_n(x) − f_n(x₀)| < ∊/3 whenever d(x, x₀) < δ. Thus, d(x, x₀) < δ implies |f(x) − f_n(x)| < ∊. ■

The next two results relate uniform convergence to interchange of the limit operation with integration and differentiation. The results can be proved in somewhat greater generality, but we have strengthened the hypotheses in order to simplify the proofs.

7.2.2THEOREM. Let f₁, f₂, ... be continuous real-valued functions on [a, b], and let α be of bounded variation on [a, b]. If ƒ_n → ƒ uniformly on [a, b], then

    Proof. By Theorem 7.2.1, ƒ is continuous on [a,b], so exists. Now

since f_n → ƒ uniformly. ■

7.2.3 THEOREM. Assume that for each n = 1, 2..., f_n has a continuous derivative on [a, b], and that for some x₀ ∈ [a, b], f_n(x₀) converges to a finite limit. If converges uniformly on [a, b], then f_n converges uniformly on [a, b] to a limit function f, and for every x ∈ [a, b].

The purpose of the hypothesis that f_n(x₀) converges to a finite limit is to exclude cases such as f_n(x) = c_n for all x, where which is identically 0, converges uniformly, but ƒ_n, need not converge uniformly.

    Proof. Suppose uniformly on [a, b]. Then for a ≤ x ≤ b we have

Since f_n(x₀) converges by hypothesis and f_n(x) − f_n(x₀) converges by the above observation, f_n(x) converges; call the limit f(x). Let n → ∞ in (1) to obtain

By Theorem 6.2.4(a), f′(x) = g(x) on [a, b], so uniformly on [a, b]. To establish uniform convergence of ƒ_n, note that

The following result is occasionally useful in establishing uniform convergence.

7.2.4LEMMA. Suppose the real-valued functions f₁, f₂, ... have the uniform Cauchy property on E; that is, given ∊ > 0 there is a positive integer N (depending only on ∊) such that for n, m ≥ N we have

Then the sequence {ƒ_n} converges uniformly on E.

    Proof. For each x ∈ E, {f_n(x)} is a Cauchy sequence of real numbers; hence, f_n(x) converges to a limit ƒ(x). If |f_n(x) − f_n(x)| < ∊ for all n, m ≥ N and all x ∈ E, fix n and let m → ∞ to conclude that |f_n(x) − f(x)| ≤ ∊ for all n ≥ N and all x ∈ E. Thus, f_n → ƒ uniformly on E. ■

We have seen that a uniform limit of continuous function is continuous. Under certain conditions, pointwise convergence of a sequence of continuous functions to a continuous limit implies uniform convergence.

7.2.5Dini's Theorem

Let f₁, f₂,... be continuous real-valued functions on the compact set E, and assume the f_n form a monotone sequence (either f_n+1(x) ≤ f_n(x) for all x ∈ E and all n = 1, 2, ..., or f_n+1(x) ≥ f_n(x)for all x ∈ E and all n = 1,2, ... ). Iƒ ƒ_n → ƒ pointwise on E and f is continuous on E, then f_n → f uniformly on E.

    Proof. We may assume that the f_n form a decreasing sequence (in the increasing case, simply consider the functions − ƒ_n, which decrease). If g_n = ƒ_n − f, the g_n form a decreasing sequence of nonnegative continuous functions converging pointwise to 0. If ∊ > 0, let V_n = {x ∈ E : g_n(x) < ∊}, an open set by continuity of g_n. If x ∈ E, then g_n(x) < ∊ eventually; hence, . Since E is compact, for some N. But since the g_n decrease, we have V_n ⊆ V_n+1 for all n, and therefore . Thus, if x ∈ E, then x ∈ V_n; that is, g_N(x) < ∊. If n ≥ N, we have 0 ≤ g_n(x) < g_N(x) < ∊· Thus, g_n → 0 uniformly on E. ■

Problems for Section 7.2

    1.In Theorem 7.2.3, drop the hypothesis that for some x₀ ∈ [a.b], f_n(x₀) converges to a finite limit. Show that the result no longer holds.

    2.If {f_n} and {g_n} converge uniformly on E, show that {ƒ_n + g_n} converges uniformly on E.

    3.Give an example in which {ƒ_n } and {g_n } converge uniformly on E, but {f_ng_n} fails to converge uniformly on E.

    4.If {ƒ_n} converges uniformly on E and each ƒ_n is bounded (i.e., for some M_n > 0, |f_n(x)| ≤ M_n for every x ∈ E), show that the ƒ_n are uniformly bounded on E; that is, for some M > 0, |f_n(x)| ≤ M for all x ∈ E and all n = 1, 2, ...

    5.In Problem 4, drop the assumption that each ƒ_n is bounded, and show that the result no longer holds.

7.3THE WEIERSTRASS M-TEST AND APPLICATIONS

Uniform convergence of a series of functions means, by definition, uniform convergence of the sequence of nth partial sums Most of the time it is difficult to evaluate the sum of an infinite series, so it is useful to have sufficient conditions for uniform convergence that involve only the functions f_n. The following result is widely used.

7.3.1Weierstrass M-Test

Let f₁, f₂, … be real–valued functions on the set E. If |f_n(x)| ≤ M_n for all x ∈ E and all n = 1, 2, ..., where , then the series converges uniformly on E. Thus, if each f_n is continuous on E, then (by Theorem 7.2.1) continuous on E.

    Proof. If s_n is the nth partial sum of the series, then for m < n,

since . Thus, the sequence {s_n} has the uniform Cauchy property and therefore converges uniformly by Lemma 7.2.4. ■

7.3.2Example

Let f_n(x) = (sin nx)/n², x ∈ R. Then |f_n(x)| ≤ M_n = 1/n² for all x, and it follows that converges uniformly on R. (Unfortunately, the Weierstrass M-test does not help us find the limit function explicitly.)

We now use the Weierstrass M-test to produce a rather spectacular application of uniform convergence.

7.3.3An Everywhere Continuous, Nowhere Differentiate Function

Let f(x) = |x|, −1 ≤ x ≤ 1, and extend f periodically to all of R. In other words, the extension is required to satisfy f(x + 2) = f(x) for all x. The resulting “sawtooth” curve is sketched in Fig. 7.3.1. Now if x and y both belong to the same branch (linear segment) of the curve, then |f(y) − f(x)| = |y − x|. In general, as shown in Fig. 7.3.2,

Figure 7.3.1 f(x) = |x|, Extended Periodically

Fix the point x ∈ R and the positive integer m. Let , where the choice of + or − will be decided in a moment. Since , either there will be no integer between and 4^mx or there will be no integer between 4^mx and see Fig. 7.3.3. We choose + or − so that no integer will lie between 4^mx and 4^m(x + h_m); in Fig. 7.3.3 we would choose +. Let

We claim that

For if , an even integer. Since f has period 2, c_n = 0. If n < m, (1) gives |c_n| ≤ 4ⁿ|h_m|/|h_m| = 4ⁿ. If n = m, the fact that no integer lies between 4^mx and 4^m(x + h_m) means that these two points belong to the same branch of f, so |c_m| = |4^m(x + h_m) − 4^mx|/|h_m| = 4^m, as desired.

Figure 7.3.2 An Inequality for the Sawtooth Function

Figure 7.3.3 Determining the Choice of h_m

We are now ready for the construction of the desired function. Define

(Intuitively, as more terms are added to the sum, more “spikes” are created; see Fig. 7.3.4.) Since |f| ≤ 1 and is continuous on R by the Weierstrass M-test and Theorem 7.2.1. If x is any real number, we show that g is not differentiable at x. If m is any positive integer,

Figure 7.3.4 Constructing an Everywhere Continuous, Nowhere Differentiable Function

Now

and

hence,

Since h_m → 0 as m → ∞, it follows that g is not differentiable at x.

Problems for Section 7.3

    1.Show that the series converges uniformly on [a, ∞) for any a > 0.

    2.Show that does not converge uniformly on (0, ∞).

    3.Suppose the power series converges at x = r. If |a| < |r|, show that the series converges absolutely at x = a.

    4.Continuing Problem 3, show that if 0 < a < |r|, then the series converges uniformly on [−a, a],

    5.Continuing Problem 4, show that if the power series has radius of convergence r, the series may be integrated and differentiated term by term (as if it were an ordinary polynomial) for −r < x < r.

    6.If f₁, f₂, … are continuous real-valued functions on [a, b] and converges uniformly on [a, b], show that for any a of bounded variation on [a, b], we have

7.4EQUICONTINU1TY AND THE ARZELA-ASCOU THEOREM

We know that any bounded sequence of real numbers has a subsequence converging to a limit in R. A similar question will now be raised for sequences of functions. When will there exist a uniformly convergent subsequence? First, we prove some preliminary results.

7.4.1 LEMMA. Let A be a subset of a metric space with the property that every open covering of A has a countable subcovering. (Every subset of R^p has this property; see the proof of Theorem 2.2.5.) Then A has a countable dense subset; in other words, there is a countable set E ⊆ A such that each x ∈ A can be expressed as a limit of a sequence of points in E. Conversely, if A has a countable dense subset, then every open covering of A has a countable subcovering.

    Proof. For each n = 1, 2, …, we have hence, A is covered by a countable union of balls of radius 1/n and centers in A, say . Let E = {x_nj : n,j = 1, 2, …}. If ∊ > 0 and x ∈ A, choose n so large that 1/n < ∊. Since x ∈ B_1/n(x_nj) for some j, we have d(x, x_nj) < ∊. It follows that E is dense.

If A has a countable dense subset E, we may reproduce the first part of the proof of the Heine-Borel Theorem 2.2.5, using E instead of the rationals in R^p, to show that every open covering of A has a countable subcovering. ■

If A is an interval of reals, we may take E to consist of the rational numbers in A. But this idea does not work for an arbitrary set A ⊆ R. If, for example, A is the set of irrationals, the set of rationals of A is empty, so is certainly not dense. However, the procedure given in the proof of Lemma 7.4.1 can be used for an arbitrary A ⊆ R^p.

7.4.2 LEMMA. Let {f_n} be a sequence of real-valued functions on the countable set E = {x₁, x₂, …}, and assume that f or each j = 1, 2, …, {f_n(x_j) : n = 1, 2, …} is bounded. Then {f_n} has a subsequence that converges (to a finite limit) at each point of E.

    Proof. The sequence f₁(x), f₂(x₁), f₃(x₁), … is bounded; hence, by Theorem 2.3.2 there is a subsequence, call it {f₁₁, f₁₂, f₁₃, …}, of {f_n} such that f_1n(x₁) converges as n → ∞. Now {f₁₁(x₂), f₁₂(x₂), f₁₃(x₂), …} is bounded, so there is a subsequence f₂₁, f₂₂, f₂₃, … of {f_1n} such that f_2n(x₂) converges as n → ∞. Continue inductively in this fashion; the results may be summarized by the following array:

We form the diagonal subsequence g_n = f_nn, n = 1, 2, … To check that we have a legal subsequence, note that row k + 1 of the above array is a subsequence of row k, and therefore f_{k + 1,k + 1} appears in row k somewhere to the right of f_kk. Similarly, f_{k + 2,k + 2} appears in row k to the right of f_{k + 1,k + 1}.·Thus, {gn, n ≥ k} is a subsequence of {f_kj, j = 1, 2, …}; in particular, {g_n, n ≥ 1} is a subsequence of {f_n}. Furthermore, since f_kj(x_k) converges as j → ∞ g_n (x_k) converges as n → ∞ k = 1, 2, …). Thus, {g_n} converges pointwise on E. ■

Recall that the function f: Ω → Ω′ is uniformly continuous on Ω if for every ∊ > 0 there is a δ > 0 such that whenever x, y ∈ Ω and d(x, y) < δ, we have d(f(x), f(y)) < ∊; δ depends only on ∊ but not on the particular x and y. We now consider a generalization of this concept.

7.4.3Definition

Let {f_i} be an arbitrary (possibly uncountable) family of functions from the metric space Ω to the metric space Ω′. If E ⊆ Ω, {f_i} is said to be equicontinuous on E if for every ∊ > 0 there is a δ > 0 such that for all x, y ∈ E and all i,

Thus, not only is δ independent of the points x and y, but δ does not depend on the particular f_i in the family either.

The central result about equicontinuity is the following.

7.4.4Arzela-Ascoli Theorem

Let {f_n} be a sequence of real-valued functions that is equicontinuous on the compact set K. If the sequence is pointwise bounded (i.e., sup_n |f_n(x)| < ∞ for each x ∈ K), then there is a subsequence converging uniformly on K. Furthermore, the sequence {f_n} is uniformly bounded; in other words, sup_n,x |f_n(x)| < ∞.

    Proof. By Lemma 7.4.1, K has a countable dense subset E. Since {f_n} is pointwise bounded, the hypothesis of Lemma 7.4.2 is satisfied; hence, there is a subsequence {g_n} converging pointwise on E; we must show that {g_n} converges uniformly on K. Given ∊ > 0, choose δ > 0 so that if x, y ∊ K and d(x,y) < δ then d(f_n(x), f_n(y)) < ∊/3 for all n. If y ∈ K, then since E is dense there is a point x ∈ E such that d(x, y) < δ; that is, y ∈ Β_δ(x). It follows that , and therefore, by compactness, for some x₁, …, x_p ∈ E. Since g_n (x_i) converges as n → ∞ for each i, there is a positive integer N such that |g_n{x_i) − g_m(x_i)| < ∊/3 for all n, m ≥ N and all i = 1, …, p. If x ∈ K, then x ∈ B_δ(x_i) for some i; thus,

The first and third terms on the right side are less than ∊/3 by equicontinuity, and the second term is less than ∊/3 provided n, m ≥ N. It follows that |g_n(x) − g_m(x)| < ∊ for all n, m ≥ N and all x ∈ K. By Lemma 7.2.4, {g_n} converges uniformly on K.

To prove uniform boundedness, let h(x) = sup_n |f_n(x)|. If x, y ∈ K and d(x, y) < δ, we have, for each n = 1, 2, …,

Take the sup over n to obtain h(y) ≤ h(x) + ≥ ∊; by symmetry, h(x) ≤ h(y) + ∊; hence, |h(x) − h(y)| < ∊. Thus, h is a continuous function on the compact set K, so h is bounded; in other words, sup_n,x |f_n(x)| < ∞, as desired. ■

If the functions f_n are not real-valued but take values in an arbitrary metric space, the conclusion of Theorem 7.4.4 still holds if the point-wise boundedness hypothesis is replaced by the assumption that for each x ∈ K, {f_n(x) : n = 1, 2, …} is a subset of a compact set, and hence the sequence f₁(x), f₂(x), … has a convergent subsequence. This allows the construction given in the proof of Lemma 7.4.2 to be retained, and the proof of Theorem 7.4.4 to be essentially reproduced.

Problems for Section 7.4

    1.Give an example of a sequence of real-valued functions f_n on a countable set E = {x₁, x₂, …} such that {f_n} is pointwise bounded but not uniformly bounded.

    2.Define, for each n = 1, 2, …,

            Take f_n(0) = 0 for all n.

        (a)Sketch f_n(x), 0 ≤ x ≤ 1, for n = 1, 2.

        (b)Show that {f_n} is uniformly bounded and converges pointwise on a countable dense subset of [0, 1], but does not converge pointwise on [0, 1].

    3.Let {f_n} be an equicontinuous sequence of functions on the compact set K. If f_n → f pointwise on K, show that f_n → f uniformly on K.

    4.Let f₁, f₂, … be continuous functions on the compact set K, and assume f_n → f uniformly on K.

        (a)Show that {f_n} is equicontinuous on K.

        (b)If x₁, x₂, … ∈ K and x_n → x, show that f_n(x_n) → f(x)·

        (c)Now remove the hypothesis that K is compact. Show that (a) no longer holds, but (b) is still valid.

7.5THE WEIERSTRASS APPROXIMATION THEOREM

We are going to investigate the problem of approximating a continuous function by polynomials. The main result, the Weierstrass Approximation Theorem, is very well motivated by considerations of basic probability theory. As we go along, we will make parenthetical comments that should be helpful for those who have some familiarity with probability. The proofs, however, do not use probability considerations. We need the following two results.

7.5.1 LEMMA. If 0 ≤ x ≤ 1,

where

(If X is the number of heads in n independent tosses of a coin with probability of heads x on a given toss, E stands for expectation or average value, and Var indicates variance, the above expression is

    Proof. By the binomial theorem,

Also,

Finally, since k² = k(k − 1) + k,

Thus,

7.5.2 LEMMA. Let x ∈ [0, 1], and let n be a positive integer. If S is the set of integers k ∈ {0, 1, …, n} such that |x − k /n| ≥ n^−1/4, then

(If X is the random variable given in Lemma 7.5.1, then the probability that |X/n − x| ≤ n^−1/4 is, by Chebyshev’s inequality, ≤ E[(X/n − x)²]/n^−1/2 = x(l − x)/nn^−1/2, which is at most .)

    Proof. The sum to be estimated may be written as

But x (1 − x) is maximum at x = , and the result follows. ■

We are now ready for the main result.

7.5.3Weierstrass Approximation Theorem

Let f be a continuous real-valued function on [0, 1]. Define the Bernstein polynomials for f by

Then B_n → f uniformly on [0, 1], so that f can be uniformly approximated by polynomials.

(If X is the random variable given in Lemma 7.5.1, then B_n(x) = E[f(X/n)]. For large n, X/n will (with high probability) be close to x by the Law of Large Numbers. Since f is continuous, f(X/n) should be close to f(x). This suggests convergence in some sense of B_n to f.)

    Proof. By Theorem 4.2.4, f is uniformly continuous on [0, 1], Thus, given ∊ > 0, there is a δ > 0 such that |x − y| < δ implies |f(x) − f(y)| < ∊/2. By Corollary 4.2.2, f is bounded, say |f| ≤ M. Now

Thus,

If we sum only over those k for which |x − k/n| ≥ n^−1/4, the result is at most by Lemma 7.5.2. This can be made less than ∊/2 provided n > M²/∊². If |x − k/n| < n^−1/4 and n^−1/4 < δ (in other words, n > δ⁻⁴), then |x − k/n| < δ so | f (x) − f(k/n)| will be less than ∊/2. Thus, if we sum over those k for which |x − k/n| < n^−1/4, the result is at most ∊/2. It follows that if n > max(M²/∊², δ⁻⁴), we have |f(x) − B_n(x)| < ∊ for all x ∈ [0, 1]. ■

There is no difficulty in approximating a continuous real-valued function on an arbitrary closed bounded interval [a, b] by polynomials; see Problem 1.

The Weierstrass approximation theorem has been generalized to the following result, known as the Stone-Weierstrass theorem.

Let A be an algebra of continuous real-valued functions on the compact set K. (“Algebra” means that if f, g ∈ A then f + g ∈ A and f g ∈ A; also, if f ∈ A and c ∈ R, then cf ∈ A.) Assume that A contains all constant functions and separates points; that is, if x, y ∈ Κ, x ≠ y, there is an f ∈ A with f(x) ≠ f(y). Then for any continuous f : K → R and any ∊ > 0 there is a g ∈ A such that |f(x) − g(x)| < ∊ for all x ∈ K. Thus, f can be uniformly approximated by functions in A. (In the case of the Weierstrass approximation theorem, K = [0, 1] and A is the set of all polynomials.)

Problems for Section 7.5

    1.Show that a continuous real-valued function f on [a, b] can be uniformly approximated by polynomials.

    The following problems constitute a project involving the reversal of the order of summation of a double series.

    2.Give an example of a double sequence of real numbers a_nj, n,j = 1, 2, …, such that .

    3.Let x_k = 1/k, k = 1, 2, …, x₀ = 0, and form the set E = {x₀, x₁, x₂, …}.·Let {a_nj} be a double sequence of reals such that . (This is assumed in Problems 4, 5, 6, and 7 also.) Define

            Use the Weierstrass M-test to show that converges uniformly on E.

    4.Show that each f_n is continuous on E; hence, by Theorem 7.2.1, f_n is continuous on E.

    5.Show that for all k ≥ 1,

    6.Show that

    7.Finally, show that if , then

            (Note that the same result holds under the hypothesis that to see this, simply set b_jn = a_nj.)

    8.Let {a_nj} be a nonnegative double sequence of real numbers. Define

            Show that

Thus, the order of summation in a nonnegative double series may always be reversed. If the series diverges to ∞ when summed in one order, it will diverge to ∞ in the other order as well.

(If s is finite, the result follows from Problem 7, but a general argument not based on Problem 7 may be given.)

    9.Suppose and converge (at least) for |x| < r. Define the Cauchy product of f(x) and g(x) as

            where

            (Thus, h is obtained by multiplying the series ∑a_nxⁿ and ∑b_nxⁿ as if they are ordinary polynomials.) Show that converges to f(x)g(x) for |x| < r.

REVIEW PROBLEMS FOR CHAPTER 7

    1.Let f_n(x) = (2 + xⁿ)/(3 + xⁿ), 0 ≤ x < 1. Show that the sequence {fn} converges pointwise, and determine whether the sequence converges uniformly on [0, 1).

    2.Give an example of a uniformly convergent series of real-valued functions for which the Weierstrass M-test fails; in other words, if for each n we have |f_n(x)| < M_n for all x, then

    3.Let f_n(x) = e^{−(x − n)} if x ≥ n; f_n(x) = 0 if x < n. Show that the sequence converges pointwise on R, and determine whether the sequence converges uniformly.

    4.From the power series expansion

            we obtain, by integrating term by term from 0 to x,

            Indicate why the term-by-term integration is justified.