Chapter 7. Sequences and series of functions
7: R 16. For an equicontinuous sequence of functions on a compact set, pointwise convergence is uniform
(d : 3)
7:R17. Uniform convergence and equicontinuity for mappings into general metric spaces. (d : ?)
Like 3:R15 , this exercise asks the student to generalize a large number of results in the chapter to a more general context. As with that exercise, it is not clear whether it would be reasonable to assign this as homework, or if one did, what instructions to give; but the exercise is certainly worth thinking about.
7:R18. A uniformly convergent subsequence of a sequence of integrals. (d : 1) 7:R19. Conditions for a subset of (K ) to be compact. (d : 2. > 2:R26)
7:R25. Existence of a solution to a differential equation with initial conditions. (d : 3)
One may ask why a result on finding convergent subsequences of not-necessarily-convergent sequences should be needed to prove such a result. The method of attacking the differential equation that is described uses successive approximations; won’t that method actually converge?
In general, no! At the end of 5:R27 , Rudin noted an example of a differential equation with nonunique solution for given initial conditions. This indicates an ‘‘instability’’ in such structures, which has the consequence that when we use the method of this exercise to approximate solutions, our successive steps may approximate different solutions, so that the sequence of approximations may not converge, and we may really have to pass to a subsequence to get a solution as our limit.
Note that at the start of his Hint, where Rudin says ‘‘Let fn ...’’, you are to show that such a function exists. A few lines later, where after defining ∆n(t) he adds ‘‘except ...’’, he means that for t = xi, one defines ∆n(t) = 0. (The preceding definition is not applicable when t = xi because fn′(xi) is generally undefined; so the arbitrary value 0 is used to make ∆n(t) defined.)
7:R26. The corresponding result for a system of differential equations. (d : 3. > 7:R25)
Rudin’s hint says ‘‘Use the vector-valued version of Theorem 7.25’’. He means that you should prove such a version and use it. The vector-valued result is not hard to prove from the theorem as given.
Exercises not in Rudin:
7.6:0. Say whether each of the following statements is true or false.
(a) The set of functions {xn! n = 1, 2, ... } on [–1, 1] is uniformly bounded.
(b) The set of functions {xn! n = 1, 2, ... } on [0, 2] is uniformly bounded.
(c) Every uniformly bounded family of continuous functions on a compact set is equicontinuous.
(d) If is a family of differentiable functions on R such that the set { f′ ! f∈ } is uniformly bounded, then is equicontinuous.
7.6:1. When do boundedness at one point and equicontinuity imply pointwise boundedness?
(d : 4, 1, 4, 2, 2)
(a) Show that the following two conditions on a metric space X are equivalent:
(i) Every equicontinuous family of functions on X whose values at one point are bounded is pointwise bounded. (I.e., if is an equicontinuous family of functions on X, and if there exists x∈X such that { f (x) | f∈ } is bounded, then { f (y) | f∈ } is bounded for every y∈X.)
(ii) For every two points x , y∈X, and every δ > 0, there exist points x0, ... , xn such that x = x0, y = xn, and d(xi–1, xi) <δ for i = 1, ... , n –1.
(b) Show that X = [0,1]∪[2, 3] does not satisfy the above equivalent conditions (i) and (ii).
(c) Show that every connected metric space satisfies the above equivalent conditions.
(d) Show that if a metric space X satisfies the above equivalent conditions, then so does every dense subset of X.
...
Answers to True/False question 7.5:0. (a)F. (b)F. (c) T. (d)F.
(e) Use the results of (c) and (d) to find an open subset of R which satisfies the equivalent conditions of (a), but which is not connected.
7.6:2. Proving Example 7.20 without using a result from Chapter 11. (d : 2)
In Example 7.20, Rudin shows that the sequence of functions fn = sin n x on the interval [0, 2π] has no pointwise convergent subsequence, but to do so, he calls on a result in Chapter 11. Below, you will prove this result using only material from Chapters 1-4, and the facts that sin x is a continuous function which takes on the value 1 at x =π⁄ 2 and the value –1 at x = 3π⁄ 2, and satisfies sin (x + 2π) = sin x for all x. When we speak of an ‘‘interval [a , b]’’ below, we shall understand this to entail a < b.
(a) Show that for any interval [a , b] ⊆ R there exists an integer N such that for all n > N the function sin n x takes on both the values 1 and –1 at points of [a , b].
(b) Deduce that for any infinite set S of positive integers and any interval [a , b], there exists an n∈S, and subintervals [a′, b′] and [a′′, b′′] of [a , b], such that the function sin n x has value everywhere
≥ 1⁄2 on [a′, b′], and has value everywhere ≤ –1⁄2 on [a′′, b′′].
(c) Deduce from (b) that for any infinite set S of positive integers one can find a point x and a sequence n1 < n2 < ... in S such that the real numbers sin n1x, sin n2x, ... are alternately ≥ 1⁄2 and ≤ –1⁄2. (d) Conclude that the sequence of functions sin n x (n = 1, 2, ... ) has no pointwise convergent subsequence.
7.7. The Weierstrass Theorem, and a corollary (beginning of Rudin’s section THE STONE-WEIERSTRASS THEOREM). (pp.159-161)
Relevant exercises in Rudin:
7:R20. A continuous function on [0,1] is determined by its moments. (d : 2)
The integral on the left-hand side of the displayed equation of this exercise is called the nth moment of the function f. Add to this exercise ‘‘Deduce that if g and h are continuous functions on [0,1] such that for every n, the nth moments of g and of h are the same, then g = h’’. (This is the meaning of the title I have given this exercise.) In Rudin’s Hint for the exercise, f2(x) in the integral should be changed to | f (x)|2.
7:R22. Every integrable function is L2-approximable by polynomials. (d : 3. > 6:R12)
Suggestion: Use the exercise Rudin refers to parenthetically at the end, together with a result from this section.
7:R23. An explicit algorithm for uniformly approximating | x | by polynomials. (d : 3)
This gives a direct proof of Corollary 7.27, which is the only consequence of the Weierstrass Theorem that Rudin will use in proving the Stone-Weierstrass Theorem. (As noted above, 7.3:1 gives another proof of this result.)
Exercises not in Rudin:
7.7:0. Say whether the following statement is true or false.
(a) A complex-valued function on [ –1, 1] is continuous if and only if it can be written as the uniform limit of a sequence of polynomial functions.
7.7:1. The Weierstrass Theorem fails for functions on the whole line. (d : 2)
(a) Show that the only polynomials which, as functions on R, are bounded, are the constant functions.
(Suggestion: Use results of Chapter 3.)
(b) Deduce that, in contrast with Theorem 7.26, if a sequence of polynomials Pn converges uniformly on the whole real line R to a function f, then f is itself a polynomial.
...
Answers to True/False question 7.6:0. (a)T. (b)F. (c) F. (d)T.
7.7:2. A modified Weierstrass theorem that does work for functions on all of R. (d : 2)
Suppose f is a continuous complex-valued function on the real line. Show that there exists a sequence of polynomials Pn such that for each finite interval [a , b], the polynomials Pn converge uniformly to f on [a , b].
7.8. Algebras of Functions, Uniform Closure, and Separation of Points (middle of Rudin’s section THE STONE-WEIERSTRASS THEOREM). (pp.161-162)
Relevant exercises in Rudin: None Exercises not in Rudin:
7.8:0. Say whether each of the following statements is true or false.
(a) For every metric space X, (X ) is an algebra of functions on X.
(b) If and are uniformly closed algebras of functions on a set E, then ∩ is also a uniformly closed algebra of functions on E.
(c) The set of all monotonically increasing real-valued functions α on an interval [a, b] is an algebra.
(d) The set of all monotonic real-valued functions α (increasing and decreasing) on an interval [a, b] is an algebra.
(e) If a family of functions on a set E separates points, then so does every family of functions on E containing .
(f ) If a family of functions on a set E separates points, then so does every subset of .
(g) If a family of functions on a set E vanishes at no point of E, then so does every family of functions on E containing .
(h) If a family of functions on a set E vanishes at no point of E, then the same is true of every subset of .
7.8:1. Equicontinuous algebras are mostly uninteresting. (d : 2, 1, 2, 4)
This exercise will show that ‘‘equicontinuous’’ is not, in general, an interesting condition to impose on algebras of functions, though as (b) shows, there are some nontrivial examples.
(a) Show that there is no equicontinuous algebra of real-valued functions on [0,1] which separates points.
(b) Show that the algebra of all real-valued functions on Z is equicontinuous and separates points.
(c) Let X ⊆ { 1 ⁄ n ! n = 1, 2, 3, ... }. Does there exist an equicontinuous algebra of real-valued functions on X which separates points?
(d) Give a simple characterization of the class of metric spaces X such that there exists an equicontinuous algebra of real-valued functions on X which separates points.
7.8:2. Transporting algebras of functions from one metric space to another. (d : 1, 1, 2, 3, 3)
Throughout this exercise, let m : X→ Y be a continuous map of metric spaces. Recall that if f is a function on Y, then f°m denotes the function on X defined by ( f°m)(x) = f (m(x)).
(a) Show that if ( fn) is a sequence of functions on Y that converges uniformly to a function f, then the sequence of functions ( fn°m) on X converges uniformly to f°m.
In the remaining parts, let be an algebra of continuous functions on X, and let be an algebra of continuous functions on Y. Let m *( ) = { f°m ! f∈ }, and let m*( ) denote the set of all continuous functions f on Y such that f°m∈ . To avoid confusion with complex conjugation, let us use cl for uniform closure; e.g., cl ( ) will denote the uniform closure of .
(b) Show that m *( ) is an algebra of continuous functions on X, and that m
*( ) is an algebra of continuous functions on Y.
...
Answer to True/False question 7.7:0. (a)T.
(c) Show that m *(cl ( )) ⊆ cl (m *( )) and that m
*(cl ( )) ⊇ cl (m
*( )).
(d) Show that the reverses of the above two inequalities do not hold in general. (Suggestions: For the first inequality, let m be the inclusion map [0,1] → R, i.e., the map defined by m(x) = x, and the algebra of all polynomial functions on R. For the second inequality, let X = [–1,1], Y = [0,1], m(x) = max(0, x), and let be the algebra of all polynomial functions on [–1,1]. You may assume 7.7:1 whether or not you did it, and you may assume the fact that the only polynomial p(x) such that the equation p(x) = 0 has infinitely many solutions is the zero polynomial.)
(e) What implications, if any, hold between the statements ‘‘ is uniformly closed’’ and ‘‘m
*( ) is uniformly closed’’? Between ‘‘ is uniformly closed’’ and ‘‘m *( ) is uniformly closed’’?
(There are four possible implications – one each way for each of the indicated pairs of statements. For full credit you need to give, for each of these four implications, either a proof that it is true, or an example showing that it is false. In doing this, you may assume any of the previous parts of this exercise, whether you did them or not.)
7.8:3. The uniform closure of the algebra of Laurent polynomials. (d : 4. > 7.7:1)
A Laurent polynomial means a function which can be written in the form f (x) = Σn = – NN anzn, where N ≥ 0 and a–N, ... , aN are constants. (An example is z–2 + 3 z–1 – 7 + z2+ 5 z3, which we can get by taking N = 3 and letting a– 3 = 0, a– 2 = 1, ... a3 = 5.) A Laurent polynomial can be evaluated at any nonzero value of x; in particular, given any subset E of R not containing the point 0, the Laurent polynomials with real coefficients yield an algebra of real-valued continuous functions on E.
Given a Laurent polynomial ΣNn = – N anzn, let us call ΣNn = 0 anzn its ‘‘polynomial part’’ and Σn = – N–1 anzn its ‘‘negative-exponent part’’.
(a) Let E be the half-open interval (0, 1], and suppose that ( fk) is a sequence of Laurent polynomials which, regarded as a sequence of functions on E, converges uniformly. Show that if we write each fk as gk + hk, where gk is its polynomial part and hk its negative-exponent part, then all but finitely many of the functions hk are equal, and the sequence of functions (gk) converges uniformly.
(b) Deduce that the uniform closure of the algebra of Laurent polynomial functions on (0,1] consists of all functions which can be written as the sum of a continuous function and a ‘‘negative-exponent Laurent polynomial’’, i.e., a function of the form Σ–1n = – N anzn.
(c) Deduce that the uniform closure of the algebra of Laurent polynomial functions on (0,1] is not an algebra. This shows that a certain word cannot be omitted from the statement of Theorem 7.29 – which word?
7.9. The Stone-Weierstrass Theorem (end of Rudin’s section THE STONE-WEIERSTRASS THEOREM). (pp.162-165)
Relevant exercise in Rudin:
7:R21. The need for self-adjointness in the complex Stone-Weierstrass Theorem. (d : 2) (A more general version of this result is 7.9:7 below.)
Exercises not in Rudin:
7.9:0. Say whether each of the following statements is true or false.
(a) If is an algebra of real-valued continuous functions on [0,1], and the function f (x) = x + 1 belongs to , then the uniform closure of is the algebra of all real-valued continuous functions on [0,1].
(b) If is an algebra of real-valued continuous functions on [0,1], and the function f (x) = x –1 belongs to , then the uniform closure of is the algebra of all real-valued continuous functions on [0,1].
...
Answers to True/False question 7.8:0. (a)T. (b)T. (c) F. (d)F. (e)T. (f )F. (g) T. (h)F.
(c) The uniform closure of any algebra of polynomials on [ –1, 1] consists of polynomials.
(d) If is an algebra of real-valued functions on [0,1]2 which contains the functions defined by f (x, y) = ex and g(x, y) = (y + 1)–1, then the uniform closure of contains the function (x + 1)y.
(e) The set of all polynomial functions in one variable with complex coefficients, regarded as an algebra of functions on [0,1], is self-adjoint.
(f ) Let D = { z∈C ! | z | ≤ 1} . The set of all polynomial functions in one variable with complex coefficients, regarded as an algebra of functions on D, is self-adjoint.
(g) If K is a compact metric space, then the only uniformly closed self-adjoint algebra of complex-valued continuous functions on K which separates points and vanishes at no point of K is (K ).
7.9:1. Some examples on which to try out the Stone-Weierstrass Theorem. (d : 2)
For each of the following sets i of continuous functions (some real and some complex-valued), determine whether the uniform closure comprises all continuous functions (real or complex as the case may be) on the given domain. In each of these cases, you will either be able to show that it does by the Stone-Weierstrass Theorem, or show that it does not by finding a continuous function which is not a uniform limit of functions in the given set. (The next exercise will ask you to show that there are sets of functions whose uniform closures give all continuous functions, but which do not satisfy the conditions of the Stone-Weierstrass Theorem; but no such examples occur in this exercise.)
In cases where the uniform closure consists of all continuous functions, an answer ‘‘Yes’’ is all that you need to write down. In each case of the opposite sort, give an example of a function not in the uniform closure of i, and state at least one hypothesis of the Stone-Weierstrass Theorem which fails to hold for that set. If the condition that fails is that i be an algebra, state one of the properties defining an algebra which is not satisfied by i. For your own sake, you should also be able to show that the function you give is not in the uniform closure; but you are not asked for this verification in your homework.
(a) The set 1 of all complex-valued polynomial functions f on [0, 1] which satisfy f (0) = f (1).
(b) The set 2 of all real-valued polynomial functions f on [0, 1] that satisfy f′(1⁄2) = 0.
(c) The set 3 of all continuous real-valued functions f on R such that limx→+∞ f (x) exists.
(d) The set 4 of all complex-valued polynomial functions f on [0, 1] that satisfy f (1) = f (0)....
. (e) The set 5 of all real-valued continuous functions f on [0, 1] that satisfy f (0) + f (1⁄2) + f (1) = 0.
(f ) The set 6 of all functions on [0,1] of the form p(x), where p is a real-valued polynomial which is divisible by x –1.
(g) The set 7 of all functions on [0,1] of the form p(x), where p is a real-valued polynomial which is divisible by x – 2.
(h) The set 8 of all continuous real-valued functions f on [0, 1] that satisfy (∃ ε > 0) (∀x∈[0,ε] ) f (x) = f (0).
7.9:2. The hypothesis of the Stone-Weierstrass Theorem is sufficient but not necessary. (d : 2)
Give an example of a set of real-valued functions on a metric space K which does not satisfy all the hypotheses of Theorem 7.32, but such that the uniform closure of does consist of all continuous real-valued functions on K.
7.9:3. An even or odd function is uniformly approximable by even or odd polynomials. (d : 2)
(a) Let be the algebra of all even polynomial functions on [ –1, 1] (polynomial functions p satisfying p( – x) = p(x)). Show that the uniform closure of consists of all even continuous functions.
(The easier direction is ‘‘⊆’’. Suggestion for ‘‘⊇’’: Apply the Stone-Weierstrass Theorem to the restrictions of these functions to [0, 1].)
(b) Let be the set of all odd polynomial functions on [ –1, 1] (polynomial functions p satisfying p( – x) = – p(x)). Show that the uniform closure of consists of all odd continuous functions.
(Suggestion for ‘‘⊇’’: Approximate such a function f by polynomials pn, break each pn into the sum of an odd and an even polynomial, and show that the odd summands also approximate f.)
7.9:4. Functions with value zero at 0 are uniformly approximable by polynomials with value zero at 0.
(d : 2)
Let be the algebra of all polynomial functions p on [ –1, 1] satisfying p(0) = 0. Show that the uniform closure of consists of all continuous functions f satisfying f (0) = 0. (As in the previous exercise, the easier direction is ‘‘⊆’’. Suggestion for ‘‘⊇’’: See Rudin’s proof of Corollary 7.27.)
7.9:5. Intersections of uniform closures, and uniform closures of intersections. (d : 4, 2)
Let be the set of restrictions to [0,1] of even polynomial functions, i.e., polynomial functions satisfying p( – x) = p(x), and let be the set of restrictions to [0,1] of polynomial functions satisfying p(2 – x) = p(x).
(a) Show that ∩ consists only of constant functions.
(b) Let us write cl ( ) instead of ...
for the uniform closure of , to avoid confusion with complex conjugation. Show that cl ( ) = cl ( ) = ([0,1]). Deduce that cl ( ) ∩ cl ( ) ≠ cl ( ∩ ).
7.9:6. Uniform closures of algebras of continuous real functions not satisfying the hypotheses of the Stone-Weierstrass Theorem. (d : 1, 2)
Let be an algebra of continuous real-valued functions on a compact set K, but let us not assume that separates points or vanishes at no point of K. Rather, given any continuous real-valued function f on K, let us say that f ‘‘separates no points not separated by ’’ if for all x, y∈K, we have
((∀ h∈ ) h(x) = h(y)) ⇒ f (x) = f (y),
and let us say that f ‘‘vanishes wherever vanishes’’ if for all x∈K, we have ((∀ h∈ ) h(x) = 0) ⇒ f (x) = 0.
(You might find these conditions easier to think about in contrapositive form:
f (x) ≠ f (y) ⇒ (∃ h∈ ) h(x) ≠ h(y), respectively, f (x) ≠ 0 ⇒ (∃ h∈ ) h(x) ≠ 0.) Then I claim that
The uniform closure of consists of all continuous real-valued functions f that separate no points not separated by , and vanish wherever vanishes.
(a) Show (by arguments and/or quoting results from Rudin) that all functions f in the uniform closure of are indeed continuous, separate no points not separated by , and vanish wherever vanishes.
To prove the converse, suppose f is a continuous real-valued function on K which separates no points not separated by and vanishes wherever vanishes. We must show that f is uniformly approximable by members of . This can be done by a small change in the proof of Theorem 7.32.
Steps 1 and 2 of that proof do not use anything about separating points or not vanishing, and so need no change. In the statements of Steps 3 and 4, the only change needed is to add, after ‘‘a real function f, continuous on K ’’, the words ‘‘which separates no points not separated by , and vanishes wherever vanishes’’. All assertions in the proofs of those two steps then become true, except for the first sentence of the proof of Step 3, which is used to justify the second sentence. So –
(b) Prove that second sentence (the one beginning ‘‘Hence’’ and ending with display (55)) under the above hypotheses. (You will need to consider different cases, depending on whether separates x and y, and whether it vanishes on one or both of these points.)
This completes the proof of the result stated in italics above.
(If you are careful, you will see the need for one small condition in the definition of an algebra that Rudin accidentally omitted: That it be nonempty, i.e., contain at least one function. Assume this.)
7.9:7. The Stone-Weierstrass theorem fails for non-self-adjoint algebras of complex-valued functions.
(d : 3)
Let K be any compact subset of the complex plane which contains the origin 0 and the unit circle ...
Answers to True/False question 7.9:0. (a)T. (b)F. (c) F. (d)T. (e)T. (f )F. (g) T.