Uniform Convergence and Polynomial Approximation

PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION

SAMEER CHAVAN

Abstract. These are the lecture notes prepared for the participants of IST to be conducted at BP, Pune from 3rd to 15th November, 2014.

Contents 1. Pointwise and Uniform Convergence 1 2. Lebesgue’s Proof of Weierstrass’ Theorem 2 3. Bernstein’s Theorem 3 4. Applications of Weierstrass’ Theorem 4 5. Stone’s Theorem and its Consequences 5 6. A Proof of Stone’s Theorem 7 7. Müntz-SzászTheorem 8 8. Nowhere Differentiable Continuous Function 9 References 10

1. Pointwise and Uniform Convergence n Exercise 1.1 : Consider the function fn(x) = x for x ∈ [0, 1]. Check that {fn} converges pointwise to f, where f(x) = 0 for x ∈ [0, 1) and f(1) = 1.

n Exercise 1.2 : Consider the function fm(x) = limn→∞(cos(m!πx)) for x ∈ R. Verify the following:

(1) {fm} converges pointwise to f, where f(x) = 0 if x ∈ R \ Q, and f(x) = 1 for x ∈ Q. (2) f is discontinuous everywhere, and hence non-integrable.

Remark 1.3 : If f is pointwise limit of a sequence of continuous functions then the set of continuties of f is everywhere dense [1, Pg 115].

Consider the vector space B[a, b] of bounded function from [a, b] into C. Let fn, f be such that fn − f ∈ B[a, b]. A sequence {fn} converges uniformly to f if

kfn − fk∞ := sup |fn(x) − f(x)| → 0 as n → ∞. x∈[a,b]

1 2 SAMEER CHAVAN

Theorem 1.4. Let {fn} be a sequence of continuous functions. If {fn} converges uniformly to f on [a, b] then f is continuous. Moreover, Z b Z b lim fn(x)dx = f(x)dx. n→∞ a a

Proof. Let N ≥ 1 be such that kfN − fk∞ < /3. Recall that fN is uniformly continuous on [a, b], that is, for some δ, |fN (x)−fN (y)| < /3 whenever |x−y| < δ. Finally, for all x, y such that |x − y| < δ,

|f(x) − f(y)| ≤ |f(x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − f(y)|

≤ kfN − fk∞ + /3 + kfN − fk∞ ≤ /3 + /3 + /3. The remaining part follows from Z b Z b | fn(x)dx − f(x)dx| ≤ kfn − fk∞(b − a). a a That’s the end of the proof.

Corollary 1.5. Let P = {f ∈ B[a, b]: ∃ {pn} such that kpn − fk∞ → 0}. Then P is contained in C[a, b]. A remarkable result of Weierstrass asserts indeed that P = C[a, b]. In particular, any continuous function on [a, b] can be approximated uniformly by a sequence of polynomials.

Theorem 1.6. For any f ∈ C[a, b], there exists a sequence {pn} of polynomials such that kf − pnk∞ → 0 as n → ∞. There are several proofs of different flavors of this great theorem (apart from Weierstrass’ original proof). Perhaps the most elementary proof is due to Lebesgue (which relies on the conclusion of Weierstrass’ theorem for the function |x|). A constructive proof is due to Bernstein (which relies on a Chebyshev’s version of Bernoulli’s law of large numbers). We will also discuss two remarkable generaliza- tions of Weierstrass’ Theorem: Stone’s Theorem and Müntz-Szász’sTheorem.

Exercise 1.7 : Let {pn} be a sequence of polynomials of degree dn. Suppose that kpn − fk∞ → 0 as n → ∞ for some continuous function f ∈ C[a, b]. If f is not a polynomial then dn → ∞ as n → ∞.

Hint. The subspace of polynomials of degree at most d is ﬁnite-dimensional, and hence closed in C[a, b].

2. Lebesgue’s Proof of Weierstrass’ Theorem 2 n Exercise 2.1 : Consider the function fn(x) = nx(1 − x ) for x ∈ [0, 1]. Verify:

(1) {fn} converges pointwise to f, where f(x) = 0 for all x ∈ [0, 1]. R 1 R 1 (2) limn→∞ 0 fn(x)dx 6= 0 f(x)dx, and hence {fn} can not converge uniformly to f.

Here is a partial converse to Theorem 1.4. UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 3

Theorem 2.2. Let {fn} be a sequence in C[a, b] converging pointwise to a continuous function f. If {fn(x)} decreasing for all x ∈ [a, b] then {fn} converges uniformly to f.

Proof. Let gn := fn − f ≥ 0. For > 0, consider the closed subset Kn := {x ∈ [a, b]: gn(x) ≥ } of [a, b]. Since gn ≥ gn+1,Kn+1 ⊆ Kn. In particular, ﬁnite intersection of Kn’s is non-empty if every Kn is non-empty. If x ∈ [a, b] then since gn(x) → 0, x∈ / Kn for suﬃciently large n. If each Kn is non-empty then we ∞ must have ∩n=1Kn 6= ∅ (Exercise), and hence KN is empty for some N. That is, 0 ≤ gn(x) < for n ≥ N. Here is an important special case of Weierstrass’ Theorem.

Corollary 2.3. Deﬁne a sequence {pn} of polynomials by p0(x) = 0, and 2 pn+1(x) := pn(x) + (x − pn(x) )/2 (n ≥ 0). 2 If qn(x) := pn(x ) then the {qn} converges uniformly to f(x) = |x| on [−1, 1]. 2 √ Proof. Let y = 1−x (x ∈ [−1, 1]) then f(x√) = 1 − y for y ∈ [0, 1]. Thus it suﬃces to check that {pn}√converges uniformly to x on [0, 1]. One may verify inductively that 0 ≤ pn(x) ≤ x for x ∈ [0, 1]. It follows that pn(x) ≤ pn+1(√x) for all n ≥ 0 and all x ∈ [0, 1]. In particular, {pn(x)} converges pointwise to x. Now apply Theorem 2.2 to fn := −pn. Exercise 2.4 : Show that the function g : R → R given by g(x) = 0 for x ≤ 0, = x for x ≥ 0. Show that for any positive number α, g can be uniformly approximated by polynomials on [−α, α].

1 Hint. Note that g(x) = 2 (x + |x|). Lebesgue’s Proof of Weierstrass’ Theorem. Since f is uniformly continuous on [a, b], there exists a positive integer N such that |f(x) − f(y)| < /2 whenever |x − y| < 1/N. For xi := a + (b − a)(i/N)(i = 0, ··· ,N), consider the function h(x) with graph a polygon of vertices at

(a, f(a)), (x1, f(x1)), ··· , (xn−1, f(xn−1)), (b, f(b)).

Check that kf − gk∞ ≤ /2. On the other hand, N−1 X h(x) = f(a) + cig(x − xi)(x ∈ [a, b]) i=0 for some scalars c0, ··· , cN−1. The result now follows from Exercise 3.2. 3. Bernstein’s Theorem

For f ∈ C[0, 1], deﬁne the nth Bernstein polynomial Bn(f) by n X n B (f)(x) := f(k/n) xk(1 − x)n−k (x ∈ [0, 1], n ≥ 1). n k k=0

Remark 3.1 : Bn(1) = 1 (Probability distribution) and Bn(nt) = nx (Mean of 2 2 2 the distribution), and Bn(n t ) = n(n − 1)x + nx (Variance of the distribution). 4 SAMEER CHAVAN

Pn 2n k n−k Exercise 3.2 : If gn(x) := k=0(x − k/n) k x (1 − x) (x ∈ [0, 1]) then verify that kgnk∞ ≤ 1/4n.

Hint. Use Remark 3.1.

Theorem 3.3. If f ∈ C[0, 1] then kBnf − fk∞ → 0 as n → ∞. Remark 3.4 : Since φ(x) = (1 − x)a + xb is a homeomorphism from [0, 1] onto [a, b], Weierstrass’ Theorem follows from Bernstein’s theorem.

The proof of Bernstein’s Theorem depends on the following variant of Bernoulli’s law of large numbers due to Chebyshev. Lemma 3.5. Given δ > 0 and x ∈ [0, 1], let F denote the set

F := {k ∈ N : 0 ≤ k ≤ n such that |k/n − x| ≥ δ}. Then, for every x ∈ [0, 1], X n 1 xk(1 − x)n−k ≤ . k 4nδ2 k∈F

|k/n−x|2 Proof. Note that for any x ∈ [0, 1] and k ∈ F, δ2 ≥ 1. It follows that X n 1 X n xk(1 − x)n−k ≤ (k/n − x)2 xk(1 − x)n−k k δ2 k k∈F k∈F n 1 X n ≤ (k/n − x)2 xk(1 − x)n−k. δ2 k k=0 The desired estimate follows from Exercise 3.2. Proof of Bernstein’s Theorem. Since f is uniformly continuous, there exists an integer N ≥ 1 such that |f(x) − f(y)| < /2 whenever |x − y| < δ = 1/N. Let us estimate kBn(f) − fk∞. Let F be as in Lemma 3.5, and note that n X n |B (f)(x) − f(x)| = (f(k/n) − f(x)) xk(1 − x)n−k n k k=0 X n ≤ |f(k/n) − f(x)| xk(1 − x)n−k k k∈F X n + |f(k/n) − f(x)| xk(1 − x)n−k k k∈ /F 2kfk ≤ ∞ + /2, 4nδ2

kfk∞ where we used Remark 3.1. Choose now an integer n ≥ 1 so that nδ2 < , and note that kBn(f) − fk∞ < .

4. Applications of Weierstrass’ Theorem Exercise 4.1 : Use Weierstrass’ Theorem to show that C[a, b] is separable. UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 5

Exercise 4.2 : For 0 ≤ x < y ≤ 1, consider the indicator function χ[x,y] : [0, 1] → R. Show that there exists a sequence {pn} of polynomials such that Z 1 |pn(x) − χ[x,y](x)|dx → 0 as n → ∞. 0 Conclude that ﬁnite linear combination of indicator functions of subintervals of [0, 1] can be approximated uniformly by polynomials.

1 Hint. Approximate χ[x,y] by continuous functions in the L norm.

R b n Exercise 4.3 : Let f ∈ C[a, b] be such that a t f(t)dt = 0 for all non-negative integers n. Show that f(t) = 0 for every t ∈ [a, b].

Hint. Get a sequence {pn} such that kpn − fk∞ → 0, and apply Theorem 1.4.

Exercise 4.4 : Let f :[a, b] → R be a continuouly diﬀerentiable function. Show that there exists a sequence {rn} of polynomials such that 0 0 krn − fk∞ → 0 and krn − f k∞ → 0 as n → ∞. 1 0 Conclude that C [a, b] is separable with norm kfk := kfk∞ + kf k∞.

0 0 Hint. Let g(x) = f(x) − f(a) and note that g = f . Find a sequence {qn} of 0 R x 0 polynomials such that kqn −g k∞ → 0. Set pn(x) := a qn(t)dt. Note that pn = qn, 0 0 and hence kpn − g k∞ → 0. Also, Z x Z x 0 0 |pn(x) − g(x)| = qn(t)dt − g (t)dt ≤ (b − a)kqn − g k∞. a a

Let rn(x) := pn(x) + f(a).

Exercise 4.5 : Let f :[a, b] → R be a Riemann-integrable function. Show that there exists a sequence {pn} of polynomials such that Z b 2 |pn(x) − f(x)| dx → 0 as n → ∞. a Hint. Approximate any Riemann-integrable function by a continuous function, and then apply Weierstrass Theorem.

Exercise 4.6 : Let K be a compact subset of Rn and let A be a subalgebra of C(K). If f ∈ A then show that |f| ∈ A , where |f|(x) = |f(x)|.

Hint. Let a := supx∈K |f(x)| and let > 0. Let p :[−a, a] → R be a polynomial such that kp − |t|k∞ < . Since p(f) ∈ A , we must have sup |p(f)(x) − |f(x)|| < . x∈K

5. Stone’s Theorem and its Consequences Recall that the Weierstrass’ Theorem says that P = C[a, b]. It is interesting to note that P is an algebra which enjoys the following properties: If x 6= y ∈ [a, b] then trivially id(x) 6= id(y), and if x ∈ [a, b] then 1(x) 6= 0, where id(x) = x and 1(x) = 1. It turns out these properties of subalgebras A of C[a, b] ensure that 6 SAMEER CHAVAN

A = C[a, b]. This remarkable fact was ﬁrst discovered by Stone in much more generality.

Theorem 5.1. Let K be a compact subset of Kn, where K is either R or C. Let A be an algebra of continuous functions f : K → R with the following properties: (1) If x 6= y ∈ K, then there exists f ∈ A such that f(x) 6= f(y). (2) For every x ∈ K, there exists f ∈ A such that f(x) 6= 0.

Then A = CR(K). Before we present a proof of Stone’s Theorem, it is advisable to understand it through its wide range of applications/consequences. Exercise 5.2 : For positive integer n, consider the polynomial subalgebra A of n CR[0, 1] generated by 1 and x . Show that A = CR[0, 1]. Exercise 5.3 : Show that the algebra P of analytic polynomials on the closed unit D satisﬁes assumptions of Theorem 5.1, however, P ( C(D). What goes wrong ?

Hint. z belongs to C(D) but z∈ / P. Corollary 5.4. Let K be a compact subset of Kn. Let A be an algebra of continuous functions f : K → C with the following properties: (1) If x 6= y ∈ K, then there exists f ∈ A such that f(x) 6= f(y). (2) For every x ∈ K, there exists f ∈ A such that f(x) 6= 0. (3) For every f ∈ A , f ∈ A , where f(x) = f(x). Then A = C(K).

Proof. Let AR denote the algebra functions in A which are real-valued, and let

Exercise 5.6 : Let f : T → R be a continuous function such that fˆ(n) := R 2π iθ inθ 0 f(e )e dθ = 0 for all n ∈ Z. Show that f is identically 0.

Hint. Use the last exercise. Exercise 5.7 : For an integer n ≥ 1, consider the polynomial subalgebra A of C[a, b] generated by x. If 0 ∈/ [a, b] then show that A = C[a, b].

Corollary 5.8. Let A be as in Theorem 5.1 except that (2) is not given. If f ∈ A whenever f ∈ A , then there exists a ∈ K such that A = {f ∈ C(K): f(a) = 0}. Proof. If (2) does not hold true then there exists a ∈ K such that f(a) = 0 for every f ∈ A , that is, A ⊆ {f ∈ C(K): f(a) = 0}. Since {f ∈ C(K): f(a) = 0} is closed in C(K), we must have A ⊆ {f ∈ C(K): f(a) = 0}. Now let f ∈ C(K) be such that f(a) = 0. By Stone’s Theorem, the algebra B generated by A and 1 is dense in C(K). Thus there exists a sequence {fn} in B such that kfn − fk∞ → 0 UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 7 as n → ∞. Note that fn = gn + αn for gn ∈ A . Also, fn(a) → f(a) = 0, and hence αn → 0. It follows that kgn − fk∞ ≤ kfn − fk∞ + |αn| → 0 as n → ∞. Exercise 5.9 : Let A := {x2p(x): p is a polynomial}. Show that A = {f ∈ C[0, 1] : f(0) = 0}.

Exercise 5.10 : Show that the subalgebra A := {p(x2): p is a polynomial} is not dense in C[−1, 1].

Hint. If possible then there exists sequence {pn} of polynomials such that 2 kpn(x ) − xk∞ → 0 as n → ∞. In particular, Z 1 Z 1 Z 1 2 2 2 pn(x )dx = pn(x )dx → xdx = 0. 0 −1 −1 R 1 2 R 1 However, 0 pn(x )dx → 0 xdx = 1, which is absurd. Exercise 5.11 : Show that for any a > 0, A := {p(|x|): p is a polynomial} is not dense in C[−a, a].

Corollary 5.12. Let f ∈ C(K) be such that f(x) > 0 for all x ∈ K. Let A be a subalgebra of C(K) that contains f. If f is injective then A = C(K).

Proof. Note that f ∈ A satisﬁes (1) and (2) of Stone’s Theorem. Remark 5.13 : Let X = [a, b]. Consider the algebra A of functions of the form p(f), where p is a polynomial such that p(0) = 0. The last corollary is applicable to f(x) = xα for α > 0, f(x) = ex, f(x) = 1/x if 0 ∈/ [a, b].

6. A Proof of Stone’s Theorem

We start with some elementary properties of subalgebras of CR(K).

Exercise 6.1 : Let A be a subalgebra of CR(K). If f1, ··· , fk ∈ A then so are max{f1, ··· , fk} and min{f1, ··· , fk}.

f1+f2 |f1−f2| Hint. max{f1, f2} = 2 + 2 ∈ A by Exercise 4.6. Now apply ﬁnite induction.

Lemma 6.2. Under the assumptions of Theorem 5.1, for distinct points x1, x2 of K and (real) scalars c1, c2, there exists f ∈ A such that f(x1) = c1 and f(x2) = c2.

Proof. By assumptions there exist g, h, k ∈ A such that g(x1) 6= g(x2) and h(x1) 6= 0 6= h(x2). Then u := (g − g(x1))k and v := (g − g(x2)h belong to A , and f := c1v + c2u does the job. v(x1) u(x2) Proof of Stone’s Theorem. Fix x ∈ K and > 0. For x and every y ∈ K, by the preceding lemma, there exists fyA such that fy(x) = f(x) and fy(y) = f(y). Note that fy − f is continuous such that (fy − f)(x) = 0. Thus there exists an open neighbourhood Uy of y such that fy(t)−f(t) > − for every t ∈ Uy. Now {Uy} is an k open cover of K and K is compact. Thus, for some y1, ··· , yk ∈ K,K ⊆ ∪i=1Uyi .

Also, for gx := max{fy1 , ··· , fyk } ∈ A (Exercise 6.1) and

(6.1) gx(t) > f(t) − for every t ∈ K. 8 SAMEER CHAVAN

Now we vary x. Note that gx(x) = max{fy1 (x), ··· , fyk (x)} = f(x). Hence, by the continuity of gx, there exists an open neighbourhood Vx of x such that gx(t)−f(t) < for every t ∈ Vx. Now {Vx} is an open cover of K and K is compact. Thus, for l some x1, ··· , xl ∈ K,K ⊆ ∪i=1Vxi . Also, for h := min{gx1 , ··· , gxl } ∈ A and h(t) < f(t) + for every t ∈ K.

Also, by (6.1), h(t) > f(t) + for every t ∈ K. That is, kh − fk∞ < .

7. Muntz-Sz¨ asz´ Theorem In this section, we see another remarkable generalization of Weierstrass’ Theo- rem, which relates the topological property of density of polynomials with diver- gence of certain Harmonic series of positive scalars.

Theorem 7.1. Suppose 0 < λ1 < λ2 < ··· . Then the closed linear span of ∞ {1, tλ1 , tλ2 , ···} equals C[0, 1] if and only if P 1 = ∞. k=1 λk

Remark 7.2 : The choice λk = k gives the conclusion of Weierstrass’ Theorem. Note that for any integer l ≥ 1, the closed linear span of {1, tλl , tλl+1 , ···} equals C[0, 1] provided P∞ 1 = ∞. k=1 λk Corollary 7.3. The closed linear span of the constant function 1 and monomials {tp : p is a prime number} equals C[0, 1]. P Proof. This follows from the fact that p 1/p = ∞, where the sum is taken over the set of prime numbers.

Corollary 7.4. Suppose 0 < λ1 < λ2 < ··· . Then the closed linear span of ∞ {tλ1 , tλ2 , ···} equals {f ∈ C[0, 1] : f(0) = 0} provided P 1 = ∞. k=1 λk

Proof. Clearly, the closed linear span of {tλ1 , tλ2 , ···} is contained in {f ∈ C[0, 1] : f(0) = 0}. Let f ∈ C[0, 1] be such that f(0) = 0. Then, by Müntz-SzászTheorem, λ1 λ2 there exists a sequence {pn} in the closed linear span of {1, t , t , ···} such that kpn − fk∞ → 0 as n → ∞. But then pn(0) → f(0) = 0, and hence qn(t) := λ1 λ2 pn(t) − pn(0) in the linear span of {t , t , ···} converges uniformly to f. We will only sketch the outline of the proof of the sufficiency part. Let us first collect all the ingredients required for a proof of Müntz-SzászTheorem. Lemma 7.5. Let M be a closed subspace of C[0, 1]. If M 6= C[0, 1] then there exists a non-zero bounded linear map φ : C[0, 1] → C such that φ(f) = 0 for every f ∈ M. Proof. Let f ∈ C[0, 1] \ M and let Y = M + {λf : λ ∈ C}. Define ψ(g + αf) = αd∞(f, M) for g ∈ M and α ∈ C. Clearly, ψ = 0 on M and φ(f) = d∞(f, M) > 0. The desired conclusion now follows from Hanh-Banach Extension Theorem.

Lemma 7.6. Every bounded linear map φ : C[0, 1] → C is given by φ(f) = R [0,1] f(t)dµ(t) for a complex Borel measure µ on [0, 1].

Lemma 7.7. If f is bounded holomorphic function deﬁned on the open unit disc D P∞ with zeros α1, α2, ··· then n=1(1 − |αn|) = ∞ implies that f(z) = 0 for all z ∈ D. UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION 9

Proof of Muntz-SzászTheorem.¨ Suppose there exists a bounded linear map φ : C[0, 1] → C such that φ(f) = 0 for every f ∈ M. By Lemma 7.6, for some complex λk R λk Borel measure µ on [0, 1], φ(t ) = [0,1] t dµ(t) = 0 for every positive integer k. In view of Lemma 7.5, it suffices to check that φ = 0. By Weierstrass’ Theorem, it k R k suffices to check that φ(t ) = [0,1] t dµ(t) = 0 for every positive integer k. We now define a function g on the open right half plane H by setting Z z log t g(z) = e dµ(t)(z ∈ H). (0,1] Since |ez log t| = elog t

Hint. Use Lemmas 7.5 and 7.6.

8. Nowhere Differentiable Continuous Function By Weierstrass’ Theorem, any continuous function on [0, 1] is a uniform limit of infinitely differentiable functions. It is quite striking that there exists a nowhere differentiable continuous function on [0, 1]. In particular, uniform limit of infinitely real differentiable functions could be nowhere differentiable. Theorem 8.1. Let φ(x) = |x| for x ∈ [−1, 1], which is extended periodically (with period 2) to R by setting φ(x + 2) = φ(x)(x ∈ R). Then the function f : R → R given by ∞ n X 3 f(x) = φ(4nx) 4 n=0 is continuous. However, for every x ∈ R, there exists a sequence {δm} converging to 0 such that

|(f(x + δm) − f(x))/δm| → ∞ as m → ∞. Proof. Clearly, φ is a continuous function such that 0 ≤ φ(x) ≤ 1 (x ∈ R). Also, k n ∞ n ∞ n X 3 X 3 X 3 f(x) − φ(4nx) ≤ φ(4nx) ≤ → 0. 4 4 4 n=0 n=k+1 n=k+1 10 SAMEER CHAVAN

Pk 3 n n Thus, n=0 4 φ(4 x) converges uniformly to f on R, and hence f is continuous. 1 −m Let x ∈ R and let m be a positive integer. Set δm = ± 2 4 , where the sign is m m so chosen that no integer lies between 4 x and 4 x + δm. Deﬁne n n γn := (φ(4 (x + δm)) − φ(4 x))/δm. n n n−m n If n > m then φ(4 (x + δm)) = φ(4 x + 4 /2) = φ(4 x), and hence γn = 0. m Note that |γm| = 4 . When 0 ≤ n < m, n n n ||4 (x + δm)| − |4 x|| |4 δm| n |γn| = ≤ = 4 . |δm| |δm| Now we complete the argument. Note that ∞ n m n m−1 n X 3 X 3 X 3 |(f(x + δ ) − f(x))/δ | = γ = γ = γ + 3m m m 4 n 4 n 4 n n=0 n=0 n=0 m−1 X 1 ≥ 3m − 3n = (3m + 1), 2 n=0 which blows up as m → ∞. References [1] R. Boas, A Primer of Real Functions, The mathematical association of America, 1981. [2] J. Burkill, Lectures on Approximation by Polynomials, Lecture Notes, Tata Institute of Fundamental Research, Bombay, 1959. [3] N. Carothers, Real Analysis, Cambridge Univ. Press, 2000. [4] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976. [5] W. Rudin, Real and Complex Analysis, McGraw-Hill Book Co. New York, 1987.