Starlike Univalent Functions Bounded on the Real Axis

THEOREM 1. Let f e 5 0 be an odd function with \f\ < oo. 77zeAz, /or some 0 G [0, 277), / w subordinate to Received December 16, 1988. The author is grateful to Prof. Q. I. Rahman for his interest and valuable discussions. This work was supported by an ACSAIR (Quebec) grant. 642

Theorem 1 admits several corollaries. We obtain:

COROLLARY 1.1. Letf e S0 andf'(0) = 1. Then \f\ â 1/2 and equality is possible if and only if z f(z) = ~v K\-W where X e [0, 1]. (1 + izf\\ - iz)2(l A)

COROLLARY 1.2. Let oo f{z) = 2 anzn e S0 n = \ with l/l < oo. Then \an\ ^ 2\f\n for n = 1, 2, . . . . //"« = 1, equality is possible if and only if (1 + /z)2X(l - /zf 1-^ where a e C #«d X G [0, 1], 7/^w > 1, equality is possible if and only if f(z) = J^?-

COROLLARY 1.3. Let oo f(z) = 2 aln_xzln~x e S0 W/7/Î l/l < oo. 77ze« |«2/i-il — ^l/l />A* « = 1,2,... . Equality is possible for some n = I if and only if OLZ f(z) = ~ where a e C 1 + z2 If M > 0 is a given positive number, there is no estimate of the type |/(x)| ^M\f\\x\, x e ( - 1 , 1 )

THEOREM 2. Letf e S0 with \f\ < oo. Then

PROPOSITION 1. Let p G ^ ( r ) a«J x \p(x) 1 = (-+ ) r2x2y2 r 2 I ( - 1, 1). Then and equality is possible for some x e (— 1, 1) if and only if 1 where p{z) = p(0)(\ + riz)k(\ - riz)n~k \p(Q) | = _ _ andO ^ k ^ n. FK J (1 4-

LEMMA 1. Let <p e [0, 2m) andf e S0 such that Then f(z) = -.—^ :—^—pr where X e [0, 11. JK } (1 - ie*zf\\ + iel<pz)2(l~X) L J

LEMMA 2. Let <p e [0, 2ir) andf e S0 as in Lemma 1. TTien | / | = 1/2 */ and only if el(p = ± 1.

LEMMA 3. Let x e (— 1, 1). There exists a polynomial q(z) e @n(r) such that (\ + r 2 x V 2 \p(x) | = \q(x) | ^ |

LEMMA 4. Let q{z) EE p(l - rz)*(l + rzf~* G ^ ( r ) , where 0 < p 0«d 0 ^ k ^ «. TTiew inf \p(x)\ < \q(x)\, ifx e ( - 1 , 1). pePn(r)

LEMMA 5. Let x e (— 1, 1) and q e ^ ( r ) w/Y/z |^(x) | = inf \p(x) |. r * ^ | ? ( i ) | = \q(-i)\ = l.

LEMMA 6. Let q(z) = p(\ - rzf\\ + rz)k>q(z) G <£(r) where 0 < p, 0 < min^j, k2) and q(z) is a polynomial of degree n — kx — k2. Then inf \p(x) | < \q(x) |, for all x G ( - 1, 1).

Lemma 5 and Lemma 6, the extremal polynomial q(z) must be (15) q(z) (1 ± rzfjl ± réazf\\ ± re~'az)n (1 + rf\\ + reia\"~k where 0 ë k < n, a is a real number such that sin(a) # 0 and (HHÏ re 1 4- re" H-fc = 1. In order to complete the proof of Proposition 1 we need to establish a

LEMMA 7. For all x e (— 1, 1), (1 4 r 2 x V 2 1TT7 < (£rf 1 + re'ax 1 + re'" n-A:

Theorem 2. The result of Theorem 2 is sharp as can be seen from the functions az f(Z) = (1 + izf\\ - izfV-K> with a G C and \ e [0, 1]. We believe that equality can be attained only for these functions but we cannot prove it due to a limiting process used in the proof. It is however not difficult to prove that given r G. [0, 1] and x e (— 1, 1), the equality in (19) is possible within the class S0 if and only if the function / is as above, with rational X e [0, 1]. Conclusion. 1° Let 5 = { / 6 S

PROPOSITION 2. /i + r2x2\n/2 P g &n°(r) =» \p(x) I ^ ( 1 + r 2 ) , ace (-1,1).

Proposition 1 and Theorem 2 were. The proofs are omitted:

PROPOSITION 3. Let p(z) be a polynomial of degree n ^ 1 having all its zeros outside the open unit disc. Then min( \p(\) |, \p(- 1) | )\p(x) | ^ 2"/2\p(0) |2(1 + x2fn for all x Œ (— 1, 1). Equality is possible if and only if p(z) = p(0)(\ + iz)k(\ ~ iz)n~k where 0 ^ k ^ n.

THEOREM 3. Letf e S0 with \f\ < oo. Then \f'(0)\2 \x\ 2|/| 1 + x ^ \f(X)\, X G ( - 1 , 1 ) . REFERENCES 1. M. Avriel, Nonlinear programming: Analysis and methods (Prentice Hall, Engelwood Cliffs, 1976). 2. L. de Branges, A proof of the Bieberbach conjecture, ACTA Math. 154 (1985), 137-152. 3. P. L. Duren, Univalent functions (Springer- Verlag, New York, 1983). 4. Q. I. Rahman and St. Ruscheweyh, Markov's inequality for typically real polynomials, to appear in Journal of Mathematical Analysis and