GROWTH OF THE MAXIMUM MODULUS OF POLYNOMIALS WITH PRESCRIBED ZEROS M. S. PUKHTA. 1. Introduction and Statement of Result
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1 Journal of Classical Analysis Volume 5, Number 2 (204, 07 3 doi:0.753/jca GROWTH OF THE MAXIMUM MODULUS OF POLYNOMIALS WITH PRESCRIBED ZEROS M. S. PUKHTA Abstract. If p(z n a j j is a polynomial of degree n satisfying p(z 0in z <, then for M(p,. In this paper we obtain R. Ankeny Rivlin [ proved that M(p,R ( R n + 2 some results in this direction by considering polynomials of degree n 2, having all its zeros on z k, k which is an improvement of the result recently proved by M. S. Pukhta (203 [Progress in Applied Mathematics, 6 (2, Introduction Statement of Result For an arbitrary entire function p(z,letm( f,rmax f (z. Then for a polynomial p(z of degree n, it is a simple consequence of maximum modulus principle (for z r reference see [4, Vol. I, p. 37, Problem III, 269 that M(p,R R n M(p,, for R. (. The result is best possible equality holds for p(zλz n,where λ, R. If we restrict ourselves to the class of polynomials having no zeros in z <, then inequality (. can be sharpened. In fact it was shown by Ankeny Rivlin [ that if p(z 0in z <, then (. can be replaced by ( R n + M(p,R M(p,, R (.2 2 The result is sharp equality holds for p(zα + β z n,where α β. While trying to obtain inequality analogous to (.2 for polynomials not vanishing in z < k, k, K.K. Dewan Arty Ahuja [2 proved the following result. THEOREM A. If p(z n a j z j is a polynomial of degree n having all its zeros on z k, k, then for every positive integer s ( k {M(p,R} s n ( + k+(r ns k n + k n {M(p,} s, R (.3 Mathematics subject classification (200: 30A0, 30C0, 30D5, 4A7. Keywords phrases: Derivative, Polynomial, Inequality, Zeros. c D l,zagreb Paper JCA
2 08 M. S. PUKHTA THEOREM B. If p(z n a j z j is a polynomial of degree n having all its zeros on z k, k, then for every positive integer s {M(p,R} s [ n an {k n ( + k 2 +k 2 (R ns + a n {2k n + R ns } k n 2 a n + n a n ( + k 2 {M(p,} s, R. (.4 In this paper, we not only improve Theorem A Theorem B but also improve the results recently proved by M.S. Pukhta [7. More precisely, we prove THEOREM. If p(z c n z n + n νμ c n ν z n ν, μ < n is a polynomial of degree n having all its zeros on z k, k, then for every positive integer s R, {M(p,R} s kn 2μ+ ( + k μ +(R ns k n 2μ+ + k n μ+ {M(p,} s s a Rns 2 {M(p,} s, if n > 2 (.5 {M(p,R} s kn 2μ+ ( + k μ +(R ns k n 2μ+ + k n μ+ {M(p,} s s a Rns {M(p,} s, if n 2 (.6 ns ns If we choose μ in Theorem, we get the following result recently proved by M. S. Pukhta [7. COROLLARY. If p(z n a j z j is a polynomial of degree n 2 having all its zeros on z k, k, then for R, {M(p,R} s kn ( + k+(r ns k n + k n {M(p,} s s a Rns 2 {M(p,} s, if n > 2 (.7 {M(p,R} s kn ( + k+(r ns k n + k n {M(p,} s s a Rns {M(p,} s, if n 2 (.8 ns ns Next we prove the following result which is a refinement of Theorem.
3 GROWTH OF THE MAXIMUM MODULUS OF POLYNOMIALS WITH PRESCRIBED ZEROS 09 THEOREM 2. If p(zc n z n + n c n ν z n ν, μ < n is a polynomial of degree n having all its zeros on z k, k, then for every positive integer s νμ R, {M(p,R} s [ k n μ+ ( n cn {k n μ+ (k μ + k 2μ +k 2μ (R ns } + c n μ {μ(k n + k n μ+ + k μ (R ns } μ c n μ (k μ + +n c n (k μ + k 2μ ( R {M(p,} s ns s a Rns 2 {M(p,} s, ( n cn {k {M(p,R} s [ n μ+ (k μ + k 2μ +k 2μ (R ns } + c n μ {μ(k n + k n μ+ + k μ (R ns } k n μ+ μ c n μ (k μ + +n c n (k μ + k 2μ {M(p,} s ( R ns s a ns if n > 2 (.9 Rns {M(p,} s, if n 2 (.0 ns If we choose μ in Theorem 2, we get the following result. COROLLARY 2. If p(z n a j z j is a polynomial of degree n 2 having all its zeros on z k, k, then for every R {M(p,R} s k n [ {M(p,R} s k n [ ( n cn {k n ( + k 2 +k 2 (R ns } + c n {( + k n +(R ns } 2 c n + n c n ( + k 2 ( R ns s a Rns 2 ( n cn {k n ( + k 2 +k 2 (R ns } + c n {(k + k n + k(r ns } 2 c n μ + n c n ( + k 2 ( R ns s a Rns ns ns {M(p,} s {M(p,} s, if n > 2 (. {M(p,} s {M(p,} s, if n 2 (.2
4 0 M. S. PUKHTA 2. Lemmas For the proof of these theorems, we need the following lemmas. LEMMA. If p(zc n z n + n c n ν z n ν, μ < n is a polynomial of degree νμ n having all its zeros on z k, k,then n max z p (z k n 2μ max p(z. (2. + kn μ+ z The above lemma is due to Govil [3. LEMMA 2. If p(zc n z n + n c n ν z n ν, μ < n is a polynomial of degree νμ n having all its zeros on z k, k,then max p (z n [ n c n k 2μ + c n μ k μ k n μ+ n c n (k 2μ + k μ +μ c n μ (k μ + The above lemma is due to Dewan Mir [5. max p(z (2.2 z LEMMA 3. If p(z n a j z j is a polynomial of degree, then for all R max z R p(z Rn M(p, (R n R n 2 p(0, if n > (2.3 max p(z RM(p, (R p(0, if n (2.4 z R The above lemma is due to Frappier, Rahman Ruscheweyh [6. 3. Proof of the theorems Proof of Theorem. We first consider the case when polynomial p(z is of degree n > 2. Since p(z is a polynomial of degree n having all its zeros on z k, k, therefore, by Lemma, we have n max z p (z k n 2μ+ M(p, (3. + kn μ+ Now applying inequality (. to the polynomial p (z which is of degree n noting (3., it follows that for all r 0 θ < 2π p (re iθ nr n k n 2μ+ M(p, (3.2 + kn μ+
5 GROWTH OF THE MAXIMUM MODULUS OF POLYNOMIALS WITH PRESCRIBED ZEROS Also for each θ,0 θ < π R, we obtain This implies {p(re iθ } s {p(e iθ } s d dt {p(teiθ } s dt s{p(te iθ } s p (te iθ e iθ dt. {p(re iθ } s {p(e iθ } s s p(te iθ s p (te iθ dt (3.3 Since p(z is a polynomial of of degree n > 2, the polynomial p (z which is of degree n 2, hence applying inequality (2.3ofLemma3to p (z,wehaveforr 0 θ < 2π p (re iθ r n M(p, (r n r n 3 p (0 (3.4 Inequality (3.4 in conjunction with inequalities (3.3 (., yields for n > 2 for R {p(re iθ } s {p(e iθ } s s (t n M(p, s [t n M(p, (t n t n 3 p (0 dt s t ns {M(p,} s M(p, (t ns t ns 3 {M(p,} s p (0 dt [ R ns s {M(p,} s M(p, ns Rns 2 {M(p,} s p (0 On applying Lemma to inequality (3.5, we get for n > 2, {p(re iθ } s {p(e iθ } s R ns k n 2μ+ {M(p,}s + kn μ+ s Rns 2 {M(p,} s p (0 This gives {M(p,R} s kn 2μ+ ( + k μ +(R ns k n 2μ+ + k n μ+ {M(p,} s s a Rns 2 {M(p,} s (3.5 This completes the proof of inequality (.5.
6 2 M. S. PUKHTA The proof of inequality (.6 follows on the same lines as that of inequality (.5, but instead of using inequality (2.3 of Lemma 3 we use inequality (2.4ofLemma3. Proof of Theorem 2. The proof of Theorem 2 follows on the same lines as that of Theorem. But for the sake of completeness we give a brief outline of the proof. We first consider the case when polynomial p(z is of degree n > 2, then the polynomial p (z is of degree (n 2, hence applying inequality (2.3 oflemma3to p (z, we have for r 0 θ < 2π p (re iθ r n M(p, (r n r n 3 p (0 (3.6 Also for each θ,0 θ < 2π R, we obtain This implies {p(re iθ } s {p(e iθ } s d dt {p(teiθ } s dt s{p(te iθ } s p (te iθ e iθ dt. {p(re iθ } s {p(e iθ } s s p(te iθ s p (te iθ dt (3.7 Inequality (3.6 in conjunction with inequalities (3.6 (.,yields for n > 2, {p(re iθ } s {p(e iθ } s s (t n M(p, s [t n M(p, (t n t n 3 p (0 dt s t ns {M(p,} s M(p, (t ns t ns 3 {M(p,} s p (0 dt [ R ns s {M(p,} s M(p, ns Rns 2 {M(p,} s p (0 Which on combining with Lemma 2, yields for n > 2 {p(re iθ } s {p(e iθ } s [ Rns n c n k 2μ + c n μ k μ k n μ+ n c n (k 2μ + k μ +μ c n μ (k μ {M(p,} s + s Rns 2 {M(p,} s p (0 from which we get the desired result. The proof of inequality (.0 follows on the same lines as that of inequality (.9, but instead of using inequality (2.3 of Lemma 3 we use inequality (2.4ofLemma3.
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