A radius of convexity problem

Theorem 1. Let p(s ) tPn(A,B) . Then Re $$p(z) \ge \frac{1-A\phi(r)}{1-B\phi(r)}$$ , $A + B \ge 0$ , where <)>(r) = r ((r+|e |)/(l+r| e |)) awd \o \ = \b \/(A-B) . The bound is sharp. Proof. p(z) € PnU, B) implies that p(z) = [l+Aw(z))/ [l+Bw(z)) , °° k -1 < S < 0 < /I 5 1 , w{z) = £ e 3 being analytic in the unit disc satisfying \w(z)\ < |3 | . Now Re $$p(z) = \frac{\text{Re}\{1+A\omega(z)\}\{1+B\omega(z)\}}{|1+B\omega(z)|^2}$$ $$= \frac{1+(A+B)\text{Re}\omega(z)+AB|\omega(z)|^2}{|1+B\omega(z)|^2}$$ $$\geq \frac{1-(A+B)|\omega(z)|+AB|\omega(z)|^2}{(1-B|\omega(z)|)^2} = \frac{1-A|\omega(z)|}{1-B|\omega(z)|}$$ when /i + 5 > 0 . Ihe function ^(x) = (1-Ax)/(l-Bx) where x = \w(z)\ is monotonic decreasing with respect to x since g'(x) = [-(A-B))/(1-Sx) < 0 . Therefore Re p(z) > [l-A\$(r))/ (l-B(J)(r)) using (i) of Lemma 1. Taking w{z) = -3 ((| e |+s)/(l+| e \z}) we see that the bound is sharp at z = r on |z | = r» < 1 .$

Theorem 2. Let p(z) € P U , B) . Then $$\left|\frac{zp'(z)}{p(z)}\right| \leq \frac{r(A-B)\phi'(r)}{\{1-A\phi(r)\}\{1-B\phi(r)\}}, \quad A+B \geq 0,$$ where <j>(r) = rn [[r+\a |)/(l+r|e |)) and \c | = \b \/[A-B) . The bound is sharp. Proof. Since p(z) (. P (A, B) , p(z) = [l+Aw(z)) / (l+Bw(z)) , -15B<O<4S 1 where w{z) is analytic in the unit disc satisfying w(0) = 0 , |u(a)| 5 \z\n there . Therefore $$\frac{zp'(z)}{p(z)} = \frac{(A-B)zw'(z)}{(1+Aw(z))(1+Bw(z))}$$ and $$\left|\frac{zp'(z)}{p(z)}\right| \leq \frac{(A-B)|zw'(z)|}{(1-A|w(z)|)(1-B|w(z)|)} \quad \text{when} \quad A+B \geq 0.$$ Using the lemma from [J] , we get $$\left|\frac{zp'(z)}{p(z)}\right| \leq \frac{(A-B)r\phi'(r)}{\left\{1-\phi^2(r)\right\}} \frac{1-\left|w(z)\right|^2}{\left\{1-A\left|w(z)\right|\right\}\left(1-B\left|w(z)\right|\right\}}$$ Conside r ?z(x ) = (l-a ; )/(l-Ax){l-Bx ) , x = \w(z) \ , 0 5x51 . The ⁿ h'(x) = {(A+B)x2 -2{l+AB)x+(A+B))/{l-Ax)2 (l-Bx)2 . This shows that h'{x) > 0 if a; 5 x and then i t becomes negative in (x , l) where x = {(l+AB)-\/(l-A2 ) [l-B2 )) / {A+B) . Clearly 0 5 x < 1 and h(z) i increasing in [o, xS\ and decreasing in (x , l) . We can therefore use x = |w(z)| 5 <J>(r) in 0 5 \zp'{z)/p(z)\ . Thus 5 x to get the upper bound of $$\left|\frac{zp'(z)}{p(z)}\right| \leq \frac{(A-B)r\phi'(r)}{\left(1-A\phi(r)\right)\left(1-B\phi(r)\right)}.$$ Taking w(z) = -s ⁿ (s+| c |) > 3 = r on I2I = r < 1 . , we see that the bound is sharp at

Theorem 3. Let f(z) € Sn{A, B) and e^ as in Lerma 1. Then f(z) is convex in the disc \z\ < r\ where r\* = min(r , r ) , r being the smallest positive root of r" + |e \r - \a \x r - x = 0 and r the smallest positive root of [l-At\$>(r)) - (A-B)r<k'(r) = 0 . The estimate is sharp for r± < rQ .* Proof. f{z) is convex in the unit disc if and only if Re{l+[zf"(z)/f'(z))} 2 0 . Now p{z) = zf'(z)/f(z) € P^A, B) since f € Sn(A, B) . Now $$1 + \frac{zf''(z)}{f'(z)} = p(z) + \frac{zp'(z)}{p(z)}$$ and $$\operatorname{Re}\left\{1 + \frac{zf''(z)}{f'(z)}\right\} \ge \operatorname{Re} p(z) - \left|\frac{zp'(z)}{p(z)}\right|$$ Using Theorems 1 and 2, we therefore get $$\operatorname{Re}\left\{1 + \frac{zf''(z)}{f'(z)}\right\} \ge \frac{1 - A\phi(r)}{1 - B\phi(r)} - \frac{(A - B)r\phi'(r)}{\left(1 - A\phi(r)\right)\left(1 - B\phi(r)\right)}$$ whenever A + B > 0 and |u(s)| 5 x. . Thus /(z ) is convex if and only i f \l>(r) = (l-A\$(r))² - (A-B)r<\$>'(r) > 0 with |u(3) | < xQ . Let r be the smallest positive root of the equation ty(r) = 0 . Further \w(z)\ S x implies rn [r+\a \) < x fl+r|c |) . Let r be ; the smallest positive root of the equation <J>(r) = r + r \a \ - r\a \x - x = 0 . Let r\ = min(r , r ) . Then f{z) is convex in the disc \z\ < r\ . The estimate is sharp can be seen by taking $$w(z) = \frac{-z^n(z+|c_n|)}{(1+z|c_n|)}$$ at $z = r$ on $|z| = r < 1$ . REMARK. Without loss of generality we can take a > 0 because otherwise we can consider the function $$w = e^{i\alpha/n} f(ze^{-i\alpha/n})$$ where arg $a_{n+1} = \alpha$ . If we now put A=1, B=-1 and replace $|c_n|$ by $na_{n+1}/(A-B)=na_{n+1}/2$ , we get $\phi(r)=r^n\big(r+\big(na_{n+1}/2\big)\big)/\big(1+\big(na_{n+1}/2\big)r\big)$ and $x_0=1$ . Therefore in [0,1], $\mathrm{Re}\big\{1+\big(zf''(z)/f'(z)\big)\big\}\geq 0$ if and only if $\big(1-\phi(r)\big)^2-2r\phi'(r)\geq 0$ . That is $$\left( \left[ 1 + \frac{rna_{n+1}}{2} \right] - r^n \left( r + \frac{na_{n+1}}{2} \right) \right)^2 - 2r^n \left[ n \left( r + \frac{na_{n+1}}{2} \right) \left( 1 + r \frac{na_{n+1}}{2} \right) + r \left( 1 - \frac{n^2 a_{n+1}^2}{4} \right) \right] \ge 0 .$$ This leads to the equation $$4r^{2n+2} + 4na_{n+1}r^{2n+1} + n^2a_{n+1}^2r^{2n} - 4n(n+1)a_{n+1}r^{n+2} - 2r^{n+1}\left(8+4n+n^3a_{n+1}^2\right) - 4n(n+1)a_{n+1}r^n + n^2a_{n+1}^2r^2 = 0.$$ This gives us the sharp estimate obtained by Mogra and Juneja [3] for analytic starlike functions in the unit disc with missing coefficients where the initial non-vanishing coefficient is fixed. We are at present unable to obtain sharp estimates of radius of convexity when A+B<0 .$