Abstract
We prove the Brannan conjecture for particular values of the parameter. The basic tool of the study is an integral representation published in a recent work (Barnard et al., in Proc Am Math Soc 143(5):2117–2128, 2015).
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1 Introduction
We consider the following Mac-Laurin development
where \(\alpha>0, \ \beta >0, \ x=\mathrm{e}^{i\theta }, \ \theta \in [-\pi ,\pi ],\) and \(z\in {U.}\) The radius of convergence of the series (1) is equal to 1. The author conjectured in [5] that if \(\alpha>0, \ \beta >0\) and \(|x|=1,\) then
where n is a natural number. Partial results regarding this question were already proved in [1, 2, 5, 8].
The case \(\beta =1,\) \(\alpha \in (0,1)\) is still open. Regarding this case, partial results were obtained in [3, 4, 6, 7]. We will prove some partial results regarding the case \(\beta =1,\) and \(\alpha \in (0,1).\) We will use an integral representation proved in [3], and the fact that the conjecture was proved for \(n\le 26\) in [6].
2 Preliminaries
We introduce the notation:
It is easily seen that
We denote
Lemma 1
For \(\alpha \in (0,1),\) the following equality holds:
where \(F'(t)=-t^{-1-\alpha }(1-t)^{\alpha -1},\) and \(F(1)=0.\)
Proof
We use two known equalities to prove the assertion of the lemma. The first one is the following:
which has been deduced in [3], while the second one is the well-known equality \(B(p,q)=\int _0^1t^{p-1}(1-t)^{q-1}\mathrm{d}t=\frac{\Gamma (p)\Gamma (q)}{\Gamma (p+q)}.\) Replacing \(p=\alpha \) and \(q=1-\alpha ,\) we get
This equality and (3) imply that
or equivalently
Now integrating by parts and using that \(\lim _{t\rightarrow 0}F(t)(\frac{t}{\alpha }+\sum _{k=1}^n\frac{(-tx)^k}{k})=0\) and \(\lim _{t\rightarrow 1}F(t)(\frac{t}{\alpha }+\sum _{k=1}^n\frac{(-tx)^k}{k})=0,\) we infer that
We get the desired equality replacing n by \(2n-1\) and \(x=\mathrm{e}^{i\theta }\).
\(\square \)
Remark 1
We note that the condition \(\alpha \in (0,1)\) implies the existence of the integrals in the previous lemma and its proof.
We denote
Since \(|\Phi (\theta )|^2\in {\mathbb {R}},\) it follows that
Lemma 2
(a) Let \(f,g:[0,1]\rightarrow {\mathbb {R}}\) be two continuous functions. If f is a decreasing function, and if there is a point \({t^*}\in (0,1)\) such that \(g(t)\ge 0, \ t\in (0,t^*)\) and \(g(t)\le 0, \ t\in (t^*,1),\) then
(b) The Chebyshev inequality for integrals. If f and g are monotonic functions with different monotony, then
and in case of the same monotony we have
Proof
We have
\(\square \)
Lemma 3
If \(\theta \in [0,\frac{\pi }{2}],\) and \(n\ge 27,\) then
Proof
According to Lemma 2 (b), we have
We use assertion (a) of Lemma 2 putting \(f(t)=\frac{F(t)}{1+t^2+2t\cos \theta }\) and \(g(t)=\frac{\frac{1}{2}-t-2t^{2n}}{1+t}.\) If \(\theta \in [0,\frac{\pi }{2}],\) then the mapping \(f:[0,1]\rightarrow [0,\infty )\) is strictly decreasing and we get
where \(t_n\) denotes the unique root of the equation \(\frac{1}{2}-t-2t^{2n}=0,\) in the interval \( \ t\in (0,1).\) The following equality holds:
The equality (10) and the inequality (9) imply that
We use again assertion (a) of Lemma 2 putting \(f(t)=\frac{F(t)}{1+t^2+2t\cos \theta }\) and \(g(t)={t^2-2t^{2n}-t^{2n-1}}.\) If \(\theta \in [0,\frac{\pi }{2}],\) then f is strictly decreasing and we get
To finish the proof of the second inequality, we take notice of the fact that in case \(\theta \in [0,\frac{\pi }{2}]\) each member of the following sum is positive:
Thus, we get
and the proof is done. \(\square \)
Lemma 4
If \(\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],\) then
Proof
The inequality (13) is equivalent to
The function f has a minimum point at \(t^*=-\frac{65}{115}\cos \theta +\frac{27}{115}\in (0,1),\) for every \(\theta \in [\frac{\pi }{2},\frac{2\pi }{3}].\) Thus, we get
Consequently, to prove (13) we have to show that
Using the notation \(x=\frac{27-65\cos \theta }{23},\) we get
The equality 14 implies that
where \(g(x)=-\frac{1}{5}+\frac{5}{13}x-\frac{1}{10}x^2,\) and consequently (13) holds. \(\square \)
Lemma 5
If \(\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],\) and \(n\ge 27,\) then
Proof
We use assertion (a) of Lemma 2, putting \(f(t)=\frac{F(t)}{(1+t)(1+t^2+2t\cos \theta )}\) and \(g(t)={\frac{27}{50}-t-2t^{2n}}.\) If \(\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],\) the mapping \(f:[0,1]\rightarrow [0,\infty )\) is strictly decreasing and we get
where \(t_n\) denotes the unique root of the equation \(\frac{27}{50}-t-2t^{2n}=0,\) in the interval \( \ (0,1).\)
The equality (18) and the inequality (17) imply (15).
To prove the second inequality, we remark that
and
If \(\cos (2n-1)\theta \) and \(\cos 2n\theta \) have different signs, then we have \(|\cos (2n-1)\theta +t\cos (2n\theta )|\le 1.\) If \(\cos (2n-1)\theta \) and \(\cos 2n\theta \) have the same sign, then it follows that \(|\cos (2n-1)\theta +t\cos (2n\theta )|\le |\cos (2n-1)\theta +\cos (2n\theta )|=2|\cos (2n-1/2)\theta |\cos \theta /2\le 2\cos \theta /2\le \sqrt{2}\) for every \(\theta \in [\frac{\pi }{2},\frac{2\pi }{3}].\) Thus, we infer
Let \(g(t)=1-2t+2t^2-t^3-4(4+\sqrt{2})t^{2n-1}\) and \(f(t)=\frac{F(t)}{2(1+t^3)}.\) The functions f and g are strictly decreasing on [0, 1] and according to the assertion (a) of Lemma 2 there is a point \(t^*\in (0,1)\) ( the unique root of the equation \(g(t)=0\)) such that the following inequality holds:
Now, inequalities (20)–(22) imply
Finally inequality (23), Lemma 4 and equality (19) imply (16), and the proof is done. \(\square \)
Lemma 6
The following inequality holds:
Proof
We introduce the notation
We will use the following implication \(\alpha>0, a>b>1\Rightarrow \frac{b}{a}<\frac{b+\alpha }{a+\alpha }.\) Using this implication, we get
and
Multiplying (24), (25), and (26), we get
Thus to finish the proof, we have to check the inequality
which is immediate. \(\square \)
3 Main Result
Theorem 1
If n is a natural number, \(n\ge 27\) and \(\alpha \in (0,1)\) then the following inequality holds:
Proof
According to (4) and (5), the inequality (27) is equivalent to
We denote
The equality (5) implies that
Since \(\Phi ^2(0)-|\Phi (\theta )|^2\) is an even function with respect to \(\theta ,\) we have to prove (28) only for \(\theta \in [0,\frac{\pi }{2}].\)
Lemma 3, (a) implies that in case \(\theta \in [0,\frac{\pi }{2}],\) the following inequality holds: \(\int _0^1F(t)(B(t,0)-{B}(t,\theta ))\mathrm{d}t\ge 0.\) Thus, the condition \(\alpha \in (0,1),\) implies that
On the other hand, we have
This inequality implies
The inequality between the arithmetic and geometric means leads to
We apply Lemma 3 and get
The Cauchy–Schwarz inequality for integrals implies
Putting \(a_1^2=\frac{\frac{1}{2}(1-\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )},\)\(b_1^2=\frac{(1+t)(1+\cos {\theta })}{1+t^2+2t\cos \theta }\) and so on in the inequality
we get
On the other hand, we have
Finally, the inequalities (30)–(35) imply that
and consequently inequalities (28) and (27) hold in case \(\theta \in [-\frac{\pi }{2},\frac{\pi }{2}].\) \(\square \)
Theorem 2
If n is a natural number, \(n\ge 27\) and \(\alpha \in (0,1),\) then the following inequality holds:
Proof
Equality (29) can be rewritten as follows:
Since \(2\Big (\Phi ^2(0)-|\Phi (\theta )|^2\Big )\) defines an even function to prove (36), it is enough to show that
The inequality between the arithmetic and geometric means and the condition \(\alpha \in (0,1]\) imply
Lemma 5 and inequality (38) lead to
We apply twice the Cauchy–Schwarz inequality just like in the proof of the previous theorem and we get
The following inequality obviously holds:
and is equivalent to
Finally, (39)–(41) imply \(\Phi ^2(0)-|\Phi (\theta )|^2\ge 0, \theta \in [\frac{\pi }{2},\frac{2\pi }{3}],\) and the proof is done. \(\square \)
Theorem 3
If n is a natural number, \(n\ge 27\) and \(\alpha \in [\frac{1}{3},1),\) then the following inequality holds:
Proof
We will use the Taylor formula with an integral remainder. Let \(f:(-1,1)\rightarrow {\mathbb {R}}\) be a 2n times derivabile function. If \(x\in (-1,1),\) then
Let f be the function defined by \(f:(-1,1)\rightarrow {\mathbb {R}}, \ \ f(x)=(1+x)^\alpha , \ \ \alpha \in (0,1)\) and we get
Since the mapping \(g:U\rightarrow {\mathbb {C}}, \ \ g(z)=(1+z)^\alpha \) is well defined (we take the principal branch of the multi valued function), it follows that the equality
holds for every \(z\in {U}.\) The mapping \(g:U\rightarrow {\mathbb {C}}, \ \ g(z)=(1+z)^\alpha \) is radially continuous and so we infer that
Taking the absolute value, this equality implies that
On the other hand, we have
Thus to prove (42), we have to show that
We denote \(x=-\cos \theta ,\) and the inequality (43) will be equivalent to
This inequality can be rewritten in the following form:
It is easily seen that the expression
is decreasing with respect to n and increasing with respect to x; moreover, the expression \(\frac{(2-2x)^\frac{\alpha }{2}-1}{\frac{\alpha }{2}}\) is negative. Consequently, to prove (44) it is enough to show that the following inequality holds:
A simple calculation shows that (45) is equivalent to
Since the left hand side is increasing and the right hand side is decreasing provided that \(\alpha \in (0,1),\) it follows that to prove inequality (46) for \(\alpha \in [\frac{1}{3},1]\) it is enough to be checked in case \(\alpha =\frac{1}{3},\) that is,
According to Lemma 6, this inequality holds and the proof is done. \(\square \)
4 Concluding Remarks
The following two corollaries are the proof of the Brannan conjecture in two different particular cases.
Theorems 1, 2 and the result of [6] imply the following corollary.
Corollary 1
If \(x\in {\mathbb {C}}\) with \(|\arg {x}|\le \frac{2\pi }{3},\) and \(|x|=1,\) then the inequality
holds for every \(\alpha \in (0,1), \ n\in {\mathbb {N}}^*.\)
Theorems 1, 2 and 3 and the result of [6] imply the following corollary. This corollary is the solution of the Brannan conjecture in case \(\alpha \in [\frac{1}{3},1).\)
Corollary 2
The inequality
holds for every \(\alpha \in [\frac{1}{3},1),\) \(n\in \mathbb {N^*}\) and \(x\in {\mathbb {C}},\) with \(|x|=1.\)
The Brannan’s conjecture holds if \(\alpha \in [\frac{1}{3},1)\) and remains an open question in case \(\alpha \in (0,\frac{1}{3}).\)
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Szász, R. On Brannan’s Conjecture. Mediterr. J. Math. 17, 38 (2020). https://doi.org/10.1007/s00009-019-1469-9
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DOI: https://doi.org/10.1007/s00009-019-1469-9