• Korean Journal of Mathematics
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• v.30 no.4
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• pp.629-642
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• 2022

In a recent paper, Sakar and Güney introduced a new class of bi-close-to-convex functions and determined the estimates for the general Taylor-Maclaurin coefficients of functions therein. The main purpose of this note is to give a generalization of this class. Also we point out the proof by Sakar and Güney is incorrect and present a correct proof.

• Kyungpook Mathematical Journal
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• v.46 no.1
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• pp.131-137
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• 2006

The object of the present paper is to obtain new conditions for analytic function to be strongly starlike and strongly convex function defined in the open unit disk.

• Bulletin of the Korean Mathematical Society
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• v.49 no.4
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• pp.775-785
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• 2012

Some errors in literatures are pointed out and corrected. A generalization of Ostrowski type inequalities for functions whose derivatives in absolute value are s-convex in the second sense is established. Special cases are discussed.

• Journal of the Chungcheong Mathematical Society
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• v.27 no.3
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• pp.433-438
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• 2014

A semi-infinite optimization problem involving a quasi-convex objective function and infinitely many convex constraint functions with data uncertainty is considered. A surrogate duality theorem for the semi-infinite optimization problem is given under a closed and convex cone constraint qualification.

• Kyungpook Mathematical Journal
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• v.45 no.4
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• pp.491-502
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• 2005

The main purpose of this paper is to extend the exponentially convex functions which are defined and exponentially convex on a cylinderical neighborhood in the Heisenberg group. They are expanded in terms of an integral transform associated to the sub-Laplacian operator. Extension of such functions on abelian Lie group are studied in [15].

• Korean Journal of Mathematics
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• v.29 no.4
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• pp.687-704
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• 2021

The refinement of an inequality provides better convergence of one quantity towards the other one. We have established the refinements of Hadamard inequalities for Riemann-Liouville fractional integrals via strongly (α, m)-convex functions. In particular, we obtain two refinements of the classical Hadamard inequality. By using some known integral identities we also give refinements of error bounds of some fractional Hadamard inequalities.

• Journal of the Korean Mathematical Society
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• v.37 no.6
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• pp.991-1005
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• 2000

We consider a zeta function of the classical dynamics in the exterior of several convex bodies. The main result is that the poles of the zeta function cannot converge to the line of absolute convergence if the abscissa of absolute convergence of the zeta function is positive.

• Korean Journal of Mathematics
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• v.27 no.4
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• pp.879-898
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• 2019

In this paper we prove some interesting inclusions concerning the numerical range of some operators and the numerical range of theirs ranges with a convex function. Further, we prove some inequalities for the numerical radius. These inclusions and inequalities are based on some classical convexity inequalities for non-negative real numbers and some operator inequalities.

• Communications of the Korean Mathematical Society
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• v.36 no.4
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• pp.801-815
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• 2021

Ellipses have a lot of interesting geometric properties. It is quite natural to ask whether such properties of ellipses and some related ones characterize ellipses. In this paper, we study some chord properties and area properties of ellipses. As a result, using the curvature and the support function of a strictly convex curve, we establish four characterization theorems of ellipses and hyperbolas centered at the origin.

• Journal of Applied and Pure Mathematics
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• v.5 no.1_2
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• pp.95-108
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• 2023

The Hadamard type inequalities for fractional integral operators of convex functions are studied at very large scale. This paper provides the Hadamard type inequalities for refined (α,h-m)-convex functions by utilizing Liouville-Caputo fractional (L-CF) derivatives. These inequalities give refinements of already existing (L-CF) inequalities of Hadamard type for many well known classes of functions provided the function h is bounded above by ${\frac{1}{\sqrt{2}}}$.