Reverses of Féjer's Inequalities for Convex Functions

Document Type : Research Articles

Author

College of Engineering \& Science Victoria University, PO Box 14428 Melbourne City, MC 8001, Australia

Abstract

Let

f

be a convex function on

I

and

a,

b \in I

with

a < b .

p :

is Lebesgue integrable and symmetric, namely

p (b + a - t) = p (t)

for all

t \in [a, b],

then we show in this paper that

\begin{aligned} 0 & \leq \frac{1}{2} \int_{a}^{b} | t - \frac{a + b}{2} | p (t) d t [f_{+}^{'} (\frac{a + b}{2}) - f_{-}^{'} (\frac{a + b}{2})] \\ \leq \int_{a}^{b} p (t) f (t) d t - (\int_{a}^{b} p (t) d t) f (\frac{a + b}{2}) \\ \leq \frac{1}{2} \int_{a}^{b} | t - \frac{a + b}{2} | p (t) d t [f_{-}^{'} (b) - f_{+}^{'} (a)] \end{aligned}

and

\begin{aligned} 0 & \leq \frac{1}{2} \int_{a}^{b} [\frac{1}{2} (b - a) - | t - \frac{a + b}{2} |] p (t) d t [f_{+}^{'} (\frac{a + b}{2}) - f_{-}^{'} (\frac{a + b}{2})] \\ \leq (\int_{a}^{b} p (t) d t) \frac{f (a) + f (b)}{2} - \int_{a}^{b} p (t) f (t) d t \\ \leq \frac{1}{2} \int_{a}^{b} [\frac{1}{2} (b - a) - | t - \frac{a + b}{2} |] p (t) d t [f_{-}^{'} (b) - f_{+}^{'} (a)] . \end{aligned}

Keywords