Reverses of Féjer's Inequalities for Convex Functions

Let $f$ be a convex function on $I$ and $a,$ $b\in I$ with $a<b.$ If $p:% \left[ a,b\right] \rightarrow \lbrack 0,\infty )$ is Lebesgue integrable and symmetric, namely $p\left( b+a-t\right) =p\left( t\right) $ for all $t\in %
\left[ a,b\right] ,$ then we show in this paper that
\begin{align*}
0& \leq \frac{1}{2}\int_{a}^{b}\left\vert t-\frac{a+b}{2}\right\vert p\left(
t\right) dt\left[ f_{+}^{\prime }\left( \frac{a+b}{2}\right) -f_{-}^{\prime
}\left( \frac{a+b}{2}\right) \right]  \\
& \leq \int_{a}^{b}p\left( t\right) f\left( t\right) dt-\left(
\int_{a}^{b}p\left( t\right) dt\right) f\left( \frac{a+b}{2}\right)  \\
& \leq \frac{1}{2}\int_{a}^{b}\left\vert t-\frac{a+b}{2}\right\vert p\left(
t\right) dt\left[ f_{-}^{\prime }\left( b\right) -f_{+}^{\prime }\left(
a\right) \right]
\end{align*}
and
\begin{align*}
0& \leq \frac{1}{2}\int_{a}^{b}\left[ \frac{1}{2}\left( b-a\right)
-\left\vert t-\frac{a+b}{2}\right\vert \right] p\left( t\right) dt\left[
f_{+}^{\prime }\left( \frac{a+b}{2}\right) -f_{-}^{\prime }\left( \frac{a+b}{%
2}\right) \right]  \\
& \leq \left( \int_{a}^{b}p\left( t\right) dt\right) \frac{f\left( a\right) +f\left( b\right) }{2}-\int_{a}^{b}p\left( t\right) f\left( t\right) dt \\ & \leq \frac{1}{2}\int_{a}^{b}\left[ \frac{1}{2}\left( b-a\right)
-\left\vert t-\frac{a+b}{2}\right\vert \right] p\left( t\right) dt\left[
f_{-}^{\prime }\left( b\right) -f_{+}^{\prime }\left( a\right) \right] .
\end{align*}.