Let $f$ be an absolutely continuous and satisfies the condition $$ |f(x+\delta)+f(x-\delta)-2f(x)|\leq \text{const}\frac{\delta}{(\log\frac{1}{\delta})^{\epsilon}}, \,\,\,\delta\in (0,1), \,\,\epsilon\in (0, 1]. $$ My question is that for any $I=[a, b],$ which is $|I|<1$ the following inequality \begin{equation} \int_{a}^{b}\Big|\frac{f(x)-f(a)}{x-a}-\frac{f(b)-f(x)}{b-x}\Big|dx\leq \text{const}\frac{|I|}{(\log\frac{1}{|I|})^{\epsilon}} \end{equation} is true? If it is how can I show?

Let $f$ be an absolutely continuous, periodic with period 1 and satisfies the condition $$ |f(x+\delta)+f(x-\delta)-2f(x)|\leq \text{const}\frac{\delta}{(\log\frac{1}{\delta})^{\epsilon}}, \,\,\,\delta\in (0,1), \,\,\epsilon\in (0, 1]. $$ My question is that for any $I=[a, b],$ which is $|I|<1$ the following inequality \begin{equation} \int_{a}^{b}\Big|\frac{f(x)-f(a)}{x-a}-\frac{f(b)-f(x)}{b-x}\Big|dx\leq \text{const}\frac{|I|}{(\log\frac{1}{|I|})^{\epsilon}} \end{equation} is true? If it is how can I show?