Let $V$ be a open convex set. We will say that a function $m$ has the order of smoothness $p$ on $V$ with $p=l+\gamma$, where $l \geq0$ is an integer and $0<\gamma\leq1$ and will write $m\in H_{p}(V)$, if (i) derivatives $m^{(u)}(x)$ exists and are continuous for all $x \in V$ and $u\leq l$, (ii) there exists a constant $C>0$ such that
\begin{equation} \mid m^{(l)}(x_1)-m^{(l)}(x_2) \mid\leq C \mid x_1 -x_2 \mid ^{\gamma}, \end{equation}
for all $x_1,x_2\in V$.
Assume the density $f(x)$ of $x$ is positive on $V$ and $f\in H_{p_{1}}(V)$, and $p_{1}=1+\gamma$, $\gamma>0$. Set $p_n (x)=\delta_n \int_{[-1,1]}f(x+\delta_nt)dt$, where $\delta_n=n^{-\kappa}$, and $\kappa\in(0,1)$.
Also, we use notation $r_{n}(x)=O_{L_{2}}(a_n)$ meaning that, as $n\rightarrow \infty$, $E(r_{n}(x)/a_{n})^2$ is bounded.
My Question is: for $x \in V$, why do we have that
\begin{equation} \frac{p_n(x)}{2 \delta_n}=f(x)+O_{L_{2}}(\delta^{1+\gamma})? \end{equation}
It is straightforward, I guess, to get a $O_{L_{2}}(\delta)$ term by taylor expansion. But how to get a smaller term $O_{L_{2}}(\delta^{1+\gamma})$? Thank you!