2}$ by $f^\alpha$ (and $f^{3/2}$ by $f^{\alpha +1}$) in the definition of $\hat f_n(t;h)$ the only $\alpha$ for which $\int_{-B}^Bw^2g''(0)dw=0$ for all $f$ twice differentiable with $f(t)\ne0$ is $\alpha=1/2$.\]. Thus, we have $$\int_{-B}^Bw^2g^{(i)}_{t,w}(0)dw=0\ \ {\rm for}\ \ i=1,2,3,$$ and we conclude, from this, (\[ftilde\]), (\[taylor\]) and (\[firstt\]), that $$\label{ftilde2}
\tilde f(t;h)=\int_{-B}^Bg_{t,w}(hw)dw=f(t)+\frac{h^4}{4!}\int_{-B}^Bw^4E_\tau g^{(4)}(\tau h w)dw.$$ Using the formula for $g^{(4)}$ in (\[deriv\]), integrating by parts and collecting terms, it is tedious but straightforward to check that $$\begin{aligned}
\label{rem}
&&\int_{-B}^Bw^4g_{t,w}^{(4)}(0)dw\\
&&~~~~~~~~=\left[\frac{24(f')^4(t)}{f^5(t)}-\frac{36(f')^2(t)f''(t)}{ f^4(t)}+\frac{8f'(t)f'''(t)+6(f'')^2(t)}{f^3(t)}-\frac{f^{(4)}(t)}{ f^2(t)}\right]\int  v^4K(v)dv,\notag\end{aligned}$$ and to note that $$\label{rem2}
\sup_{f\in {\cal D}_{C,z}}\sup_{t\in D_r}\left|\int_{-B}^Bw^4g_{t,w}^{(4)}(0)dw\right|<\infty.$$

Now, the boundedness and uniform continuity of $K$ and its four derivatives and the facts that, for $f\in{\cal D}_{C,z}$, $f$ and its first three derivatives are Lipschitz with common constant $C$ and the fourth derivatives $f^{(4)}$ have all the same modulus of continuity $z$ at all $t$, and that $f$ is bounded away from zero in a neighborhood of $D_r$, imply that $$\label{comp}
\lim_{n\to\infty}\sup_{f\in {\cal D}_{C,z}}\sup_{0\le \tau\le 1}\sup_{w\in[-B,B]}\sup_{t\in D_r}|g_{t,w}^{(4)}(\tau h_nw)-g_{t,w}^{(4)}(0)|=0.$$ Therefore, $$\lim_{n\to\infty}\sup_{f\in {\cal D}_{C,z}}\sup_{t\in D_r}\left|\int_{-B}^Bw^4E_\tau (g_{t,w}^{(4)}(\tau h_nw)-g_{t,w}^{(4)}(0))dw\right|=0$$ and we have from this and (\[ftilde2\]) that $$\begin{aligned}
&&\sup_{f\in {\cal D}_{C,z}}\sup_{t\in D_r} \left|h_n^{-4}(\tilde f(t;h_n)-f(t)) -{1\over 4!}\int_{-B}^Bw^4g_{t,w}^{(4)}(0)dw\right|\\
&&~~~~~~~~~~~=\sup_{f\in {\cal D}_{C,z}}\sup_{t\in D_r}\left| {1\over 4!}E_\tau\int_{-B}^Bw^4 (g_{t,w}^{(4)}(\tau h_nw)-g_{t,w}^{(4)}(0))dw\right|\to
0