let $Y_i = m(X_i) + \eta_i$, $W_j = X_j + U_j$, $E[\eta_i | X_i] = 0$ with $X_i \sim f_X$, $U_i \sim f_U$ be an errors-in-variable problem and $K_{U}(x) = \dfrac{1}{2 \pi} \int \mathrm{e}^{-itx} \dfrac{\phi_K(t)}{\phi_{U}(t/h)} \, dt$ with $\phi_K$ characteristic function of a Kernel and $U$ the characteristic function of the error-variable $U$. I already solved the integral \begin{align} E \left[ \dfrac{1}{h} K_{U} \left(\dfrac{w - W_i}{h} \right)\right] &= \int E \left[ \mathrm{e}^{-it(w - W)/h}\right] \dfrac{\phi_K(t)}{\phi_{U}(t/h)} \,dt \\ &=\dfrac{1}{2 \pi} \int E \left[\mathrm{e}^{-iz(w - W)} \right] \dfrac{\phi_K(hz)}{\phi_{U}(z)} \, dz \\ &= \dfrac{1}{2 \pi} \int \phi_X(z) \phi_U(z) \mathrm{e}^{-izw} \dfrac{\phi_K(hz)}{\phi_{U}(z)} \, dz \\ &= \dfrac{1}{h} \int K\left( \dfrac{u - x}{h} \right) f_X(u) \, du \\ &= \int K(v) f_X(x + hv) \, dv \end{align} I substituted in the second line with $t = hz$, $dt = h \, dz$ and used Plancherels theorem in the fourth line.
Now I want to show that $$ E \left[ \dfrac{1}{h} K_{U} \left( \dfrac{x - W_i}{h} \right) \, (Y_i - m(x)) \right] = \int (m(x + hv) - m(x)) \, K(v) f_X(x + hv) \, dv,$$ which is nearly the same as above. Unfortunately I don't know how to handle the $Y_i$.