Im considering kernels of the form

$$K_s(u) = A(s)k_s(u)I[\lvert u \lvert \leq 1]$$ and $$k_s(u) = (1-u^2)^s$$

with $r$'th derivative

$$K_s^r(u) = A(s)\frac{d^r k_s(u)}{du^r}I[\lvert u \lvert \leq 1].$$

I guess this must imply that $$\lim_{a\rightarrow \infty} K^r_s(a)f(a) = \lim_{b\rightarrow -\infty} K^r_s(b)f(b) = 0$$

Let us assume that $\lim f(t)$ exist for $t\rightarrow \infty$ and for $t \rightarrow -\infty$ then the limit must be $0$ because $f(t)$ is a density. The limits of $K_s^r(t)$ is also $0$ and therefore the above identity must hold.

Considering integration, using integration by parts, it follows that

$$\int_{-\infty}^\infty \frac{1}{h}K_s^r\left(\frac{t-x}{h}\right)f(t) dt = \left[f(t) K_s^{r-1}\left(\frac{t-x}{h}\right)\right]_{-\infty}^\infty - \int_{-\infty}^{\infty}f^1(t) K_s^{r-1}\left(\frac{t-x}{h}\right) dt= - \int_{-\infty}^{\infty}f^1(t) K_s^{r-1}\left(\frac{t-x}{h}\right) dt$$

because $$\left[f(t) K_s^{r-1}\left(\frac{t-x}{h}\right)\right]_{-\infty}^\infty =0$$

and using $f^1(t)$ as first derivative of density.

This can then be generalised, by application of integration by parts $s$ times, to

$$\int_{-\infty}^\infty \frac{1}{h^s}K_s^r\left(\frac{t-x}{h}\right)f(t) dt = (-1)^s\int_{-\infty}^{\infty}f^s(t) K_s\left(\frac{t-x}{h}\right) $$

if in each step it can be assumed that the limits $f^r(t)$ exist. Or am I missing something?


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