Prove that $\int k(w)o(h^2w^2)dw=o(h^2)$ for $\int k(w)dw=1$ 
Suppose that $k$ is nonnegative real-valued function satisfying
  $$
\int k(w)dw=1,\quad\int wk(w)dw=0,\quad\int w^2k(w)dw=\kappa_2<\infty.\tag{$\star$}
$$
  (The limits of the integrals are all from $-\infty$ to $+\infty$.) Can you please teach me a rigorous argument to justify that
  $$
\int k(w)[\underbrace{o(w^2h^2)}_{|wh|\to 0}]dw=o(h^2)\text{ as }|h|\to 0.
$$

Context: I'm starting to study kernel density estimation. The bias for the Rosenblatt-Parzen estimator $\hat{f}(x)=\frac{1}{nh}\sum_{i=1}^nk[(X_i-x)/h]$ ($n$ is the sample size, $X_1,\ldots,X_n$ are i.i.d. draws from some unknown density $f$ that we are trying to estimate) is:
$$
E[\hat{f}(x)]-f(x)=\underbrace{\cdots}_{\text{some simplications}}=\int k(w)[f(x+hw)-f(x)]dw.
$$
If $f$ is twice differentiable on an interval containing $x$, then we may write
$$
f(x+hw)-f(x)=hwf'(x)+\frac{h^2w^2}{2}f''(x)+o(h^2w^2).\tag{$\star\star$}
$$
Then, using ($\star$) and ($\star\star$), we may simplify the bias above as
$$
E[\hat{f}(x)]-f(x)=\frac{h^2}{2}f''(x)\kappa_2+\int k(w)o(h^2w^2)dw\overset{?}=\frac{h^2}{2}f''(x)\kappa_2+o(h^2).
$$
The ? step above is claimed by my instructor but I only vaguely understand it: treat $o(h^2w^2)$ as $o(h^2)$, then factor it out, and use $\int k(w)dw=1$. But I find this unsatisfactory and would like to learn a proper way to get $\frac{h^2}{2}f''(x)\kappa_2+o(h^2)$. Thank you!
 A: Generally, this is not true: the assumption $o(w^2h^2)$ as $wh\to 0$ does not yield the conclusion. For example, let $k(w)=c/(1+w^4)$ with  $c>0$ chosen so that the integral of $k$ is $1$. The function $w^4h^4$ satisfies the assumption  $o(w^2h^2)$ as $wh\to 0$, but 
$$\int k(w) w^4h^4\,dw = \infty \quad \text{ for all } h>0$$

So, you need some global control. If $k$ is compactly supported, there is no issue. Indeed, your integral is of the form 
$$\int_{|w|\le M} k(w) g(w^2h^2)\,dw$$ where $g(t)/t\to 0$ as $t\to 0$. For every $\epsilon>0$ there is $\delta$ such that $|g(t)/t|<\epsilon$ when $|t|<\delta$. When $h<\sqrt{\delta}/M$ we have $w^2h^2  < \delta$, hence
$$\frac{1}{h^2}\left|\int_{|w|\le M} k(w) g(w^2h^2)\,dw\right| \le \frac{\epsilon}{h^2} \int_{|w|\le M} k(w) h^2 w^2\,dw \le \kappa_2 \epsilon$$ and since $\epsilon$ was arbitrary, the conclusion follows. 

If $k$ is not compactly supported, you need the assumption  $g(t)=O(t)$ as $t\to\infty$ where $g$, as above, means the factor by which you multiply $k$. Under this assumption, the proof proceeds as follows. We have $g(t)\le Ct$ for all $t$. Given $\epsilon>0$, there is $M$ such that 
$$\int_{|w|>  M}w^2 k(w)\,dw < \epsilon $$
because $w^2 k(w)$ is integrable. Hence, 
$$\int_{|w|>  M} k(w) g(h^2w^2)\,dw \le  C\int_{|w|>  M} k(w)  h^2w^2 \,dw <Ch^2 \epsilon$$
And the contribution of $|w|\le M$ is $o(h^2)$, as was shown above.
