This is a standard result and should be easy to prove. Yet I cannot figure out the final step.
Consider the Kernel density estimator:
$\hat f_h=\frac{1}{nh}\sum K\left(\frac{X_i-x}{h}\right)$
where $X_i \sim iid$ with density $f$ K integrable with $\int K(u) du=1$, $K(u)=K(-u)$,
To compute its expectation apply
$E\left(\hat f_h\right) =\int \frac{1}{h}K\left(\frac{u-x}{h}\right)f(u)du \\ = \int K\left(u\right)f(uh+x)du\\ = \int K(u)(f(x)+f'(x)uh+\frac{1}{2}f''(\gamma(u))(uh)^2)du\\ = f(x)+\int K(u) \frac{1}{2}f''(x)(uh)^2 du + \int K(u) \frac{1}{2}(f''(\gamma(u))- f''(x)) (uh)^2 du$
where $\gamma$ is between x and $uh$, by the standard Taylor expansion argument. $\int K(u)u=0 $ by symmetry of $K$. It remains to show $\int K(u) \frac{1}{2}(f''(\gamma(u))- f''(x)) (uh)^2 du=o(h^2)$.
Most sources prove it by assuming bounded support of $K$, but many kernels do not satisfy this. Other sources ignore the dependence of $\gamma$ on $u$ altogether. How do I prove it without assuming bounded support of K?
