Let's say the kernel density estimator is given by
$$\hat f(x) = \frac{1}{nh_n} \sum_{i=1}^n K\left(\frac{X_i-x}{h_n}\right),$$ where $h_n \to 0$, $nh_n \to \infty$, $K$ a symmetric probability distribution function with finite variance, and $\int K^2 < \infty$, and the true pdf $f$ is such that $f'$(instead of $f''$) is bounded and continuous (And samples $X_1, \ldots, X_n$ comes from $f$).
Normally, the bias for KDE is $$\frac{1}{2} h_n^2 (\sigma_K^2 f''(x) + o(1)),$$ where $\sigma_K^2 = \int u^2 K(u) du$.
But with the assumption that $f'$ is bounded instead of $f''$ is bounded, does the bias change?
Each $K\left(\frac{X_i-x}{h_n}\right)$ is i.i.d, so consider $$E\left(\frac{1}{h_n} K\left(\frac{X_i-x}{h_n}\right) \right) = \int_{-\infty}^\infty \frac{1}{h_n} K\left(\frac{y-x}{h_n}\right)f(y) dy.$$
Let $u = \frac{y-x}{h_n}$, then $du = \frac{1}{h_n} dy$, $y = x+u h_n$ so we have $\int_{-\infty}^\infty K(u) f(x+uh_n)$.
So by the mean value theorem(since $f$ is continuous and differentiable too) for some $x_1 \in (x, x+uh_n)$, $f(x+uh_n) - f(x) = f'(x_1)(u h_n)$.
Now, $x_1$ is a function of $u$.
Write $$f(x+u h_n) = f(x) + h_n u f'(x) - h_n u(f'(x) - f'(x_1)).$$
Then $$\int K(u) f(x+ u h_n) = f(x) + h_n \int u K(u) (f'(x) - f'(x_1)) du.$$
Now, $u K(u) (f'(x) - f'(x_1))$ is dominated since $f'$ is bounded, and its limit as $n \to \infty$ is $0$ since $x_1 \to x$ as $n \to \infty$.
Hence, the expected valus is $f(x) + h_n o(1)$, and the bias is $h_n o(1) = o(h_n) = o(1)$?
This seems to be wrong, but I don't know where.
The proof was inspired since in class, we wrote $$f(x-h_n u) - f(x) = -h_n u f'(x) + \frac{1}{2} h_n^2 u^2 f''(x) + \frac{1}{2} h_n^2 u^2 (f''(x-\lambda h_n u) - f''(x)),$$ where $0 \leq \lambda \leq 1$, and evaluated the integral of each term, and we said $\frac{1}{2} h_n^2 u^2$ ($f''(x-\lambda h_n u) - f''(x)) K(u)$ was dominated since $f''$ was bounded.
Am I applying the idea from this proof wrong?
