= Solution
A <kernel for density estimation> is a bounded nonnegative integrable function $K$ with $\int_{\mathbb R}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the <kernel density estimator> is
$$
\widehat f_n(x)=\frac1n\sum_{i=1}^nK_h(x-X_i).
$$
Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,
$$
\mathbb EV_1^2
\leq\frac{\lVert K\rVert_\infty^2}{h^2}
\int_{x-h}^{x+h}f(y)\,dy
=\frac{2\lVert K\rVert_\infty^2}{h}f_h(x).
$$
The <Cauchy-Schwarz inequality> and <variance additivity for independent random variables> give
$$
\mathbb E|\widehat f_n(x)-\mathbb E\widehat f_n(x)|
\leq\lVert K\rVert_\infty
\frac{2^{1/2}f_h(x)^{1/2}}{(nh)^{1/2}}.
$$
Nonnegativity also gives $\mathbb E|V_1-\mathbb EV_1|\leq2\mathbb EV_1\leq4\lVert K\rVert_\infty f_h(x)$. Taking the better estimate at each $x$ and applying <Tonelli theorem> proves
$$
\mathbb E\int_{-\infty}^{\infty}|\widehat f_n-\mathbb E\widehat f_n|
\leq\lVert K\rVert_\infty\int_{-\infty}^{\infty}
\min\left\{
\frac{2^{1/2}f_h^{1/2}}{(nh)^{1/2}},4f_h
\right\}.
$$
For nonnegative $a,b$ and $0\leq\theta\leq1$, $\min{a,b}\leq a^\theta b^{1-\theta}$. Taking $\theta=2\rho$ yields
$$
\mathbb E\int|\widehat f_n-\mathbb E\widehat f_n|
\leq\frac{2^{2-3\rho}\lVert K\rVert_\infty}{(nh)^\rho}
\int f_h^{1-\rho}.
$$
For $\rho>0$, the <Holder inequality> with conjugate exponents $(1-\rho)^{-1}$ and $\rho^{-1}$ gives
$$
\int f_h^{1-\rho}
\leq C_{\rho,\delta}^{\rho}
\left(\int(1+|x|)^\delta f_h(x)\,dx\right)^{1-\rho}.
$$
If $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has <probability density function> $f_h$. Since
$$
1+|X_1+hU|\leq(1+h)(1+|X_1|),
$$
the last integral is at most $(1+h)^\delta\int(1+|x|)^\delta f(x)\,dx$. Substitution proves the second displayed bound. The case $\rho=0$ is the first bound integrated using $\int f_h=1$ and follows directly.
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