[proofplan]
We compute the expectation and variance directly from the one-sample summand in the kernel estimator. The expectation becomes a kernel average of $f(x-hu)$ after the change of variables $y=x-hu$, and a second-order Taylor expansion near $x$ gives the bias; symmetry of $K$ removes the first-order term. For the variance, independence reduces the computation to the variance of one summand, whose second moment has leading order $h^{-1} f(x)R(K)$, while the square of the mean is lower order after division by $n$.
[/proofplan]
[step:Rewrite the expectation as a kernel average of the density near $x$]
Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space from the statement, and let $X_1,\dots,X_n: (\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ be the independent identically distributed real-valued random variables with common density $f$. For each $h>0$, define the measurable [random variable](/page/Random%20Variable)
\begin{align*}
Y_{h,1}: \Omega \to \mathbb{R}, \qquad Y_{h,1}(\omega)=\frac{1}{h}K\left(\frac{x-X_1(\omega)}{h}\right).
\end{align*}
Then
\begin{align*}
\hat f_h(x) = \frac{1}{n}\sum_{i=1}^{n} Y_{h,i},
\end{align*}
where $Y_{h,i}$ is defined from $X_i$ in the same way as $Y_{h,1}$.
Since $X_1$ has density $f$ with respect to $\mathcal{L}^1$, the expectation of $Y_{h,1}$ is
\begin{align*}
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} \frac{1}{h} K\left(\frac{x-y}{h}\right) f(y)\,d\mathcal{L}^1(y).
\end{align*}
Apply the change of variables $u=(x-y)/h$, equivalently $y=x-hu$. The one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) transforms as $d\mathcal{L}^1(y)=h\,d\mathcal{L}^1(u)$, and the domain $\mathbb{R}$ maps onto $\mathbb{R}$. Hence
\begin{align*}
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} K(u)f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
Linearity of expectation and identical distribution of the $Y_{h,i}$ give
\begin{align*}
\mathbb{E}[\hat f_h(x)]
=
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} K(u)f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
[guided]
The estimator is an average of independent copies of a single kernel summand, so the first task is to compute the mean of that one summand. Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space from the statement, and let $X_1,\dots,X_n: (\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ be the independent identically distributed real-valued random variables with common density $f$. Define
\begin{align*}
Y_{h,1}: \Omega \to \mathbb{R}, \qquad Y_{h,1}(\omega)=\frac{1}{h}K\left(\frac{x-X_1(\omega)}{h}\right).
\end{align*}
For each $i \in \{1,\dots,n\}$, define $Y_{h,i}$ by replacing $X_1$ with $X_i$. Then
\begin{align*}
\hat f_h(x) = \frac{1}{n}\sum_{i=1}^{n}Y_{h,i}.
\end{align*}
Because $X_1$ has probability density function $f: \mathbb{R}\to[0,\infty)$ with respect to $\mathcal{L}^1$, expectation against $X_1$ is integration against $f(y)\,d\mathcal{L}^1(y)$. Therefore
\begin{align*}
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} \frac{1}{h}K\left(\frac{x-y}{h}\right) f(y)\,d\mathcal{L}^1(y).
\end{align*}
Now use the substitution $u=(x-y)/h$, so $y=x-hu$. Since $h>0$, this affine change of variables maps $\mathbb{R}$ bijectively onto $\mathbb{R}$ and transforms Lebesgue measure by
\begin{align*}
d\mathcal{L}^1(y)=h\,d\mathcal{L}^1(u).
\end{align*}
Substituting gives
\begin{align*}
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} K(u) f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
Finally, expectation is linear and all $Y_{h,i}$ have the same distribution, so
\begin{align*}
\mathbb{E}[\hat f_h(x)]
=
\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}[Y_{h,i}]
=
\mathbb{E}[Y_{h,1}]
=
\int_{\mathbb{R}} K(u) f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
[/guided]
[/step]
[step:Expand the kernel average to obtain the pointwise bias]
Let $U_x\subset\mathbb{R}$ be an open interval containing $x$ on which $f$ has two continuous derivatives. Choose $\delta>0$ such that $(x-\delta,x+\delta)\subset U_x$. Since $f''$ is continuous at $x$, define the remainder function $\rho: (-\delta,\delta) \to \mathbb{R}$ as follows. For $t\ne0$, set
\begin{align*}
\rho(t)=\frac{f(x+t)-f(x)-t f'(x)-\frac{t^2}{2}f''(x)}{t^2}.
\end{align*}
Set $\rho(0)=0$.
Then $\rho(t)\to 0$ as $t\to 0$. For $|hu|<\delta$,
\begin{align*}
f(x-hu)
=
f(x)-hu f'(x)+\frac{h^2u^2}{2}f''(x)+h^2u^2\rho(-hu).
\end{align*}
Integrating this identity over $\{|u|<\delta/h\}$ against $K(u)\,d\mathcal{L}^1(u)$ gives the local contribution. The constant term over the truncated region is
\begin{align*}
f(x)\int_{|u|<\delta/h}K(u)\,d\mathcal{L}^1(u),
\end{align*}
which equals $f(x)\int_{\mathbb{R}}K(u)\,d\mathcal{L}^1(u)$ up to the tail $f(x)\int_{|u|\ge\delta/h}K(u)\,d\mathcal{L}^1(u)$. This constant-tail term satisfies
\begin{align*}
\left|f(x)\int_{|u|\ge\delta/h}K(u)\,d\mathcal{L}^1(u)\right|
\le
|f(x)|\int_{|u|\ge\delta/h}|K(u)|\,d\mathcal{L}^1(u)
\le
\frac{|f(x)|h^2}{\delta^2}\int_{|u|\ge\delta/h}u^2|K(u)|\,d\mathcal{L}^1(u)
=
o(h^2).
\end{align*}
The function $u\mapsto uK(u)$ is integrable on the bounded interval $\{|u|<\delta/h\}$ because $K$ is Lebesgue-integrable and $|u|\le \delta/h$ there. The first-order term vanishes on the symmetric interval $\{|u|<\delta/h\}$ because symmetry of $K$ makes $uK(u)$ odd. The second-order term over the truncated region equals
\begin{align*}
\frac{h^2 f''(x)}{2}\int_{|u|<\delta/h}u^2K(u)\,d\mathcal{L}^1(u),
\end{align*}
which differs from
\begin{align*}
\frac{h^2 f''(x)}{2}\int_{\mathbb{R}}u^2K(u)\,d\mathcal{L}^1(u)
=
\frac{h^2\mu_2(K)}{2}f''(x)
\end{align*}
by a tail error bounded by
\begin{align*}
\frac{h^2|f''(x)|}{2}\int_{|u|\ge \delta/h}u^2|K(u)|\,d\mathcal{L}^1(u)
=
o(h^2).
\end{align*}
The Taylor remainder satisfies
\begin{align*}
\left|h^2\int_{|u|<\delta/h}u^2K(u)\rho(-hu)\,d\mathcal{L}^1(u)\right|
\le
h^2\int_{\mathbb{R}}u^2|K(u)|\,|\rho(-hu)|\mathbb{1}_{\{|u|<\delta/h\}}\,d\mathcal{L}^1(u).
\end{align*}
Because $\rho(t)\to0$ as $t\to0$ and $\rho(0)=0$, after decreasing $\delta$ if necessary there is a finite constant
\begin{align*}
A_\rho := \sup_{|t|<\delta}|\rho(t)| < \infty.
\end{align*}
The integrand divided by $h^2$ converges pointwise to $0$ and is dominated by $A_\rho u^2|K(u)|$, which is integrable. Hence the [Dominated Convergence Theorem](/theorems/4) gives that this term is $o(h^2)$.
It remains to control the part of the expectation over $\{|u|\ge \delta/h\}$. Since $f$ is bounded, let
\begin{align*}
M := \sup_{y\in\mathbb{R}} f(y) < \infty.
\end{align*}
Then
\begin{align*}
\left|\int_{|u|\ge \delta/h}K(u)f(x-hu)\,d\mathcal{L}^1(u)\right|
\le
M\int_{|u|\ge \delta/h}|K(u)|\,d\mathcal{L}^1(u)
\le
\frac{Mh^2}{\delta^2}\int_{|u|\ge \delta/h}u^2|K(u)|\,d\mathcal{L}^1(u)
=
o(h^2).
\end{align*}
Combining the local expansion and the tail estimate yields
\begin{align*}
\mathbb{E}[\hat f_h(x)]-f(x)
=
\frac{h^2\mu_2(K)}{2}f''(x)+o(h^2).
\end{align*}
[guided]
The bias calculation is a local Taylor expansion plus a tail estimate. Let $U_x\subset\mathbb{R}$ be an open interval containing $x$ on which $f$ has two continuous derivatives. Choose $\delta>0$ such that $(x-\delta,x+\delta)\subset U_x$. Since $f''$ is continuous at $x$, [Taylor's theorem](/theorems/827) with Peano remainder gives a function $\rho: (-\delta,\delta)\to\mathbb{R}$ such that $\rho(t)\to0$ as $t\to0$ and, whenever $|t|<\delta$,
\begin{align*}
f(x+t)=f(x)+t f'(x)+\frac{t^2}{2}f''(x)+t^2\rho(t).
\end{align*}
Taking $t=-hu$ gives, for $|hu|<\delta$,
\begin{align*}
f(x-hu)=f(x)-hu f'(x)+\frac{h^2u^2}{2}f''(x)+h^2u^2\rho(-hu).
\end{align*}
We integrate this identity only over $\{|u|<\delta/h\}$ because precisely on that set the point $x-hu$ remains in the neighbourhood where the Taylor expansion is valid.
The constant term contributes $f(x)\int_{|u|<\delta/h}K(u)\,d\mathcal{L}^1(u)$. Since $\int_{\mathbb{R}}K(u)\,d\mathcal{L}^1(u)=1$, replacing the truncated integral by the full integral introduces only the tail $f(x)\int_{|u|\ge\delta/h}K(u)\,d\mathcal{L}^1(u)$. This tail is small at the needed scale because
\begin{align*}
\left|f(x)\int_{|u|\ge\delta/h}K(u)\,d\mathcal{L}^1(u)\right|
\le
|f(x)|\int_{|u|\ge\delta/h}|K(u)|\,d\mathcal{L}^1(u)
\le
\frac{|f(x)|h^2}{\delta^2}\int_{|u|\ge\delta/h}u^2|K(u)|\,d\mathcal{L}^1(u)
=
o(h^2).
\end{align*}
The first-order term is
\begin{align*}
-h f'(x)\int_{|u|<\delta/h}uK(u)\,d\mathcal{L}^1(u).
\end{align*}
The function $u\mapsto uK(u)$ is integrable on $\{|u|<\delta/h\}$ because $K$ is Lebesgue-integrable and $|u|\le\delta/h$ on this bounded interval. The interval $\{|u|<\delta/h\}$ is symmetric about $0$, and symmetry of $K$ makes $uK(u)$ odd, so this integral is $0$. The second-order term is
\begin{align*}
\frac{h^2 f''(x)}{2}\int_{|u|<\delta/h}u^2K(u)\,d\mathcal{L}^1(u).
\end{align*}
Extending this truncated integral to $\mathbb{R}$ gives the main term
\begin{align*}
\frac{h^2 f''(x)}{2}\int_{\mathbb{R}}u^2K(u)\,d\mathcal{L}^1(u)=\frac{h^2\mu_2(K)}{2}f''(x),
\end{align*}
and the discarded second-moment tail is $o(h^2)$ because $u^2|K(u)|$ is integrable.
For the Taylor remainder, we estimate
\begin{align*}
\left|h^2\int_{|u|<\delta/h}u^2K(u)\rho(-hu)\,d\mathcal{L}^1(u)\right|
\le
h^2\int_{\mathbb{R}}u^2|K(u)|\,|\rho(-hu)|\mathbb{1}_{\{|u|<\delta/h\}}\,d\mathcal{L}^1(u).
\end{align*}
Because $\rho(t)\to0$ as $t\to0$ and $\rho(0)=0$, after decreasing $\delta$ if necessary the constant
\begin{align*}
A_\rho := \sup_{|t|<\delta}|\rho(t)|
\end{align*}
is finite. After dividing by $h^2$, the integrand converges pointwise to $0$ and is dominated by the integrable function $A_\rho u^2|K(u)|$. The [Dominated Convergence Theorem](/theorems/4) therefore gives an $o(h^2)$ remainder.
It remains to control the part of the expectation with $|u|\ge\delta/h$. With $M:=\sup_{y\in\mathbb{R}}f(y)<\infty$,
\begin{align*}
\left|\int_{|u|\ge \delta/h}K(u)f(x-hu)\,d\mathcal{L}^1(u)\right|
\le
M\int_{|u|\ge \delta/h}|K(u)|\,d\mathcal{L}^1(u)
\le
\frac{Mh^2}{\delta^2}\int_{|u|\ge \delta/h}u^2|K(u)|\,d\mathcal{L}^1(u)=o(h^2).
\end{align*}
Combining the Taylor expansion, the vanishing first-order term, and the tail estimates gives
\begin{align*}
\mathbb{E}[\hat f_h(x)]-f(x)=\frac{h^2\mu_2(K)}{2}f''(x)+o(h^2).
\end{align*}
[/guided]
[/step]
[step:Compute the leading second moment of one kernel summand]
Using the density of $X_1$,
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\int_{\mathbb{R}}\frac{1}{h^2}K\left(\frac{x-y}{h}\right)^2 f(y)\,d\mathcal{L}^1(y).
\end{align*}
With the same substitution $u=(x-y)/h$, so $y=x-hu$ and $d\mathcal{L}^1(y)=h\,d\mathcal{L}^1(u)$, this becomes
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\frac{1}{h}\int_{\mathbb{R}}K(u)^2 f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
Since $f$ is continuous at $x$, $f(x-hu)\to f(x)$ for each fixed $u\in\mathbb{R}$. Since $f$ is bounded by $M$ and $K^2$ is integrable, the [Dominated Convergence Theorem](/theorems/4) gives
\begin{align*}
\int_{\mathbb{R}}K(u)^2 f(x-hu)\,d\mathcal{L}^1(u)
\to
f(x)\int_{\mathbb{R}}K(u)^2\,d\mathcal{L}^1(u)
=
f(x)R(K).
\end{align*}
Therefore
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\frac{f(x)R(K)}{h}+o\left(\frac{1}{h}\right).
\end{align*}
[guided]
The variance of the average will be controlled by the second moment of one summand, so we compute that moment explicitly. Since
\begin{align*}
Y_{h,1}
=
\frac{1}{h}K\left(\frac{x-X_1}{h}\right),
\end{align*}
and $X_1$ has density $f$, we have
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\int_{\mathbb{R}}\frac{1}{h^2}K\left(\frac{x-y}{h}\right)^2 f(y)\,d\mathcal{L}^1(y).
\end{align*}
Use the same change of variables as in the expectation calculation: $u=(x-y)/h$, equivalently $y=x-hu$. The measure transforms as $d\mathcal{L}^1(y)=h\,d\mathcal{L}^1(u)$, and the full real line maps onto the full real line. Thus
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\frac{1}{h}\int_{\mathbb{R}}K(u)^2 f(x-hu)\,d\mathcal{L}^1(u).
\end{align*}
The factor $1/h$ is the source of the leading variance order. It remains to identify the limit of the integral. For each fixed $u\in\mathbb{R}$, the condition $h\to0$ gives $x-hu\to x$, and continuity of $f$ at $x$ gives
\begin{align*}
f(x-hu)\to f(x).
\end{align*}
To pass the limit through the integral, use the [Dominated Convergence Theorem](/theorems/4). The domination is valid because $f$ is bounded: with
\begin{align*}
M := \sup_{y\in\mathbb{R}} f(y) < \infty,
\end{align*}
we have
\begin{align*}
|K(u)^2 f(x-hu)| \le M K(u)^2.
\end{align*}
The function $u\mapsto M K(u)^2$ is integrable because $R(K)<\infty$. Hence
\begin{align*}
\int_{\mathbb{R}}K(u)^2 f(x-hu)\,d\mathcal{L}^1(u)
\to
f(x)\int_{\mathbb{R}}K(u)^2\,d\mathcal{L}^1(u)
=
f(x)R(K).
\end{align*}
Multiplying by $1/h$ gives
\begin{align*}
\mathbb{E}[Y_{h,1}^2]
=
\frac{f(x)R(K)}{h}+o\left(\frac{1}{h}\right).
\end{align*}
[/guided]
[/step]
[step:Subtract the squared mean and divide by $n$]
From the bias expansion,
\begin{align*}
\mathbb{E}[Y_{h,1}]
=
f(x)+O(h^2),
\end{align*}
so
\begin{align*}
(\mathbb{E}[Y_{h,1}])^2 = O(1).
\end{align*}
Therefore
\begin{align*}
\operatorname{Var}(Y_{h,1})
=
\mathbb{E}[Y_{h,1}^2]-(\mathbb{E}[Y_{h,1}])^2
=
\frac{f(x)R(K)}{h}+o\left(\frac{1}{h}\right),
\end{align*}
because the bounded term $O(1)$ is $o(1/h)$ as $h\to0$.
The random variables $Y_{h,1},\dots,Y_{h,n}$ are independent because $X_1,\dots,X_n$ are independent and each $Y_{h,i}$ is a measurable function of $X_i$. Hence
\begin{align*}
\operatorname{Var}(\hat f_h(x))
=
\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}Y_{h,i}\right)
=
\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(Y_{h,i})
=
\frac{1}{n}\operatorname{Var}(Y_{h,1}).
\end{align*}
Substituting the single-summand variance gives
\begin{align*}
\operatorname{Var}(\hat f_h(x))
=
\frac{f(x)R(K)}{nh}+o\left(\frac{1}{nh}\right).
\end{align*}
Together with the bias expansion above, this proves both asserted asymptotic formulas.
[guided]
We now assemble the variance from the one-summand calculation. The bias expansion already proved that
\begin{align*}
\mathbb{E}[Y_{h,1}]=f(x)+O(h^2),
\end{align*}
so the squared mean satisfies
\begin{align*}
(\mathbb{E}[Y_{h,1}])^2=O(1).
\end{align*}
The second-moment computation gave
\begin{align*}
\mathbb{E}[Y_{h,1}^2]=\frac{f(x)R(K)}{h}+o\left(\frac{1}{h}\right).
\end{align*}
Since $h\to0$, every bounded quantity is $o(1/h)$. Therefore subtracting the squared mean does not change the leading order:
\begin{align*}
\operatorname{Var}(Y_{h,1})=\mathbb{E}[Y_{h,1}^2]-(\mathbb{E}[Y_{h,1}])^2=\frac{f(x)R(K)}{h}+o\left(\frac{1}{h}\right).
\end{align*}
Because $X_1,\dots,X_n$ are independent and each $Y_{h,i}$ is a measurable function of $X_i$, the random variables $Y_{h,1},\dots,Y_{h,n}$ are independent. Variance is additive for independent random variables, so
\begin{align*}
\operatorname{Var}(\hat f_h(x))=\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}Y_{h,i}\right)=\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(Y_{h,i})=\frac{1}{n}\operatorname{Var}(Y_{h,1}).
\end{align*}
Substituting the one-summand asymptotic gives
\begin{align*}
\operatorname{Var}(\hat f_h(x))=\frac{f(x)R(K)}{nh}+o\left(\frac{1}{nh}\right).
\end{align*}
Together with the bias formula, this proves both asserted asymptotic statements.
[/guided]
[/step]
