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For a normalized kernel for density estimation $K$, with $\int K=1$, the kernel density estimator isIt replaces each atom of the empirical measure by a translated, scaled copy of $K$: averaging these small probability-density bumps gives a smooth approximation to the sampling density. Integral one is part of the probability-kernel convention; without it, the estimator generally targets $f(x)\int K$.

The printed bandwidth assumption is insufficient for the claimed normal limit. The additional condition $n{h}_n\to \infty$ is needed at points with positive density. For a counterexample, take a normal distribution density $f$, $x=0$, the unit-width box kernel and $h_n=n^{-2}$. Then $n{h}_n^3\to 0$, but the probability that any observation lies in the window is at most $n\Vert f{\Vert }_{\infty }{h}_n\to 0$. On the complementary event the estimator is zero, and $\sqrt{n{h}_n}({\hat{f}}_{n,h_n}(0)-f(0))=-f(0)/\sqrt{n}\to 0$. The limit in probability is therefore zero, whereas the stated normal limit has positive variance.

We now prove the intended pointwise central limit theorem for a kernel density estimator under $h=h_n\to 0$, $nh\to \infty$ and $n{h}^3\to 0$. The first condition also follows from the last one. A mean value theorem bound, with $L=\Vert f^{\prime }{\Vert }_{\infty }$, gives the bias of a kernel density estimatorThus the bias multiplied by $\sqrt{nh}$ tends to zero.

The Lindeberg-Feller central limit theorem says that independent centered variables in each row of a triangular array have sum converging in distribution to $N(0,{\sigma }^2)$ if the sum of their variances tends to ${\sigma }^2>0$ and, for every $\eta >0$, the sum of their truncated second moments $\mathbb{E}[Z_{n,i}^2{1}_{\{|Z_{n,i}|>\eta \}}]$ tends to zero. Put $Y_{n,i}=K((x-X_i)/h)$ and $Z_{n,i}=(Y_{n,i}-\mathbb{E}{Y}_{n,i})/\sqrt{nh}$. The independence hypothesis holds within each row. A change of variable givesContinuity of $f$ at $x$ and the compact support and boundedness of $K$ justify the limit. Also $|Z_{n,i}|\le 2\Vert K{\Vert }_{\infty }/\sqrt{nh}\to 0$, so the Lindeberg condition is eventually identically zero for every $\eta$. The sum is exactly $\sqrt{nh}({\hat{f}}_{n,h}(x)-\mathbb{E}{\hat{f}}_{n,h}(x))$. Combining the centered limit with the negligible bias by the Slutsky theorem provesIf $f(x)=0$, the centered variance tends to zero, and the Chebyshev inequality proves the corresponding degenerate limit directly.

For $f(x)>0$, let $q_{\beta }$ be the supplied $\beta$-quantile of this limiting normal distribution. Inverting the limiting central probability gives the confidence intervalContinuity of the limiting distribution gives asymptotic coverage $1-\alpha$. Equivalently, its half-width is $z_{1-\alpha /2}\sqrt{f(x)\Vert K{\Vert }_2^2/(n{h}_n)}$ under the stipulated oracle quantiles. A degenerate limit at $f(x)=0$ does not by itself justify this nominal-coverage normal interval.