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For a strictly positive probability density function with Lipschitz bound $L$, $p_x(r)={\int }_{x-r}^{x+r}f(y)dy$ satisfies $|p_x(r)-2rf(x)|\le L{r}^2$. If $0<s<1$ and $s\le f(x{)}^2/L$, monotonicity gives $p_x^{-1}(s)\le s/f(x)$ and hence $|2f(x)p_x^{-1}(s)-s|\le L{s}^2/f(x{)}^2$. The Lipschitz density height bound prevents extending this local argument to every $s\in (0,1)$ by assuming $f(x)\ge \sqrt{L}$ for a strictly positive density.

Let $\varphi$ be the scaling function and $\psi$ an orthonormal wavelet, with ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$. For $f\in L^2(\mathbb{R})$, the wavelet series at a coarse level $j_0$ isThe series converges in the L2 norm. Its coarse part captures broad variation and successive wavelet levels add finer detail. Truncating at levels $j<J$ produces the orthogonal projection onto $V_J$; Parseval identity makes its squared approximation error the sum of the omitted squared coefficients. Merely square-integrable functions need not have pointwise convergence.

For the Haar wavelet, use the cell convention $I_{j,k}=(k{2}^{-j},(k+1)2^{-j}]$ and let $I_j(x)$ be the unique such interval containing $x$. The Haar scaling functions are $2^{j/2}{1}_{I_{j,k}}$, so the Haar projection isFor a merely locally integrable function this cell-average formula remains defined, even if a global $L^2$ orthogonal projection is unavailable. If $\Vert f^{\prime }{\Vert }_{\infty }=L$, the mean value theorem gives Lipschitz continuity and, since $|t-x|\le 2^{-j}$ on the cell,Thus the requested bound holds with $\boxed{c=\Vert f^{\prime }{\Vert }_{\infty }}$, by the Haar projection error for a Lipschitz function.

To estimate a density, replace each scaling or wavelet coefficient $\int f{\varphi }_{j,k}$ or $\int f{\psi }_{j,k}$ by the sample average of that basis function. The finite-resolution Haar density estimator is thereforeIt is the histogram on the dyadic cells, is nonnegative, and integrates to one. Its expectation is $K_{j}f(x)$. A density with bounded derivative is bounded: the Lipschitz density height bound gives $f(x{)}^2\le L$ by integrating the triangular lower envelope $(f(x)-L|t-x|{)}_{+}$. The case $L=0$ cannot be a probability density on the whole real line. Hence $B=\Vert f{\Vert }_{\infty }<\infty$.

The cell probability is $p_j(x)={\int }_{I_j(x)}f\le B{2}^{-j}$. The sample count has a binomial distribution, soThe Cauchy-Schwarz inequality bounds the mean centered absolute deviation by the square root of this variance. Adding the projection bias givesChoose $j_n=\lfloor ({\log }_{2}n)/3\rfloor$, so $2^{j_n}\asymp n^{1/3}$. The two terms have the same order, andThis is a bias-variance tradeoff: refining the cells decreases the averaging bias but increases the sampling variance.

The Haar projection averages this normal distribution density on each cell; the finer cells follow its variation more closely. A Haar density estimator replaces these exact cell probabilities by empirical frequencies and therefore also introduces sampling noise.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.