= Solution
Let $\phi$ be the <scaling function> and $\psi$ an <orthonormal wavelet>, with $\phi_{j,k}(x)=2^{j/2}\phi(2^jx-k)$ and $\psi_{j,k}(x)=2^{j/2}\psi(2^jx-k)$. For $f\in L^2(\mathbb R)$, the <wavelet series> at a coarse level $j_0$ is
$$
f=\sum_{k\in\mathbb Z}\langle f,\phi_{j_0,k}\rangle\phi_{j_0,k}
+\sum_{j\ge j_0}\sum_{k\in\mathbb Z}\langle f,\psi_{j,k}\rangle\psi_{j,k}.
$$
The 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 $I_{j,k}=(k2^{-j},(k+1)2^{-j}]$ and let $I_j(x)$ be the unique such interval containing $x$. The <Haar scaling functions> are $2^{j/2}\mathbf1_{I_{j,k}}$, so the <Haar projection> is
$$
K_jf(x)=2^j\int_{I_j(x)}f(t)\,dt.
$$
For a merely locally integrable function this cell-average formula remains defined, even if a global $L^2$ orthogonal projection is unavailable. If $\|f'\|_\infty=L$, the <mean value theorem> gives <Lipschitz continuity> and, since $|t-x|\le2^{-j}$ on the cell,
$$
|K_jf(x)-f(x)|\le2^j\int_{I_j(x)}L|t-x|\,dt\le L2^{-j}.
$$
Thus the requested bound holds with $\boxed{c=\|f'\|_\infty}$, by the <Haar projection error for a Lipschitz function>.
To estimate a density, replace each scaling or wavelet coefficient $\int f\phi_{j,k}$ or $\int f\psi_{j,k}$ by the sample average of that basis function. The finite-resolution <Haar density estimator> is therefore
$$
\widehat f_j^W(x)=\sum_k\left(\frac1n\sum_{i=1}^n\phi_{j,k}(X_i)\right)\phi_{j,k}(x)
=\frac{2^j}{n}\sum_{i=1}^n\mathbf1_{I_j(x)}(X_i).
$$
It is the <histogram> on the dyadic cells, is nonnegative, and integrates to one. Its expectation is $K_jf(x)$. 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 $B=\|f\|_\infty<\infty$.
The cell probability is $p_j(x)=\int_{I_j(x)}f\le B2^{-j}$. The sample count has a <binomial distribution>, so
$$
\operatorname{Var}(\widehat f_j^W(x))=\frac{2^{2j}}n p_j(x)(1-p_j(x))\le\frac{B2^j}n.
$$
The <Cauchy-Schwarz inequality> bounds the mean centered absolute deviation by the square root of this variance. Adding the projection bias gives
$$
\mathbb E|\widehat f_j^W(x)-f(x)|\le L2^{-j}+\sqrt{\frac{B2^j}n}.
$$
Choose $j_n=\lfloor(\log_2n)/3\rfloor$, so $2^{j_n}\asymp n^{1/3}$. The two terms have the same order, and
$$
\boxed{\mathbb E|\widehat f_{j_n}^W(x)-f(x)|=O(n^{-1/3}).}
$$
This is a <bias-variance tradeoff>: refining the cells decreases the averaging bias but increases the sampling variance.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-33-haar-projections.png]
{title=Haar projections of the standard normal density at three resolutions}
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.
Back to article page
