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A dyadic interval is one cell in a dyadic partition. Its children are the left and right half intervals. Disjoint dyadic intervals at a fixed level produce orthogonal indicator functions and independent Brownian motion increments.

A dyadic partition of $[0,1]$ at level $J$ consists of $2^J$ equal cells of length $2^{-J}$. Each cell splits into two cells at the next level. Use left-closed, right-open cells, adding the endpoint $1$ to the final cell when a pointwise representative is needed. These nested partitions underlie the Haar scaling functions and Haar wavelets.

For an $L$-Lipschitz function $f$ on $[0,1]$, its slopes on the dyadic partition, evaluated at a uniform distribution input, form a bounded martingale for the dyadic filtration. The uniformly integrable martingale convergence theorem yields a bounded terminal random variable $g(X)$, with $g$ chosen as a Borel measurable function by the Doob-Dynkin lemma. The terminal conditional expectation identity says that each slope is the average of $g$ over the corresponding dyadic interval. Integrating over adjacent dyadic intervals and taking a telescoping sum proves the displayed representation first at dyadic endpoints. The Lipschitz condition for $f$ and the bound $|g|\le L$ extend it to every endpoint by continuity. This proves absolute continuity of a function without presupposing a derivative.

In the Gaussian white noise model, observe the entire pathwhere $W$ is standard Brownian motion and the deterministic drift belongs to L2 space on $[0,1]$. Equivalently, $dY(t)=f(t)dt+n^{-1/2}dW(t)$. For every deterministic $h\in L^2[0,1]$, the observed stochastic integral satisfiesThe noise $W(h)=\int hdW$ is an isonormal Gaussian process. In particular its variance is $\Vert h{\Vert }_2^2$. Gaussian white noise is interpreted through these integrals, rather than as an ordinary random function with a pointwise value at every time. No smoothness of $f$ is needed.

Here is the Gaussian maximum bound without independence. Put $M={\max }_{1\le i\le N}|g_i|$. It is integrable since it is bounded by ${\sum }_i|g_i|$. For every $\lambda >0$,The second inequality uses only the moment-generating function of the standard normal distribution, so independence is unnecessary. Jensen inequality gives $e^{\lambda EM}\le E{e}^{\lambda M}$, henceMinimizing at $\lambda =\sqrt{2\log (2N)}$ proves

For the dyadic partition, let $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$, $0\le k<2^J$, putting the endpoint $1$ into the final interval. The Haar scaling functionsform an orthonormal basis of the space $V_J$ of functions constant on each interval. The Haar wavelets can be written asFor $J\ge 1$, the constant function ${\varphi }_{0,0}=1$ together with ${\psi }_{j,k}$, $0\le j<J$, is another orthonormal basis of $V_J$. Indeed, each successive level splits a cell's two constants into their sum and difference; the number of basis elements is $1+{\sum }_{j=0}^{J-1}{2}^j=2^J$. Thus the Haar projection is the orthogonal projectionOn cell $I_{J,k}$ its value is $2^J{\int }_{I_{J,k}}f(t)dt$. The first formula also applies to $J=0$.

Estimate each coefficient by its observed stochastic integral:This is the Haar projection estimator in Gaussian white noise. The noise integrals over disjoint intervals form a Gaussian vector with zero off-diagonal covariance. The principle that uncorrelated jointly Gaussian variables are independent then makes these coefficients independent. ThereforeTaking expectations shows that ${\hat{\Pi }}_{V_J}f$ is an unbiased estimator of ${\Pi }_{V_J}f$, both coefficientwise and pointwise for the stated step-function representatives.

Only one scaling function is nonzero in each cell, so the supremum norm of the error has the exact formThe endpoint convention ensures the same identity at $1$. Apply the Gaussian maximum bound without independence with $N=2^J$ to obtain the supremum norm risk of a Haar projection estimator:The factor $2^{J/2}$ comes from cellwise scaling, while the extra square root of $J+1$ comes from taking the maximum over $2^J$ Gaussian errors.

The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition: