= Solution
For an <orthonormal wavelet> construction, set
$$
\phi_{j,k}(x)=2^{j/2}\phi(2^jx-k),\qquad \psi_{j,k}(x)=2^{j/2}\psi(2^jx-k),\qquad k\in\mathbb Z.
$$
The <scaling function> $\phi$ generates the approximation spaces $V_j$, while $\psi$ generates the detail spaces $W_j$ in the <multiresolution analysis>, with $V_{j+1}=V_j\oplus W_j$. At coarse resolution zero, the family $\{\phi_{0,k}\}_k\cup\{\psi_{j,k}\}_{j\geq0,k}$ is an <orthonormal basis> of $L^2(\mathbb R)$.
For the given square-integrable function the coefficients and <wavelet series> are
$$
a_{0,k}=\int_{\mathbb R}m(x)\phi_{0,k}(x)dx,\qquad b_{j,k}=\int_{\mathbb R}m(x)\psi_{j,k}(x)dx,
$$
$$
\boxed{m=\sum_{k\in\mathbb Z}a_{0,k}\phi_{0,k}+\sum_{j=0}^\infty\sum_{k\in\mathbb Z}b_{j,k}\psi_{j,k}.}
$$
For real <wavelets> these are the real <L2 inner products>; use complex conjugates for a complex basis. The series converges in the <L2 norm>, and its coefficients satisfy the <Parseval identity for a Hilbertian basis>
$$
\|m\|_2^2=\sum_k|a_{0,k}|^2+\sum_{j\geq0,k}|b_{j,k}|^2.
$$
An $L^2$ hypothesis alone does not assert pointwise convergence.
The resolution-$J$ <wavelet projection> is the <orthogonal projection> onto $V_J$:
$$
\boxed{P_Jm=\sum_ka_{0,k}\phi_{0,k}+\sum_{j=0}^{J-1}\sum_kb_{j,k}\psi_{j,k}=\sum_ka_{J,k}\phi_{J,k},\qquad a_{J,k}=\langle m,\phi_{J,k}\rangle.}
$$
The two expressions coincide by iterating $V_{j+1}=V_j\oplus W_j$. <Orthogonality> gives $\|m-P_Jm\|_2^2=\sum_{j\geq J,k}|b_{j,k}|^2$, which tends to zero. One can equivalently write the projection with kernel $A_J(x,y)=\sum_k\phi_{J,k}(x)\phi_{J,k}(y)$, interpreted as the same <orthogonal projection>.
In <fixed-design nonparametric regression>, the observations satisfy
$$
\boxed{Y_i=m(x_i)+\varepsilon_i,\qquad x_i\text{ deterministic},\qquad \mathbb E\varepsilon_i=0,\quad \operatorname{Var}(\varepsilon_i)=\sigma^2,}
$$
with independent errors. In the usual Gaussian version $\varepsilon_i\sim N(0,\sigma^2)$. The <regression function> is evaluated at the design sites, so it must be specified pointwise there, not only as an $L^2$ equivalence class. For $x_i=i/n$, the observations concern $m$ on $[0,1]$. They cannot identify its values outside this interval. Use an <interval-adapted wavelet basis> on $[0,1]$, consisting of coarse functions $\phi_{0,k}$ and detail functions $\psi_{j,k}$; a <Haar wavelet> basis is one explicit choice.
Here is a <wavelet regression estimator> valid for every $n$. Divide $[0,1]$ into cells $I_i=((i-1)/n,i/n]$, with the value at zero assigned arbitrarily, and form the observed step function $\widetilde Y_n(t)=Y_i$ on $I_i$. For any normalized basis function $b_\lambda$, estimate its coefficient by
$$
\widehat c_\lambda=\int_0^1\widetilde Y_n(t)b_\lambda(t)dt=\sum_{i=1}^nw_{\lambda i}Y_i,\qquad w_{\lambda i}=\int_{I_i}b_\lambda(t)dt.
$$
The <wavelet regression estimator> at resolution $J=J_n$ is therefore
$$
\boxed{\widehat m_J(t)=\sum_k\widehat a_{0,k}\phi_{0,k}(t)+\sum_{j=0}^{J-1}\sum_k\widehat b_{j,k}\psi_{j,k}(t)=P_J\widetilde Y_n(t).}
$$
Only finitely many interval basis functions occur. Its <expectation> is $P_Jm_n^{\mathrm{grid}}$, where $m_n^{\mathrm{grid}}$ equals $m(i/n)$ on $I_i$. Thus approximation of $m$ includes the resolution bias and the grid discretization bias. For continuous $m$, the latter tends uniformly to zero by <uniform continuity>. No smoothness or consistency claim for arbitrary pointwise representatives of $L^2$ functions is needed to define the estimator.
The common point-evaluation version replaces each weight by $b_\lambda(i/n)/n$, giving
$$
\widehat a_{0,k}=\frac1n\sum_iY_i\phi_{0,k}(i/n),\qquad \widehat b_{j,k}=\frac1n\sum_iY_i\psi_{j,k}(i/n).
$$
These are Riemann-sum approximations to the coefficient integrals. For <Haar wavelets>, $n=2^L$, and $J\leq L$, they equal the integrated-weight coefficients exactly if the dyadic cells and pointwise basis versions are chosen right-closed. Each retained basis function is then constant on every sampling cell. This gives a concrete implementation by the <discrete wavelet transform>. A sensible finest resolution has $2^J\leq n$; beyond the grid scale there are details the data cannot resolve.
For <wavelet coefficient thresholding>, retain the coarse coefficients but discard detail coefficients dominated by noise. A hard-threshold version is
$$
\boxed{\widehat m^{\mathrm{thr}}_J=\sum_k\widehat a_{0,k}\phi_{0,k}+\sum_{j<J,k}\widehat b_{j,k}\mathbf1_{\{|\widehat b_{j,k}|>\tau_{j,k}\}}\psi_{j,k}.}
$$
Alternatively, <soft thresholding> replaces each detail coefficient by $\operatorname{sgn}(\widehat b)(|\widehat b|-\tau)_+$. The unthresholded <wavelet regression estimator> retains all details at the chosen resolution; thresholding retains evidence for local features while suppressing small fluctuations.
The thresholds can be derived from a <simultaneous wavelet coefficient noise bound>. Write $q_\lambda^2=\sum_iw_{\lambda i}^2$. Independence gives
$$
\operatorname{Var}(\widehat c_\lambda)=\sigma^2q_\lambda^2,\qquad q_\lambda^2\leq\frac1n\int_0^1b_\lambda(t)^2dt=\frac1n,
$$
where the inequality applies the <Cauchy-Schwarz inequality> on each cell of length $1/n$. In the dyadic Haar implementation, $q_\lambda^2=1/n$ exactly. For Gaussian errors, the normalized coefficient noise is standard normal. More generally, independent centred <sub-Gaussian random variables> with parameter $\sigma$ give
$$
\mathbb P\bigl(|\widehat c_\lambda-\mathbb E\widehat c_\lambda|>t\bigr)\leq2\exp\!\left(-\frac{t^2}{2\sigma^2q_\lambda^2}\right).
$$
If $N_J$ detail coefficients are considered, a <union bound> consequently justifies
$$
\boxed{\tau_\lambda=\sigma q_\lambda\sqrt{2\log(2N_J/\delta)}.}
$$
With <probability> at least $1-\delta$, every retained detail's noise is below its corresponding threshold; independence between the coefficient errors is unnecessary. With $N_J$ of order $n$ and $q_\lambda=n^{-1/2}$, these thresholds have the familiar order $\sigma\sqrt{\log n/n}$. Coefficients larger than their noise level represent potential signal, and a coefficient zero in the discretized mean is not selected on this simultaneous event. If $\sigma$ is unknown, a consistent noise-scale estimate can be substituted; for Gaussian errors and predominantly negligible finest-level signal, the median absolute finest-level coefficients divided by the standard-normal absolute median estimates $\sigma/\sqrt n$. Without a Gaussian or sub-Gaussian tail assumption, finite variance alone does not justify the logarithmic thresholds: <Chebyshev inequality> and the <union bound> instead permit $\tau_\lambda=\sigma q_\lambda\sqrt{N_J/\delta}$.
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