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Fix the ordering of the orthonormal basis and write $c_m=⟨f,g_m⟩$, taking indices from $1$. The linear N-term approximation retains the first $N$ coefficients:Its index set is independent of $f$, so the map is an orthogonal projection and is linear.

For the nonlinear N-term approximation, let ${\Lambda }_N(f)$ be any set of $N$ indices with greatest $|c_m|$, resolving ties arbitrarily and padding with zero coefficients if necessary. Since $(c_m)\in {\ell }^2$, such a selection exists. ThenThe Parseval identity shows that for any fixed index set the displayed coefficients minimize the Hilbert space error. Choosing the largest squared magnitudes minimizes the omitted sum over every set of at most $N$ indices. This is the best N-term approximation; dependence of the selected indices on $f$ makes the operation nonlinear.

Write $c_m=⟨f,g_m⟩$. A linear N-term approximation fixes the indices independently of $f$, normally the first $N$ in a prescribed ordering:A best N-term approximation chooses the indices using $f$: retain $N$ coefficients of largest absolute value, resolving ties arbitrarily, and set $f_N^n={\sum }_{m\in {\Lambda }_N(f)}{c}_m{g}_m$. The Parseval identity shows why this choice is optimal:For any fixed index set, the orthogonal projection coefficients minimize the error; optimizing the set then means discarding the smallest squared coefficients. Thus “linear” requires a specified ordering, while “nonlinear” refers to the data-dependent selection.

Use the usual localized, compact support construction of an interval-adapted wavelet basis, including boundary wavelets with the stated vanishing moments, and order the linear N-term approximation by increasing resolution. Also interpret a piecewise polynomial function as having finitely many pieces. These conventions matter: regularity and vanishing moments alone, or an arbitrary enumeration, do not establish the asserted rates.

At scale $j$, a wavelet whose support lies in one polynomial piece has zero coefficient because $p<q$. Only a bounded number $B$ of wavelets per scale can meet a partition point. Their $L^1$ norms are bounded by $C{2}^{-j/2}$, and $f$ is bounded. Thus $|c_{j,k}|\le C_f{2}^{-j/2}$ andRetain the fixed number of coarse scaling function coefficients and every nonzero coefficient through level $J$. This uses at most $B(J+1)+B_0$ terms, leaving squared error at most $C{2}^{-J}$. The optimal best N-term approximation is no worse; choose $J$ proportional to $N$ to obtain $O({\omega }^N)$ for some $\omega \in (0,1)$. In contrast, retaining all wavelets through level $J$ costs $\asymp 2^J$ terms. Choosing the last complete level before $N$ givesThese are squared errors; the corresponding $L^2$ errors are $O({\omega }^{N/2})$ and $O(N^{-1/2})$.