= Solution
Fix the ordering of the <orthonormal basis> and write $c_m=\langle f,g_m\rangle$, taking indices from $1$. The <linear N-term approximation> retains the first $N$ coefficients:
$$
\boxed{f_N^{\mathrm{lin}}=\sum_{m=1}^Nc_mg_m.}
$$
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. Then
$$
\boxed{f_N^{\mathrm{nonlin}}=\sum_{m\in\Lambda_N(f)}c_mg_m,\qquad\|f-f_N^{\mathrm{nonlin}}\|^2=\sum_{m\notin\Lambda_N(f)}|c_m|^2.}
$$
The <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.
Back to article page
