= Solution
Use localized <tensor-product wavelets> and finitely many bounded polynomial pieces, with the <boundary wavelets> adapted as in part (b). A rectifiable smooth curve of length $\ell$ meets $O(1+\ell2^j)$ dyadic squares of side $2^{-j}$: subdividing an arclength parametrization into pieces of length at most $2^{-j}$ covers it by that many balls, each meeting only a bounded number of squares. Enlarging squares by the fixed support diameter preserves the count.
Each normalized two-dimensional <wavelet> has $L^1$ <norm> $O(2^{-j})$. A coefficient meeting the curve is therefore $O(2^{-j})$, and the total squared energy of these coefficients at level $j$ is $O(2^{-j})$. If $p<q$, all other coefficients vanish by the <vanishing moments>. Keeping the curve coefficients through level $J$ costs $O(2^J)$ and leaves squared error $O(2^{-J})$.
The printed part (e) does not repeat $p<q$. The bound still holds for any fixed polynomial degree when $q\ge1$: on a smooth piece, a <Taylor polynomial> in the variable carrying a <wavelet> gives coefficient size $O(2^{-(q+1)j})$. There are $O(4^j)$ such coefficients, so their squared energy is $O(2^{-2qj})$. Retain all coefficients through $K=\lfloor J/2\rfloor$, and curve coefficients through $J$. The cost is $O(4^K+2^J)=O(2^J)$ and the omitted squared energy is
$$
O(2^{-2qK}+2^{-J})=O(2^{-J}).
$$
The <best N-term approximation> is at least as good as this selection, proving
$$
\boxed{\|f-f_N^n\|_2^2=O(N^{-1}).}
$$
This argument covers the unqualified finite-degree clause without adding an unnecessary restriction $p<q$.
Back to article page
