Compact wavelet projections approximate smooth functions and commute asymptotically with multiplication

The finite orthogonal projection has kernel $K_L(x,y)={\sum }_k{\varphi }_{Lk}(x){\varphi }_{Lk}(y)$. At each $x$ only a uniformly bounded number of terms are nonzero, and each has $\mathrm{\parallel }{\varphi }_{Lk}{\mathrm{\parallel }}_1\le C{2}^{-L\mathrm{/}2}$. Thus $\int \mathrm{\mid }{K}_L(x,y)\mathrm{\mid }dy\le {\sum }_{k:{\varphi }_{Lk}(x)\mathrm{\ne }0}\mathrm{\mid }{\varphi }_{Lk}(x)\mathrm{\mid }\mathrm{\parallel }{\varphi }_{Lk}{\mathrm{\parallel }}_1\le C^{\mathrm{\prime }}$. A nonzero term requires $x,y$ to lie in one scaling support, giving the support bound. Also $K_L(x,x)={\sum }_k{\varphi }_{Lk}(x{)}^2\le C^{\mathrm{\prime }}{2}^L$; Cauchy–Schwarz on the scaling coefficients gives the claimed $L^2$ to $L^{\mathrm{\infty }}$ bound.

The levelwise tail satisfies $\mathrm{\parallel }{\sum }_k⟨g,{\psi }_{lk}⟩{\psi }_{lk}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{C}_g{2}^{-\alpha l}$. Summing its geometric series over $l>L$ proves uniform convergence and the claimed approximation rate. Its uniform limit agrees with its $L^2$ limit, hence with $g$ almost everywhere, and everywhere for continuous representatives.

For $h\in V_L$, $P_{L}h=h$. Therefore $P_L(ah)(x)-a(x)h(x)=\int K_L(x,y)(a(y)-a(x))h(y)dy$. Hölder continuity, the support bound, and the kernel integral bound give the asserted commutator estimate.

For composition put $u_j=P_{j}g$ and $v_j=u_j-u_{j-1}$, with the first finite low block treated as $u_0$. Levelwise derivative bounds and overlap give $\mathrm{\parallel }{v}_j^{(q)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_g{2}^{j(q-\alpha )}$ for $0\le q\le m$, and $\mathrm{\parallel }{u}_j^{(q)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_g^{\mathrm{\prime }}{2}^{jq}$ for $1\le q\le m$ by summing levels; also ${\sup }_j\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. The identity $e^{u_j}-e^{u_{j-1}}=v_j{\int }_0^1{e}^{u_{j-1}+t{v}_j}dt$ and repeated product and chain rules therefore give, writing $H_j=e^{u_j}-e^{u_{j-1}}$, $\mathrm{\parallel }{H}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^{-j\alpha }$ and $\mathrm{\parallel }{H}_j^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^{j(m-\alpha )}$. Indeed a derivative of order $q$ of the exponential factor is a finite sum of products of derivatives of its exponent of total order $q$, bounded by $C{2}^{jq}$; pairing with the derivative of order $m-q$ of $v_j$ gives $C{2}^{j(m-\alpha )}$.

The telescoping series $e^g=e^{u_0}+{\sum }_{j\ge 1}{H}_j$ converges uniformly. For a wavelet at level $l$, the terms with $j\ge l$ have coefficients bounded by $C{2}^{-l\mathrm{/}2}{\sum }_{j\ge l}{2}^{-j\alpha }\le C{2}^{-l(\alpha +1\mathrm{/}2)}$. For $j<l$, subtract the degree $m-1$ Taylor polynomial of $H_j$ about a point of the wavelet support; all its integrals vanish by the stated moments. The Taylor remainder gives $\mathrm{\mid }⟨H_j,{\psi }_{lk}⟩\mathrm{\mid }\le C{2}^{-l(m+1\mathrm{/}2)}{2}^{j(m-\alpha )}$. Sum the geometric series over $j<l$ to get the same $C{2}^{-l(\alpha +1\mathrm{/}2)}$ bound. The fixed smooth $e^{u_0}$ has the stronger order $2^{-l(m+1\mathrm{/}2)}$. This proves composition in the coefficient class directly, for every positive $\alpha$ including integers, without presuming a Hölder–wavelet equivalence.

For $0<\beta <\min (\alpha ,1)$ the levelwise derivative and supremum bounds for $g$ are respectively at most $C{2}^{j(1-\beta )}$ and $C{2}^{-j\beta }$. Given distance $0<\delta <1$, choose $J$ with $2^{-J-1}<\delta \le 2^{-J}$. The difference of the levels through $J$ is at most $C\delta {\sum }_{j\le J}{2}^{j(1-\beta )}\le C{\delta }^{\beta }$ by the derivative bound; the remaining difference is at most $C{\sum }_{j>J}{2}^{-j\beta }\le C{\delta }^{\beta }$. Thus $g$ is $\beta$-Hölder. The exponential has bounded derivative on the bounded range of $g$, so $e^g$ has the same Hölder bound.