Wavelet Approximation of Besov Functions

Why Besov Smoothness Is Visible in Wavelet Scales

A wavelet expansion separates a function into a coarse component and details at progressively finer resolutions. Smooth functions tend to have small fine-scale details, but the precise meaning of “small” depends on how coefficients are grouped within each scale. Besov smoothness makes this relationship explicit: it controls scale-wise coefficient norms, while allowing different patterns of coefficients within a scale.

This connection is useful for approximation. If a function has sufficiently rapid coefficient decay, discarding all scales above a chosen resolution gives a controlled error. The rate depends not only on the smoothness index $s$, but also on the coefficient norm index $p$. In particular, an $L^2$ error bound must account for how many coefficients occur at each fine scale.

Wavelet Expansion and Besov Coefficient Norm

Work on the unit interval with endpoints identified, so there are no boundary terms. Let $\{\phi_{j_0 k}\}$ be the scaling functions at a fixed coarse resolution $j_0$, and let $\{\psi_{jk}: j\geq j_0,\ 0\leq k<2^j\}$ be an orthonormal wavelet basis. For $f\in L^2$, write

f=\sum_{k=0}^{2^{j_0}-1} a_{j_0k}\phi_{j_0k} +\sum_{j=j_0}^{\infty}\sum_{k=0}^{2^j-1}\beta_{jk}\psi_{jk}, \qquad \beta_{jk}=\langle f,\psi_{jk}\rangle.

Assume the wavelet has regularity and vanishing moments sufficient for the smoothness range under consideration: in particular, its regularity exceeds $s$, and it has more than $s$ vanishing moments. Under the standard wavelet characterization of Besov spaces, the norm below is equivalent to the usual Besov norm for this range. We use it to make the coefficient-scale relationship explicit.

Definition: For $s>0$ and $1\leq p,q\leq\infty$, the wavelet Besov coefficient norm is $$ \|f\|_{B^s_{p,q}} \asymp \|(a_{j_0k})_k\|_p + \left\| \left( 2^{j(s+1/2-1/p)} \|(\beta_{jk})_{k=0}^{2^j-1}\|_p \right)_{j\geq j_0} \right\|_q. $$ Here $\asymp$ means bounded above and below by positive constants that do not depend on $f$. For $p=\infty$ or $q=\infty$, the corresponding sequence norm is the supremum norm.

Write $A_j=2^{j(s+1/2-1/p)}\|(\beta_{jk})_k\|_p$. The Besov norm controls the sequence $(A_j)$ in $\ell^q$. The index $p$ measures how coefficients are distributed within a scale; $q$ measures how the normalized scale sizes vary across scales. The factor $2^{j(s+1/2-1/p)}$ compensates for both smoothness and the increasing number of coefficients.

Worked Example: Reading the Scale Norm for $p=2$

Suppose $f\in B^1_{2,2}$. At scale $j$, set $\|\beta_j\|_2=(\sum_k|\beta_{jk}|^2)^{1/2}$. Since $1+1/2-1/2=1$, the normalized scale size is $A_j=2^j\|\beta_j\|_2$. Thus $B^1_{2,2}$ requires the sequence $(2^j\|\beta_j\|_2)_j$ to be square-summable, in addition to a finite coarse-scale contribution.

For example, if $\|\beta_j\|_2=2^{-2j}$, then $A_j=2^{-j}$, and $\sum_{j\geq j_0} A_j^2=\sum_{j\geq j_0}2^{-2j}<\infty$. The scale sequence therefore satisfies the $q=2$ condition. This calculation is about the norm of the full coefficient vector at each scale, not the size of one selected coefficient.

Coefficient Decay Implied by Besov Smoothness

The Besov bound immediately limits every individual coefficient. For any $p\geq1$, the absolute value of a coordinate is at most the $\ell^p$ norm of its vector. Therefore

|\beta_{jk}| \leq \|\beta_j\|_p =2^{-j(s+1/2-1/p)}A_j.

If $q<\infty$, then $(A_j)\in\ell^q$ implies $A_j\to0$. In that case the displayed coefficient bound has a scale factor that is little-$o$ of $2^{-j(s+1/2-1/p)}$. If $q=\infty$, the general conclusion is only that $A_j$ is bounded, so the bound is of the stated order and need not be little-$o$.

Worked Example: A Sparse Coefficient Pattern

At each scale $j$, let exactly one coefficient be nonzero, with value $\beta_{j,k_j}=2^{-j/2}$, and let all other coefficients vanish. Consider $s=1$, $p=1$, and $q=\infty$. Here $s+1/2-1/p=1/2$, and the scale $\ell^1$ norm is $\|\beta_j\|_1=2^{-j/2}$. Consequently,

A_j=2^{j/2}\|\beta_j\|_1 =2^{j/2}2^{-j/2}=1.

Thus the detail coefficients satisfy the $B^1_{1,\infty}$ scale condition. Their scale-wise $L^2$ norm is also $2^{-j/2}$, because there is only one nonzero coefficient. This example shows why the $p$-norm within a scale matters: many zero coefficients do not make a nonzero coefficient disappear from the scale norm.

Truncation Error Through Resolution $J$

Define the approximation that retains the coarse scaling coefficients and all wavelet scales through $J$ by

f_{\leq J} = \sum_k a_{j_0k}\phi_{j_0k} +\sum_{j=j_0}^{J}\sum_k\beta_{jk}\psi_{jk}.

The exponent in the $L^2$ tail depends on $p$. Set $t=s-(1/p-1/2)_+$, where $x_+=\max(x,0)$. If $p\geq2$, then $t=s$. If $1\leq p<2$, then $t=s+1/2-1/p$. The condition $t>0$ is the relevant $L^2$ embedding condition when $p<2$.

Theorem: Suppose $f\in L^2$ has the orthonormal wavelet expansion above, belongs to $B^s_{p,q}$, and $t=s-(1/p-1/2)_+>0$. Then, for every $J\geq j_0$, $$ \|f-f_{\leq J}\|_2 \leq \frac{2^{-(J+1)t}}{\sqrt{1-2^{-2t}}} \left\|(A_j)_{j\geq j_0}\right\|_q. $$ The same bound holds with the Besov norm in place of the sequence norm, up to the equivalence constants in the wavelet characterization.

Proof. Orthonormality gives Parseval’s identity for the omitted details: $\|f-f_{\leq J}\|_2^2=\sum_{j>J}\|\beta_j\|_2^2$. For a vector with $2^j$ coordinates, the finite-dimensional norm inequalities are $\|x\|_2\leq\|x\|_p$ when $p\leq2$, and $\|x\|_2\leq(2^j)^{1/2-1/p}\|x\|_p$ when $p\geq2$. For $p<2$, these imply $\|\beta_j\|_2\leq 2^{-j(s+1/2-1/p)}A_j=2^{-jt}A_j$. For $p\geq2$, they imply $\|\beta_j\|_2\leq2^{j(1/2-1/p)}2^{-j(s+1/2-1/p)}A_j=2^{-js}A_j=2^{-jt}A_j$. Since $\sup_j A_j\leq\|(A_j)\|_q$ for every $q\in[1,\infty]$, it follows that

\|f-f_{\leq J}\|_2^2 \leq \|(A_j)\|_q^2\sum_{j>J}2^{-2jt} =\|(A_j)\|_q^2\frac{2^{-2(J+1)t}}{1-2^{-2t}}.

Taking square roots proves the bound. The assumption $t>0$ ensures that the geometric series converges. $\square$

Worked Example: An $L^2$ Tail Bound for $B^1_{1,\infty}$

Suppose $f\in B^1_{1,\infty}$ and $\sup_j A_j\leq3$, where $A_j=2^{j/2}\|\beta_j\|_1$. Here $t=1-(1-1/2)=1/2$, so the theorem gives, for any $J\geq j_0$,

\|f-f_{\leq J}\|_2 \leq \frac{3\,2^{-(J+1)/2}}{\sqrt{1-2^{-1}}} =3\sqrt{2}\,2^{-(J+1)/2} =3\,2^{-J/2}.

For example, at $J=4$, this bound is $3/4$. It is an upper bound, not a claim that the actual error equals $3/4$. The exponent is $1/2$, rather than $1$, because $p=1$ allows the scale’s coefficient mass to be concentrated in a way that is less restrictive for $L^2$ approximation.

When the $p=2$ Rate Applies

For $p=2$, orthonormality makes the scale coefficient norm exactly the $L^2$ norm of that scale. The theorem then gives an error of order $2^{-Js}$. For $p>2$, the number $2^j$ of coefficients must be included when converting an $\ell^p$ bound into an $\ell^2$ bound; that conversion cancels the $1/p$ term in the Besov weight, again giving exponent $s$. For $p<2$, the conversion does not provide that factor, and the guaranteed exponent is $s+1/2-1/p$.

Worked Example: A Dense Pattern at Every Scale

Let $s=1$, $p=\infty$, and $q=\infty$. At every scale $j$, suppose all $2^j$ coefficients equal $2^{-3j/2}$. Then $\|\beta_j\|_\infty=2^{-3j/2}$, and

A_j=2^{j(1+1/2)}\|\beta_j\|_\infty =2^{3j/2}2^{-3j/2}=1.

The normalized scale sizes are bounded, as required for $B^1_{\infty,\infty}$. The $L^2$ norm at scale $j$ is $\|\beta_j\|_2=(2^j)^{1/2}2^{-3j/2}=2^{-j}$. Thus the scale’s $L^2$ size decays as $2^{-j}$, and the $L^2$ tail after $J$ is exactly $\big(\sum_{j>J}2^{-2j}\big)^{1/2}=2^{-(J+1)}/\sqrt{1-2^{-2}}$.

Common Pitfall: Confusing One Coefficient With a Scale Norm

A common mistake is to infer the Besov condition from the decay of a few individual coefficients, or to assume that every coefficient must have the same size at a given resolution. The Besov condition is about the $\ell^p$ norm of the entire coefficient vector at each scale, followed by an $\ell^q$ condition across scales. A scale can have one relatively large coefficient and many zeros, or many small coefficients whose combined norm is substantial.

A second mistake is to report the $2^{-Js}$ $L^2$ approximation rate for every $p$. That rate follows directly for $p\geq2$, but when $p<2$ the general bound is $2^{-J(s+1/2-1/p)}$. For instance, with $s=1$ and $p=1$, the exponent is $1/2$. The condition $s+1/2-1/p>0$ also matters: without it, the geometric tail argument does not imply convergence in $L^2$.

Key takeaway: Besov smoothness controls normalized wavelet coefficient norms scale by scale. Converting those norms into an $L^2$ truncation error requires accounting for $p$, the number of coefficients at each scale, and the positivity of the resulting decay exponent.

Check Your Understanding

Use the definitions and truncation bound above to answer these questions.

For $s=2$, $p=2$, and $q=\infty$, what scale-weighted coefficient norm must remain bounded?
Why does the $L^2$ tail bound require $s+1/2-1/p>0$ when $p<2$?
For $p=1$, $s=3/4$, and $q=\infty$, calculate the exponent $t$. Does the theorem guarantee a vanishing $L^2$ truncation bound?
In the definition, what does $q$ control, and what does $p$ control?
For $p=\infty$, explain why the number of coefficients at scale $j$ enters when converting the scale norm to an $L^2$ norm.

Wavelet Approximation of Besov Functions

What You'll Learn

Why Besov Smoothness Is Visible in Wavelet Scales

Wavelet Expansion and Besov Coefficient Norm

Worked Example: Reading the Scale Norm for \(p=2\)

Coefficient Decay Implied by Besov Smoothness

Worked Example: A Sparse Coefficient Pattern

Truncation Error Through Resolution \(J\)

Worked Example: An \(L^2\) Tail Bound for \(B^1_{1,\infty}\)

When the \(p=2\) Rate Applies

Worked Example: A Dense Pattern at Every Scale

Common Pitfall: Confusing One Coefficient With a Scale Norm

Check Your Understanding

Wavelet Approximation of Besov Functions

What You'll Learn

Why Besov Smoothness Is Visible in Wavelet Scales

Wavelet Expansion and Besov Coefficient Norm

Worked Example: Reading the Scale Norm for \(p=2\)

Coefficient Decay Implied by Besov Smoothness

Worked Example: A Sparse Coefficient Pattern

Truncation Error Through Resolution \(J\)

Worked Example: An \(L^2\) Tail Bound for \(B^1_{1,\infty}\)

When the \(p=2\) Rate Applies

Worked Example: A Dense Pattern at Every Scale

Common Pitfall: Confusing One Coefficient With a Scale Norm

Check Your Understanding

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