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**Lemma 12.** We consider $q \ge 2$ and two non-negative maps $\zeta_0, \zeta_1$ on $[0, +\infty)$. We suppose that these functions are non-increasing, compactly supported on $[0, 1]$ and with continuous derivatives up to order $q$. Moreover, $\zeta_\delta(0) = 1$, $\zeta_\delta(1) = 0$, and $\zeta_\delta^{(s)}(0) = \zeta_\delta^{(s)}(1) = 0$ for all $\delta \in \{0, 1\}$ and $s \in \{1, \dots, q-1\}$, where $\zeta_\delta^{(s)}$ denotes the $s$th derivative of $\zeta_\delta$. We consider a positive integer $r$, three positive numbers $\sigma, b_0, \ell_0, x_0 \in (0, r\ell_0]$, $\varepsilon = \sigma/r$, and set for all $k \ge 0$,

$$ 
\begin{aligned}
b_k &= b_0(1 + \varepsilon)^{-k} \\
\ell_k &= \ell_0(1 + \varepsilon)^{2k} \\
x_{k+1} &= x_k + \ell_k.
\end{aligned}
$$

We then define for all $x \ge 0$ and $\delta = (\delta_k)_{0 \le k \le r} \in \{0, 1\}^{r+1}$,

$$ 
\begin{aligned}
g_{\delta}(x) = b_0 \mathbb{1}_{[0,x_0)} &+ \frac{1}{1+\varepsilon} \sum_{k=0}^{r} b_k \left[1 + \varepsilon \zeta_{\delta_k} \left(\frac{x-x_k}{\ell_k}\right)\right] \mathbb{1}_{[x_k,x_{k+1})}(x) \\
&+ b_{r+1} \zeta_0 \left(\frac{x-x_{r+1}}{r\ell_0}\right) \mathbb{1}_{[x_{r+1},+\infty)}(x).
\end{aligned}
\quad (4.35) $$

We extend $g_\delta$ to an even function on $\mathbb{R}$.

This function $g_\delta$ has the following properties: it is unimodal and such that $g_\delta(x_k) = b_k$ for all $k \in \{0, \dots, r\}$, $\delta \in \{0, 1\}^{r+1}$. Moreover, $g_\delta$ lies in $\mathfrak{B}_{p,\infty}^\alpha(cR)$ for all $p \in (0, +\infty]$, $\alpha \in (\max\{1, 1/p - 1\}, q)$ and $R > 0$ satisfying

$$ b_0 r^{1/p-1} \le R \ell_0^{\alpha-1/p} \quad (4.36) $$

$$ b_0 \ell_0^{1/p} r^{1/p} \le R. \quad (4.37) $$

Above, c only depends on the wavelet basis, $\zeta_0, \zeta_1, p, q, \alpha$ and $\sigma$.

**Lemma 13.** Consider some $q \ge 2$. There exist two functions $\zeta_0, \zeta_1$ fulfilling the assumptions of Lemma 12. They satisfy

$$ \int_0^1 \zeta_0(x) \, dx = \int_0^1 \zeta_1(x) \, dx \quad (4.38) $$

and do not coincide almost everywhere on $[0, 1]$.

Let $\bar{r}$ be the largest integer such that

$$ \bar{r} \le R^{\beta_1} M^{\beta_2} n^{1-\gamma}. $$

We consider the smallest integer $q$ larger than $\alpha$, and the set $\mathcal{D}$ given in the first part of Lemma 11. Let then $\zeta_0, \zeta_1$ be the maps given by Lemma 13. We consider $a > 0$ and set

$$ 
\begin{aligned}
& \varepsilon = 1/\bar{r} \\
& \ell_0^{2(\alpha-1/p+1/2)} = aR^{-2}n^{-1}r^{2/p}
\end{aligned}
$$
2003.

UNIVERSITÉ JEAN MONNET, ICJ UMR5208, CNRS, ECOLE CENTRALE DE LYON, INSA LYON, UNIVERSITE
CLAUDE BERNARD LYON 1, 42023 SAINT-ETIENNE, FRANCE.

Email address: mathieu.sart@univ-st-etienne.fr
$$b_0^2 = ar^2(n\ell_0)^{-1}.$$

Note that (4.36) holds true. Moreover, $b_0 \ell_0^{1/p} \varepsilon^{-1/p}$ tends to 0 when $n$ goes to infinity when $\theta < p/2$. Hence (4.37) is true. When $\theta = p/2$,

$$b_0 \ell_0^{1/p} \varepsilon^{-1/p} \le 2^{(2\alpha p - 3 + p + 2/p)/(p + 2\alpha p - 2)} a^{\alpha/(1 + 2\alpha - 2/p)} M^{1/p}.$$

We thus also have (4.37) when $M \le R^p$ and $a$ small enough.

We consider $x_0 = r\ell_0$, $\delta \in \mathcal{D}$, and the map $g_\delta$ defined by (4.35). Let then $I$ be the value of the integral in (4.38), and

$$I_{0,2} = \int_0^1 \zeta_0^2(x) \, dx$$

$$I_{1,2} = \int_0^1 \zeta_1^2(x) \, dx.$$

Elementary maths lead to:

$$\int (g_{\delta}(x))^2 dx = \frac{2ar^3}{n} + \frac{8a(1 + 2\varepsilon I)(2 + \varepsilon)}{n(1 + \varepsilon)^2 \varepsilon^3} + \frac{8a(\bar{r}I_{1,2} + (\bar{r} + 1)I_{0,2})}{n(1 + \varepsilon)^2} \\ + \frac{ar^3}{n(1 + \varepsilon)^2 (r+1)} I_{0,2}.$$

This integral does not depend on $\delta$ and tends to 0 when $n$ goes to infinity. Besides,

$$\int (g_{\delta}(x))^{2\theta} dx \le c_1 b_0^{2\theta} \ell_0 r,$$

for all $n$ large enough and some $c_1 > 0$ only depending on $\zeta_0$, $\zeta_1$ and $\theta$. In particular,

$$\int (g_{\delta}(x))^{2\theta} dx \le c_1 a^{(1-2\theta/p+2\alpha\theta)/(1+2\alpha-2/p)} M \\ \le c_1 M,$$

when $a \le 1$. We also have $\text{supp } g_\delta \subset [-c_1 M, c_1 M]$ when $\theta = 0$.

We now apply Lemma 12 with suitable values of parameters to get a unimodal non-negative function $\zeta \in \mathfrak{B}_{p,\infty}^{\alpha}(c_2 R)$, compactly supported on $[-1, 1]$, and such that $\int (\zeta(x))^2 dx > 1$, $\int (\zeta(x))^{2\theta} dx \le M$ (up to an increase of $R, M$). We then consider $s \in (0, 1)$ and set

$$f_{\delta} = (g_{\delta} + s\zeta)^2.$$

As $x_0 > 1$ for $n$ or $M$ large enough,

$$\int f_{\delta}(x) \, dx = \int (g_{\delta}(x))^2 \, dx + 2sb_0 \int \zeta(x) \, dx + s^2 \int (\zeta(x))^2 \, dx.$$
Note that $b_0$ tends to 0 as $n$ goes to infinity. We may hence find $s \in (0, 1)$, not depending on $\delta$, such that this integral is 1.

By putting all these results together, by using Lemma 12 and Proposition 3.1, we get that $f_\delta$ is a density lying in $\mathcal{W}\mathcal{S}^\alpha_{p,\theta}(c_3R, c_4M)$ and such that $\sqrt{f_\delta} \in \mathfrak{B}^\alpha_{p,\infty}(c_3R)$. Its support is included in $[-c_1M, c_1M]$ when $\theta = 0$.

Since $g_\delta$ is unimodal, we have $g_\delta(x) \in [b_{r+1}, b_0]$ for all $x \in [-x_{r+1}, x_{r+1}]$. When $x \notin [-x_{r+1}, x_{r+1}]$, $g_\delta(x) = g_{\delta'}(x)$ for all $\delta, \delta'$. An elementary inequality yields $b_0 \le e^3 b_{r+1}$. We then deduce from Lemma 2.7 of [Tsy08], that for all $\delta \ne \delta' \in \mathcal{D}$,

$$
\begin{align*}
K(f_\delta, f_{\delta'}) &\le b_{r+1}^{-2} \int (f_\delta(x) - f_{\delta'}(x))^2 dx \\
&\le b_{r+1}^{-2} \int (g_\delta(x) - g_{\delta'}(x))^2 (g_\delta(x) + g_{\delta'}(x) + 2s\zeta(x))^2 dx \\
&\le b_{r+1}^{-2} \int (g_\delta(x) - g_{\delta'}(x))^2 (g_\delta(x) + g_{\delta'}(x))^2 dx \\
&\le 4e^6 \int (g_\delta(x) - g_{\delta'}(x))^2 dx \\
&\le \frac{8e^6}{(1+\varepsilon)^2} \varepsilon^2 \sum_{k=0}^r b_k^2 \ell_k |\delta_k - \delta_{k'}| \int (\zeta_0(x) - \zeta_1(x))^2 dx \\
&\le c_5 b_0^2 \ell_0 \varepsilon^2 r \\
&\le c_6 ar/n,
\end{align*}
$$

where $c_5, c_6$ only depend on $\zeta_0, \zeta_1$. We now choose a small enough to ensure that (4.34) holds true.

Moreover,

$$
\begin{align*}
h^2(f_\delta, f_{\delta'}) &= \frac{\varepsilon^2}{(1+\varepsilon)^2} \sum_{k=0}^{r} b_k^2 \ell_k |\delta_k - \delta_{k'}| \int_0^1 (\zeta_0(x) - \zeta_1(x))^2 \, dx \\
&\geq c_7 b_0^2 \ell_0 \varepsilon^2 \Delta(\delta, \delta'), \\
&\geq c_8 (a/n) \Delta(\delta, \delta'),
\end{align*}
$$

for some $c_7, c_8 > 0$.

We may hence apply Lemma 11 with $\eta$ of the order of $n^{-1}$. This leads to

$$
\mathcal{R}(\mathcal{F}) \geq c_9 r/n,
$$

where $\mathcal{F} = \{f_\delta, \delta \in \mathcal{D}\}$. We conclude using the definition of $r$. $\square$

<figure><img src="image_7.png" /></figure>

*Proof of Lemma 12.* The only delicate point is $g_\delta \in \mathfrak{B}^\alpha_{p,\infty}(cR)$ for some $c$. We prove this result when $p < \infty$. The proof when $p = \infty$ is obtained by making slight modifications. We suppose without loss of generality that $q \ge 2$ is the smallest integer larger than $\alpha$.
The qth order difference operator evaluated in gδ is defined for h > 0 and
x ∈ ℝ by

$$
\Delta_h^q g_\delta(x) = \sum_{j=0}^{q} \binom{q}{j} (-1)^{q-j} g_\delta(x + jh).
$$

Section 7 in Chapter 2 of [DL93] gives

$$
\Delta_h^q g_\delta(x) = h^{q-1} \int (\Delta_h^1 g_\delta)^{(q-1)} (x + th) M(t) \, dt,
$$

where M is a compactly supported density function on $[0, q - 1]$ and bounded
by 1. Therefore,

$$
\Delta_h^q g_\delta(x) = h^{q-1} \int \left[ g_\delta^{(q-1)}(x + (t+1)h) - g_\delta^{(q-1)}(x+th) \right] M(t) \, dt. \quad (4.39)
$$

We consider $k \in \{0, \dots, r\}$ and the map $\zeta_{k,\delta_k}$ defined for $x \ge 0$ by

$$
\zeta_{k, \delta_k}(x) = \left[1 + \varepsilon \zeta_{\delta_k}((x - x_k)/\ell_k)\right] \mathbb{1}_{[x_k, x_{k+1})}(x).
$$

When $k = r + 1$, we set

$$
\zeta_{r+1}(x) = \zeta_0 \left( \left( \frac{x - x_{r+1}}{r \ell_0} \right) \mathbf{1}_{[x_{r+1}, \infty)}(x) \right).
$$

When $x < 0$, we put $\zeta_{k,\delta_k}(x) = \zeta_{k,\delta_k}(|x|)$ and $\zeta_{r+1}(x) = \zeta_{r+1}(|x|)$. These maps are $q-1$ times differentiable at all points except at $-x_{r+1}, -x_{k+1}, -x_k, x_k, x_{k+1}, x_{r+1}$. We nevertheless set

$$
\begin{align*}
\zeta_{k, \delta_k}^{(q-1)}(-x_{k+1}) &= \zeta_{k, \delta_k}^{(q-1)}(-x_k) \\
&= \zeta_{k, \delta_k}^{(q-1)}(x_k) \\
&= \zeta_{k, \delta_k}^{(q-1)}(x_{k+1}) \\
&= 0
\end{align*}
$$

and $\zeta_{r+1}^{(q-1)}(x_{r+1}) = \zeta_{r+1}^{(q-1)}(-x_{r+1}) = 0$ so that

$$
g_{\delta}^{(q-1)}(x) = \frac{1}{1+\varepsilon} \sum_{k=0}^{r} b_k \zeta_{k, \delta_k}^{(q-1)}(x) + b_{r+1} \zeta_{r+1}^{(q-1)}(x) \quad (4.40)
$$

holds true on $\mathbb{R}$.

Define now

$$
\bar{\alpha} = \alpha - (q - 1) \in [0, 1).
$$

Since $\zeta_{\delta_k}^{(q-1)}$ is compactly supported with a continuous derivative on $\mathbb{R}$,

$$
|\zeta_{\delta_k}^{(q-1)}(b) - \zeta_{\delta_k}^{(q-1)}(a)| \leq c_1 |b - a|^\alpha
$$
for all $a, b \in \mathbb{R}$, $\delta_k \in \{0, 1\}$, and some $c_1$ only depending on $\zeta_0, \zeta_1, \alpha, q$. Therefore, for all $a, b \ge 0$,

$$ \left| \zeta_{k, \delta_k}^{(q-1)}(b) - \zeta_{k, \delta_k}^{(q-1)}(a) \right| \le c_1 \varepsilon \ell_k^{-\alpha} |b-a|^{\bar{\alpha}}. \quad (4.41) $$

The same result holds true when $a, b \le 0$. Suppose now that $a, b$ have opposite signs, say $a \ge 0$ and $b \le 0$. Then, $\zeta_{k, \delta_k}^{(q-1)}(b) = 0$ if $|b| \le x_k$. Otherwise, we use $\zeta_{k, \delta_k}^{(q-1)}(x_k) = 0$ to get

$$ \begin{aligned} \left| \zeta_{k, \delta_k}^{(q-1)}(b) \right| &= \left| \zeta_{k, \delta_k}^{(q-1)}(|b|) - \zeta_{k, \delta_k}^{(q-1)}(x_k) \right| \\ &\le c_1 \varepsilon \ell_k^{-\alpha} (|b| - x_k)^{\bar{\alpha}} \\ &\le c_1 \varepsilon \ell_k^{-\alpha} (|b| + a)^{\bar{\alpha}} \\ &\le c_1 \varepsilon \ell_k^{-\alpha} |b - a|^{\bar{\alpha}}. \end{aligned} $$

A similar reasoning applies to $|\zeta_{k, \delta_k}^{(q-1)}(a)|$. To sum up, (4.41) holds true for all $a, b \in \mathbb{R}$ (to within a multiplication of $c_1$ by 2). Likewise,

$$ \begin{aligned} \left| \zeta_{r+1}^{(q-1)}(b) - \zeta_{r+1}^{(q-1)}(a) \right| &\le c_2 (r\ell_0)^{-\alpha} |b-a|^{\bar{\alpha}} \\ &\le c_3 \varepsilon \ell_0^{-\alpha} |b-a|^{\bar{\alpha}} \end{aligned} $$

as $\alpha > 1$.

Consider now $x \in \mathbb{R}$, $h > 0$ and $t \in [0, q-1]$. There can only be one non-zero term in the sum of (4.40). Therefore,

$$ \left| g_{\delta}^{(q-1)}(x + (t+1)h) - g_{\delta}^{(q-1)}(x + th) \right| \le c_4 b_0 \varepsilon \ell_0^{-\alpha} h^{\bar{\alpha}}. $$

We deduce from (4.39) that $|\Delta_h^q g_\delta(x)|$ satisfies

$$ |\Delta_h^q g_\delta(x)| \le c_5 b_0 \varepsilon \ell_0^{-\alpha} h^{\bar{\alpha}}. $$

Moreover, $\Delta_h^q g_\delta(\cdot)$ is compactly supported on $[-x_{r+1} - r\ell_0 - qh, x_{r+1} + r\ell_0]$. Observe that

$$ \begin{aligned} x_{r+1} &\le x_0 + \sum_{j=0}^{r} \ell_j \\ &\le r\ell_0 + \ell_0 \sum_{j=0}^{r} (1 + \varepsilon)^{2j} \\ &\le c_6 r\ell_0. \end{aligned} $$

We hence get when $h \le r\ell_0$,

$$ \begin{aligned} \int_{-\infty}^{\infty} |\Delta_h^q g_\delta(x)|^p dx &\le c_7 r\ell_0 (b_0 \varepsilon \ell_0^{-\alpha} h^{\bar{\alpha}})^p \\ &\le c_8 R^p h^{\alpha p}, \end{aligned} $$
where the last inequality relies on (4.36). When $h \ge r\ell_0$, we bound the $\mathbb{L}^p$ (quasi) norm of $|\Delta_h^q g_\delta(\cdot)|$ as follows:

$$ 
\begin{align*}
\int_{-\infty}^{\infty} |\Delta_h^q g_\delta(x)|^p dx &\le c_9 \int_{-\infty}^{\infty} |g_\delta(x)|^p dx \\
&\le c_{10} \left[ b_0^p x_0 + \sum_{k=0}^{r} b_k^p \ell_k + b_{r+1}^p \ell_0 r \right] \\
&\le c_{10} \left[ b_0^p r \ell_0 + b_0^p \ell_0 \sum_{k=0}^{r} (1+\varepsilon)^{(2-p)k} + b_0^p \ell_0 r \right] \\
&\le c_{11} b_0^p r \ell_0.
\end{align*}
$$

We apply (4.36):

$$ \int_{-\infty}^{\infty} |\Delta_h^q g_\delta(x)|^p dx \le c_{12} R^p r^p \ell_0^{\alpha p}. $$

Since $\alpha > 1$, $r^p \le r^{\alpha p}$ and the right-hand side of the last inequality is not larger than $c_{12} R^p h^{\alpha p}$.

We may also use (4.37) in place of (4.36) to get $\|g_\delta\|_p \le c_{13} R$. The conclusion $g_\delta \in \mathfrak{B}_{p,\infty}^\alpha(c_{14}R)$ then stems from another (equivalent) definition of Besov balls. See Section 10 in Chapter 2 of [DL93] for more details. $\square$

*Proof of Lemma 13.* We introduce for all $p_1, p_2 \ge 1$ and $x \in [0, +\infty)$,

$$ f_{p_1, p_2}(x) = (1 - x^{p_1})^{p_2} \mathbb{1}_{[0,1]}(x). $$

This map is non-increasing on $[0, 1]$, $p-1$ times differentiable where $p = \min\{p_1, p_2\}$, and such that $f_{p_1, p_2}(0) = 1$, $f_{p_1, p_2}(1) = 0$, $f_{p_1, p_2}^{(s)}(0) = f_{p_1, p_2}^{(s)}(1) = 0$ for all $s \in \{1, \dots, p-1\}$. It fulfils the assumptions of Lemma 12 when $p \ge q+2$. We define the first function $\zeta_0$ by $\zeta_0 = f_{q+3, q+3}$.

We then consider $t \in [0, 1]$ and set

$$ \zeta_1 = t f_{q+3, q+2} + (1-t) f_{q+2, q+3}. $$

Since $f_{q+2, q+3} \le f_{q+3, q+3} \le f_{q+3, q+2}$, the integral of $\zeta_1$ evolves continuously from $\int f_{q+2, q+3}$ to $\int f_{q+3, q+2}$ as $t$ varies from 0 to 1. There is therefore some $t \in [0, 1]$ such that (4.38) holds true.

We conclude by noticing that $\zeta_1$ satisfies the assumptions of Lemma 12 and cannot coincide almost everywhere with $\zeta_0$ because they are polynomials of different degrees on $[0, 1]$. $\square$

### 4.8. Proof of the claims of Section 2.4

We introduce the linear estimators

$$ \tilde{f} = \sum_{k \in K_{-1}} \hat{\alpha}_{0,k} \bar{\phi}_{0,k} + \sum_{j=0}^{J} \sum_{k \in K_j} \hat{\beta}_{j,k} \bar{\psi}_{j,k}, $$

$$ \tilde{f}' = \sum_{k \in \mathbb{Z}} \hat{\alpha}_{0,k} \bar{\phi}_{0,k} + \sum_{j=0}^{J} \sum_{k \in \mathbb{Z}} \hat{\beta}_{j,k} \bar{\psi}_{j,k} $$
where

$$ \hat{\alpha}_{0,k} = \frac{1}{n} \sum_{i=1}^{n} \phi_{0,k}(X_i) \quad \text{and} \quad \hat{\beta}_{j,k} = \frac{1}{n} \sum_{i=1}^{n} \psi_{j,k}(X_i). $$

We define $\bar{f}(x)$ and $\bar{f}'(x)$ for $x \in \mathbb{R}$ by $\bar{f}(x) = \mathbb{E}[\tilde{f}(x)]$, $\bar{f}'(x) = \mathbb{E}[\tilde{f}'(x)]$. All the statements concerning the results for the $\mathbb{L}^1$ loss can be deduced from the lemma below:

**Lemma 14.** If $f \in \mathcal{B}_{1,\infty}^{\alpha}(R) \cap \mathcal{T}_{1/2}(M)$ with $\alpha \in (0, \tau)$, $R > 0$, $M \ge 1$,

$$ \max \{d_1(f, \bar{f}), d_1(f, \bar{f}')\} \le c_1 \left[ R2^{-J\alpha} + R^{1/(2\alpha+1)} M^{2\alpha/(2\alpha+1)} n^{-\alpha/(2\alpha+1)} + Mn^{-1/2} \right] \quad (4.42) $$

$$ \max \{\mathbb{E}[d_1(\tilde{f}, \bar{f})], \mathbb{E}[d_1(\tilde{f}', \bar{f}')]\} \le c_2 M 2^{J/2} n^{-1/2}, \quad (4.43) $$

where $c_1, c_2$ only depend on $\alpha$ and the wavelets.

*Proof of Lemma 14.* We have,

$$ \max \{d_1(f, \bar{f}), d_1(f, \bar{f}')\} \le c_1 \left[ \sum_{k \notin K_{-1}} |\alpha_{0,k}(f)| + \sum_{j=0}^{J} 2^{-j/2} \sum_{k \notin K_j} |\beta_{j,k}(f)| + \sum_{j=J+1}^{\infty} 2^{-j/2} \sum_{k \in \mathbb{Z}} |\beta_{j,k}(f)| \right]. $$

The condition $f \in \mathcal{B}_{1,\infty}^{\alpha}(R)$ ensures

$$ \sum_{j=J+1}^{\infty} 2^{-j/2} \sum_{k \in \mathbb{Z}} |\beta_{j,k}(f)| \le c_2 R 2^{-J\alpha}. $$

Now, $2^{-j/2}|\beta_{j,k}(f)| \le c_3 f_{j,k}$ where $f_{j,k}$ is defined by (4.3). Set

$$ \bar{Z}_{j,r} = \{k \in \mathbb{Z}, 2^{-r} < f_{j,k} \le 2^{-r+1}\}, $$

and observe that $|\bar{Z}_{j,r}| \le c_4 M 2^{r/2} 2^{j/2}$. We deduce,

$$ 2^{-j/2} \sum_{k \notin K_j} |\beta_{j,k}(f)| \le c_5 \sum_{2^r \ge n} M 2^{-r/2} 2^{j/2} \le c_6 M 2^{j/2} n^{-1/2}. $$

A similar reasoning applies for the father wavelet coefficients. Now,

$$ \sum_{j=0}^{J} 2^{-j/2} \sum_{k \notin K_j} |\beta_{j,k}(f)| \le c_7 \sum_{j=0}^{J} \min \{R 2^{-j\alpha}, M 2^{j/2} n^{-1/2}\} \\ \le c_8 R^{1/(2\alpha+1)} M^{2\alpha/(2\alpha+1)} n^{-\alpha/(2\alpha+1)}. $$
This shows (4.42). As to (4.43), we merely note

$$
\begin{align*}
\max \{ \mathbb{E}[d_1(\tilde{f}, \tilde{f})], \mathbb{E}[d_1(\tilde{f}', \tilde{f}')] \} &\le c_9 \left[ \sum_{k \in \mathbb{Z}} f_{-1,k}^{1/2} + \sum_{j=0}^{J} \sum_{k \in \mathbb{Z}} f_{j,k}^{1/2} \right] n^{-1/2} \\
&\le c_{10} M \left[ 1 + \sum_{j=0}^{J} 2^{j/2} \right] n^{-1/2} \\
&\le c_{11} M 2^{J/2} n^{-1/2}. \quad \square
\end{align*}
$$

The result concerning the bias term $d_2^2(\sqrt{f}, V)$ comes from the following lemma:

**Lemma 15.** If $\sqrt{f} \in \mathcal{B}_{2,\infty}^{\alpha}(R)$ and $f \in \mathcal{T}_{1/2}(M)$, for $\alpha \in (0, \tau)$, $R > 0$, $M \ge 1$,

$$
d_2^2(\sqrt{f}, V) \le c [R^2 2^{-2J\alpha} + \epsilon_n],
$$

where $c$ only depends on $\alpha$ and the wavelets.

*Proof of Lemma 15.* We have,

$$
d_2^2(\sqrt{f}, V) \le c_1 \left[ \sum_{k \notin K_{-1}} (\alpha_{0,k}(\sqrt{f}))^2 + \sum_{j=0}^{J} \sum_{k \notin K_j} (\beta_{j,k}(\sqrt{f}))^2 \right. \\
\left. \qquad + \sum_{j=J+1}^{\infty} \sum_{k \in \mathbb{Z}} (\beta_{j,k}(\sqrt{f}))^2 \right].
$$

The smoothness condition ensures

$$
\sum_{k \in \mathbb{Z}} (\beta_{j,k}(\sqrt{f}))^2 \le R^2 2^{-2j\alpha}.
$$

Besides, we have $(\beta_{j,k}(\sqrt{f}))^2 \le c_2 f_{j,k}$ where $f_{j,k}$ is defined in (4.3). By doing
as in the previous proof,

$$
\sum_{k \notin K_{-1}} (\alpha_{0,k}(\sqrt{f}))^2 \le c_3 M n^{-1/2} \quad \text{and} \quad \sum_{k \notin K_j} (\beta_{j,k}(\sqrt{f}))^2 \le c_4 M n^{-1/2} 2^{j/2}.
$$

Therefore,

$$
\sum_{j=0}^{J} \sum_{k \notin K_j} (\beta_{j,k}(\sqrt{f}))^2 \le c_5 \sum_{j=0}^{J} \min \{M n^{-1/2} 2^{j/2}, R^2 2^{-2j\alpha}\},
$$

hence the result.

<figure><img src="image_8.png" /></figure>
As to the $L^2$ loss, the statements ensue from the following inequalities.

**Lemma 16.** If $f \in B_{2,\infty}^{\alpha}(R)$ with $\alpha \in (0, \tau)$, $R > 0$,

$$ \max \{d_2^2(f, \bar{f}), d_2^2(f, \bar{f}')\} \le c_1 \left[ R^2 2^{-2J\alpha} + R^{2/(2\alpha+1)} n^{-2\alpha/(2\alpha+1)} + n^{-1} \right] \quad (4.44) $$

$$ \max \{\mathbb{E}[d_2^2(\tilde{f}, \bar{f})], \mathbb{E}[d_2^2(\tilde{f}', \bar{f}')]\} \le c_2 2^J/n, \quad (4.45) $$

where $c_1, c_2$ only depend on $\alpha$ and the wavelet basis.

**Proof of Lemma 16.** We have,

$$ \max \{d_2^2(f, \bar{f}), d_2^2(f, \bar{f}')\} \le c_1 \left[ \sum_{k \notin K_{-1}} (\alpha_{0,k}(f))^2 + \sum_{j=0}^J \sum_{k \notin K_j} (\beta_{j,k}(f))^2 + \sum_{j=J+1}^{\infty} \sum_{k \in \mathbb{Z}} (\beta_{j,k}(f))^2 \right]. $$

The last term can be bounded using $f \in B_{2,\infty}^{\alpha}(R)$. For the second last, we note $(\beta_{j,k}(f))^2 \le c_2 2^j f_{j,k}^2$ and do as in the proof of Lemma 14 by noticing that $|\mathbb{Z}_{j,r}| \le c_3 2^r$ as $f$ is a density:

$$ \sum_{k \notin K_j} (\beta_{j,k}(f))^2 \le c_4 \sum_{2^r \ge n} 2^{-r} 2^j \le c_5 2^j n^{-1}. $$

The reasoning is similar for the first term. Now,

$$ \sum_{j=0}^{J} \sum_{k \notin K_j} (\beta_{j,k}(f))^2 \le c_6 \sum_{j=0}^{J} \min \{2^j n^{-1}, R^2 2^{-2j\alpha}\} \\ \le c_7 R^{2/(2\alpha+1)} n^{-2\alpha/(2\alpha+1)}. $$

We put everything together to get (4.44). We now show (4.45):

$$ \max \{\mathbb{E}[d_2^2(\tilde{f}, \bar{f})], \mathbb{E}[d_2^2(\tilde{f}', \bar{f}')]\} \le c_8 \left[ \sum_{k \in \mathbb{Z}} f_{-1,k} + \sum_{j=0}^{J} 2^j \sum_{k \in \mathbb{Z}} f_{j,k} \right] n^{-1} \\ \le c_9 2^J n^{-1}. \quad \square $$
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