Appendix
Proof
of Lemma 4.2
First, denote
$$\begin{aligned} A_{N,p}\left( f,\hat{f}_N\right) = {\mathcal {K}}_{BO_p}(\hat{f}_N) \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} - {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} , \end{aligned}$$
and note that, \({\mathbb {P}}_X \otimes {\mathbb {P}}_Y\)-a.s.,
$$\begin{aligned} \left( \hat{Z}_{B} - \bar{Z}_{B}\right) ^2 = a_{B}^T\left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^TX_{O_p} X_{O_p}^TA_{N,p}\left( f,\hat{f}_N\right) a_{B} \end{aligned}$$
Then, notice that
$$\begin{aligned}&\sup _{\mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \Big ( \bar{Z}_{B} - \hat{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}}\\&\qquad \le \left( \sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} {\mathbb {E}} _{{\mathbb {P}}_Y}\left[ a_{B}^T \left( A_{N,p} \left( f,\hat{f}_p\right) \right) ^T{\mathcal {K}}_{O_p}(f)A_{N,p}\left( f,\hat{f}_N\right) a_{B} \right] \right) ^{\frac{1}{2}}\\&\qquad \le {\mathbb {E}} _{{\mathbb {P}}_Y}\left[ \sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T{\mathcal {K}}_{O_p}(f)A_{N,p}\left( f,\hat{f}_N\right) a_{B} \right] ^{\frac{1}{2}} . \end{aligned}$$
Thus,
$$\begin{aligned} \sup _{ \mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \left[ \left( \bar{Z}_{B} - \hat{Z}_{B} \right) ^2 \right] ^{\frac{1}{2}} \le {\mathbb {E}} _{{\mathbb {P}}_Y}\left[ \left\| \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T{\mathcal {K}}_{O_p}(f)A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op} \right] ^{\frac{1}{2}}. \end{aligned}$$
Furthermore, it holds, \({\mathbb {P}}_Y\)-a.s., that
$$\begin{aligned} \left\| \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T{\mathcal {K}}_{O_p}(f)A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op}&\le \left\| A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op}^2 \left\| {\mathcal {K}}_{O_p}(f) \right\| _{2,op} \end{aligned}$$
But,
$$\begin{aligned}&\left\| A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op}\\&\quad \le \left\| {\mathcal {K}}_{BO_N}(\hat{f}_N) \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} - {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} \right\| _{2,op}\\&\qquad + \left\| {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} - {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} \right\| _{2,op}\\&\quad \le \left\| \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} \right\| _{2,op} \left\| {\mathcal {K}}_{BO_p}(\hat{f}_N) - {\mathcal {K}}_{BO_p}(f) \right\| _{2,op} + \left\| {\mathcal {K}}_{BO_p}(f) \right\| _{2,op} \\&\qquad \times \left\| \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} \right\| _{2,op} \left\| \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} \right\| _{2,op} \left\| {\mathcal {K}}_{O_p}(\hat{f}_N)- {\mathcal {K}}_{O_p}(f) \right\| _{2,op}\\&\quad \le \frac{1}{m} \left\| f - \hat{f}_N \right\| _\infty + \frac{M}{m^2} \left\| f - \hat{f}_N \right\| _\infty \end{aligned}$$
Here we used the inequality
$$\begin{aligned} \left\| {\mathcal {K}}(f) \right\| _{2,op} \le \left\| f \right\| _\infty \end{aligned}$$
We get
$$\begin{aligned} \sup _{ \mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \left[ \left( \bar{Z}_{B} - \hat{Z}_{B} \right) ^2 \right] ^{\frac{1}{2}}&\le \frac{\sqrt{M}(m+M)}{m^2} {\mathbb {E}} _{{\mathbb {P}}_Y}\left[ \left\| f - \hat{f}_N \right\| _\infty ^2 \right] ^{\frac{1}{2}}.\\&\le \frac{\sqrt{M}(m+M)}{m^2} r_N \end{aligned}$$
\(\square \)
Proof
of Lemma 4.3
First, define for all \(A \subset G\), the operator \(p_{A}\) by
$$\begin{aligned} \forall i,\quad j \in G,\quad \left( p_{A}\right) _{ij} = {\mathbb {I}}_{i \in A}{\mathbb {I}}_{i \sim j}. \end{aligned}$$
To compute the rate of convergence of the bias term \(\bar{Z}_B-\tilde{Z}_B\), we can compute directly,
$$\begin{aligned}&\sup _{ \mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \big ( \bar{Z}_{B} - \tilde{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}}\\&\qquad \le \left\| {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} p_{O_p} - K_{BO}(f) \left( {\mathcal {K}}_{O}(f)\right) ^{-1} \right\| _{2,op} \left\| {\mathcal {K}}_{O}(f)\right\| _{2,op}^{\frac{1}{2}} \end{aligned}$$
Note that, since \(B \cup O = G\), we have immediately that \(\left( {\mathcal {K}}_{B \cup O}(f) \right) ^{-1} = {\mathcal {K}}\left( \frac{1}{f}\right) \).
Using a Schur decomposition, we get
$$\begin{aligned}&\sup _{\mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \big ( \bar{Z}_{B} - \tilde{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}}\\&\qquad \le \sqrt{M} \left\| {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} + \left( {\mathcal {K}}_{B}\left( \frac{1}{f}\right) \right) ^{-1} {\mathcal {K}}_{BO}\left( \frac{1}{f}\right) \right\| _{2,op}\\&\qquad \le \sqrt{M} \left\| {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} + \left( {\mathcal {K}}_{B}\left( \frac{1}{f}\right) \right) ^{-1}{\mathcal {K}}_{BO_p}\left( \frac{1}{f}\right) \right\| _{2,op}\\&\qquad \quad + \sqrt{M} \left\| \left( {\mathcal {K}}_{B}\left( \frac{1}{f}\right) \right) ^{-1} \right\| _{2,op} \left\| {\mathcal {K}}_{B(O\setminus O_p)}\left( \frac{1}{f}\right) \right\| _{2,op}\\&\qquad \le \sqrt{M} \left\| {\mathcal {K}}_{B}\left( \frac{1}{f}\right) {\mathcal {K}}_{BO_p}(f) + {\mathcal {K}}_{BO_N}\left( \frac{1}{f}\right) {\mathcal {K}}_{O_p}(f) \right\| _{2,op}\\&\qquad \quad \times \left\| \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} \right\| _{2,op} \left\| \left( {\mathcal {K}}_{B}\left( \frac{1}{f}\right) \right) ^{-1} \right\| _{2,op} + M^{\frac{3}{2}} \left\| {\mathcal {K}}_{B(O\setminus O_p)}\left( \frac{1}{f}\right) \right\| _{2,op}\\&\qquad \le \frac{M^{\frac{3}{2}}}{m} \left\| - {\mathcal {K}}_{B(O\setminus O_p)}\left( \frac{1}{f}\right) {\mathcal {K}}_{(O\setminus O_p)O_p}(f) \right\| _{2,op}\\&\qquad \quad + M^{\frac{3}{2}} \left\| {\mathcal {K}}_{B(O\setminus O_p)}\left( \frac{1}{f}\right) \right\| _{2,op} \end{aligned}$$
This leads to
$$\begin{aligned} \sup _{ \mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \big ( \bar{Z}_{B}&- \tilde{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}} \le \left( \frac{M^{\frac{5}{2}}}{m} + M^{\frac{3}{2}} \right) \left\| {\mathcal {K}}_{B(O \setminus O_p)}\left( \frac{1}{f}\right) \right\| _{2,op} \end{aligned}$$
(4)
Moreover, we can write
$$\begin{aligned} \left\| {\mathcal {K}}_{B(O \setminus O_p)}\left( \frac{1}{f}\right) \right\| _{2,op}&\le \sum _{k \ge 0} \left| \left( \frac{1}{f}\right) _k \right| \left\| \left( A^k\right) _{B(O \setminus O_p)} \right\| _{2,op}\\&\le \sum _{k \ge d_G\left( B, (O\setminus O_p)\right) } \left| \left( \frac{1}{f}\right) _k \right| \le \sum _{k \ge d_G\left( B, (G\setminus G_p)\right) } \left| \left( \frac{1}{f}\right) _k \right| \\&\le \frac{1}{d_G\left( B, (G\setminus G_p)\right) } \sum _{k \ge d_G\left( B, (G\setminus G_p)\right) } k \left| \left( \frac{1}{f}\right) _k \right| \\&\le \frac{M}{m^2}m_p \end{aligned}$$
\(\square \)
Proof
of Lemma 4.5
For sake of simplicity, let us denote \(O_p\) instead of \(O_{p(N)}\) in the whole proof.
We still denote
$$\begin{aligned} A_{N,p}\left( f,\hat{f}_N\right) = {\mathcal {K}}_{BO_p}(\hat{f}_N) \left( {\mathcal {K}}_{O_p}(\hat{f}_N)\right) ^{-1} - {\mathcal {K}}_{BO_p}(f) \left( {\mathcal {K}}_{O_p}(f)\right) ^{-1} , \end{aligned}$$
Then, notice that
$$\begin{aligned}&\sup _{\mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \Big ( \bar{Z}_{B} - \hat{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}}\\&\qquad \le \left( \sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} {\mathbb {E}} \left[ {\text {Tr}} \left( a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^TX_{O_p}X_{O_p}^TA_{N}\left( f,\hat{f}_N\right) a_{B} \right) \right] \right) ^{\frac{1}{2}}\\&\qquad \le \left( \sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} {\mathbb {E}} \left[ {\text {Tr}} \left( X_{O_p}X_{O_p}^TA_{N,p}\left( f,\hat{f}_N\right) a_{B} a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T\right) \right] \right) ^{\frac{1}{2}} \end{aligned}$$
Applying Cauchy Schwarz inequality, we get
$$\begin{aligned}&\sup _{\mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \Big ( \bar{Z}_{B} - \hat{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}}\\&\qquad \le \Bigg ( {\mathbb {E}} \left[ {\text {Tr}} \left( \left( X_{O_p}X_{O_p}^T \right) ^2 \right) \right] \\&\qquad \qquad \times \sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} {\mathbb {E}} \left[ {\text {Tr}} \left( \left( A_{N,p}\left( f,\hat{f}_N\right) a_{B} a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T\right) ^2 \right) \right] \Bigg )^{\frac{1}{4}} \end{aligned}$$
But, on the one hand, we have
$$\begin{aligned} {\mathbb {E}} \left[ {\text {Tr}} \left( \left( X_{O_p}X_{O_p}^T \right) ^2 \right) \right]&= {\mathbb {E}} \left[ {\text {Tr}} \left( \left( X_{O_p}^TX_{O_p} \right) ^2 \right) \right] \\&= {\mathbb {E}} \left[ \left( \sum _{i \in O_p} X_i^2 \right) ^2 \right] \end{aligned}$$
And on the other hand, we can write
$$\begin{aligned}&\sup _{ \mathop { \left\| a_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}) \subset {B}}} {\mathbb {E}} \left[ {\text {Tr}} \left( \left( A_{N}\left( f,\hat{f}_N\right) a_{B} a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^T\right) ^2 \right) \right] \\&\qquad \le \sup _{ \mathop { \left\| a_{B} \right\| _2 = \left\| b_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(a_{B}, b_{B}) \subset {B}}} {\mathbb {E}} \Bigg [{\text {Tr}} \bigg ( a_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^TA_{N,p}\left( f,\hat{f}_N\right) \\&\qquad \qquad \times b_{B} b_{B}^T \left( A_{N,p}\left( f,\hat{f}_N\right) \right) ^TA_{N,p}\left( f,\hat{f}_N\right) a_{B} \bigg ) \Bigg ]\\&\qquad \le \sup _{ \mathop { \left\| b_{B} \right\| _2 = 1}\limits ^{ {\text {supp}}(b_{B}) \subset {B}}} {\mathbb {E}} \Bigg [ \left\| A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op}^4 \left\| b_{B} b_{B}^T \right\| _{2,op} \Bigg ]\\&\qquad \le {\mathbb {E}} \Bigg [ \left\| A_{N,p}\left( f,\hat{f}_N\right) \right\| _{2,op}^4 \Bigg ]\\&\qquad \le \left( \frac{m+M}{m^2}\right) ^4 {\mathbb {E}} \Bigg [ \left\| f -\hat{f}_N \right\| _{\infty }^4 \Bigg ] \end{aligned}$$
Thus,
$$\begin{aligned} \sup _{ \mathop { \left\| a_B\right\| _2 = 1}\limits ^{ Z_B = a_B^TX_B}} {\mathbb {E}} \Bigg [ \Big ( \bar{Z}_{B}&- \hat{Z}_{B} \Big )^2 \Bigg ] ^{\frac{1}{2}} \le \frac{m+M}{m^2} {\mathbb {E}} \left[ \left( \sum _{i \in O_p} X_i^2 \right) ^2 \right] ^{\frac{1}{4}} r_N \end{aligned}$$
\(\square \)
