SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Let then  $(f_n)_{n\in \mathbb{N}}$  be a sequence of functions  $f_{n}\in C^{k}([-1,1]^{s};\mathbb{C})$  and  $g^{*}\in C_{2\pi}^{k}(\mathbb{R}^{s};\mathbb{C})$  such that  $f_{n}\to f$  in  $C^k ([-1,1]^s;\mathbb{C})$  and  $f_{n}^{*}\rightarrow g^{*}$  in  $C_{2\pi}^{k}(\mathbb{R}^{s};\mathbb{C})$ . We then have
$$
\begin{array}{l} \left\| f ^ {*} - g ^ {*} \right\| _ {L ^ {\infty} ([ - \pi , \pi ] ^ {s})} \leq \left\| f ^ {*} - f _ {n} ^ {*} \right\| _ {L ^ {\infty} ([ - \pi , \pi ] ^ {s})} + \left\| f _ {n} ^ {*} - g ^ {*} \right\| _ {L ^ {\infty} ([ - \pi , \pi ] ^ {s})} \\ = \| f - f _ {n} \| _ {L ^ {\infty} ([ - 1, 1 ] ^ {s}; \mathbb {C})} + \| f _ {n} ^ {*} - g ^ {*} \| _ {L ^ {\infty} ([ - \pi , \pi ] ^ {s})} \\ \leq \left\| f - f _ {n} \right\| _ {C ^ {k} ([ - 1, 1 ] ^ {s}; \mathbb {C})} + \left\| f _ {n} ^ {*} - g ^ {*} \right\| _ {C ^ {k} ([ - \pi , \pi ] ^ {s}; \mathbb {C})} \to 0 (n \to \infty). \\ \end{array}
$$
It follows  $f^{*} = g^{*}$  and the closed graph theorem yields the desired continuity.
We refer to Appendix C.2 (see Theorem C.19 and Remark C.20) for a proof of the claimed bound on the operator norm.
For a function  $f \in C^k([-1,1]^s; \mathbb{C})$  we want to express  $v_m(f^*)$  in a convenient way, involving a product of cosines. To this end, we make use of the following identity, which is a generalization of the well-known product-to-sum formula for  $\cos$ .
Lemma C.12. Let  $s \in \mathbb{N}$ . Then it holds for any  $x \in \mathbb{R}^s$  that
$$
\prod_ {j = 1} ^ {s} \cos (x _ {j}) = \frac {1}{2 ^ {s}} \sum_ {\sigma \in \{- 1, 1 \} ^ {s}} \cos (\langle \sigma , x \rangle).
$$
Proof. This is an inductive generalization of the product-to-sum formula
$$
2 \cos (x) \cos (y) = \cos (x - y) + \cos (x + y) \tag {C.5}
$$
for  $x, y \in \mathbb{R}$ , which can be found for instance in [1, Eq. 4.3.32]. The case  $s = 1$  holds since  $\cos$  is an even function. Assume that the claim holds for a fixed  $s \in \mathbb{N}$  and take  $x \in \mathbb{R}^{s + 1}$ . Writing
$x^{\prime} = (x_{1},\dots,x_{s})$  , we derive
$$
\begin{array}{l} \prod_ {j = 1} ^ {s + 1} \cos (x _ {j}) = \left(\frac {1}{2 ^ {s}} \sum_ {\sigma \in \{- 1, 1 \} ^ {s}} \cos (\langle \sigma , x ^ {\prime} \rangle)\right) \cdot \cos (x _ {s + 1}) \\ = \frac {1}{2 ^ {s}} \sum_ {\sigma \in \{- 1, 1 \} ^ {s}} \cos (\langle \sigma , x ^ {\prime} \rangle) \cos (x _ {s + 1}) \\ \stackrel {(C. 5)} {=} \frac {1}{2 ^ {s + 1}} \sum_ {\sigma \in \{- 1, 1 \} ^ {s}} [ \cos (\langle \sigma , x ^ {\prime} \rangle + x _ {s + 1}) + \cos (\langle \sigma , x ^ {\prime} \rangle - x _ {s + 1}) ] \\ = \frac {1}{2 ^ {s + 1}} \sum_ {\sigma \in \{- 1, 1 \} ^ {s + 1}} \cos (\langle \sigma , x \rangle), \\ \end{array}
$$
The following proposition states that  $v_{m}(f^{*})$  can be expressed as a linear combination of products of cosines. This representation is useful since these cosines can be interpolated by Chebyshev polynomials which in the end leads to the desired approximation result.
Proposition C.13. Let  $s \in \mathbb{N}$  and  $k \in \mathbb{N}_0$ . For any  $f \in C^k([-1,1]^s; \mathbb{C})$  and  $m \in \mathbb{N}$  the de-la-Vallee-Poussin operator given as  $f \mapsto v_m(f^*)$  with  $v_m$  as in Proposition C.7 and  $f \mapsto f^*$  as in Lemma C.11 has a representation
$$
v_{m}\left(f^{*}\right)\left(x_{1},\ldots ,x_{s}\right) = \sum_{\substack{\boldsymbol {k}\in \mathbb{N}_{0}^{s}\\ \boldsymbol {k}\leq 2m - 1}}\mathcal{V}_{\boldsymbol{k}}^{m}(f)\prod_{j = 1}^{s}\cos \left(\boldsymbol{k}_{j}x_{j}\right)
$$
for continuous linear functionals
$$
\mathcal {V} _ {\boldsymbol {k}} ^ {m}: C ^ {k} \left(\left[ - 1, 1 \right] ^ {s}; \mathbb {C}\right)\rightarrow \mathbb {C}, \quad f \mapsto 2 ^ {\| \boldsymbol {k} \| _ {0}} \cdot a _ {\boldsymbol {k}} ^ {m} \cdot \widehat {f ^ {*}} (\boldsymbol {k}),
$$
where  $\| \pmb {k}\| _0 = \# \{j\in \{1,\dots ,s\} :\pmb {k}_j\neq 0\}$  and  $a_{\pmb{k}}^{m} = \widehat{V_{m}^{s}} (\pmb {k})$  . Furthermore, if  $f\in C^k ([-1,1]^s;\mathbb{R})$  then  $\mathcal{V}_k^m (f)\in \mathbb{R}$  for every  $\pmb {k}\in \mathbb{N}_0^s$  with  $\pmb {k}\leq 2m - 1$
Proof. First of all, it is easy to see that  $v_{m}(f^{*})$  is even in each variable, which follows directly from the fact that  $f^{*}$  and  $V_{m}^{s}$  are both even in each variable. Furthermore, if we write
$$
\begin{array}{l} V _ {m} ^ {s} = \sum_ {\mathbf {k} \in \mathbb {Z} ^ {s}} a _ {\mathbf {k}} ^ {m} e _ {\mathbf {k}} \\ - (2 m - 1) \leq \mathbf {k} \leq 2 m - 1 \\ \end{array}
$$
with appropriately chosen coefficients  $a_{\mathbf{k}}^{m}\in \mathbb{R}$  , we easily see
$$
\begin{array}{l} v _ {m} \left(f ^ {*}\right) = \sum_ {\mathbf {k} \in \mathbb {Z} ^ {s}} a _ {\mathbf {k}} ^ {m} \widehat {f ^ {*}} (\mathbf {k}) e _ {\mathbf {k}}. \\ - (2 m - 1) \leq \mathbf {k} \leq 2 m - 1 \\ \end{array}
$$
Using Euler's identity and the fact that  $v_{m}(f^{*})$  is an even function, we get the representation
$$
\begin{array}{l} v _ {m} \left(f ^ {*}\right) (x) = \sum_ {\mathbf {k} \in \mathbb {Z} ^ {s}} a _ {\mathbf {k}} ^ {m} \widehat {f ^ {*}} (\mathbf {k}) \cos \left(\langle \mathbf {k}, x \rangle\right) \\ - (2 m - 1) \leq \mathbf {k} \leq 2 m - 1 \\ \end{array}
$$
for all  $x \in \mathbb{R}^s$ . Using  $\odot$  to denote the componentwise product of two vectors of the same size, i.e.,  $x \odot y = (x_i \cdot y_i)_i$ , and using the identity  $\langle \mathbf{k}, \sigma \odot x \rangle = \langle \sigma, \mathbf{k} \odot x \rangle$ , we see since  $v_m(f^*)$  is even in
every variable that
$$
\begin{array}{l} v _ {m} \left(f ^ {*}\right) (x) = \frac {1}{2 ^ {s}} \cdot \sum_ {\sigma \in \{- 1, 1 \} ^ {s}} v _ {m} \left(f ^ {*}\right) (\sigma \odot x) \\ = \frac{1}{2^{s}}\cdot \sum_{\sigma \in \{-1,1\}^{s}}\sum_{\substack{\mathbf{k}\in \mathbb{Z}^{s}\\ -(2m - 1)\leq \mathbf{k}\leq 2m - 1}}a_{\mathbf{k}}^{m}\widehat{f^{*}} (\mathbf{k})\cos \big(\langle \mathbf{k},\sigma \odot x\rangle \big) \\ = \sum_{\substack{\mathbf{k}\in \mathbb{Z}^{s}\\ -(2m - 1)\leq \mathbf{k}\leq 2m - 1}}\left(a_{\mathbf{k}}^{m}\widehat{f^{*}} (\mathbf{k})\frac{1}{2^{s}}\sum_{\sigma \in \{-1,1\}^{s}}\cos (\langle \sigma ,\mathbf{k}\odot x\rangle)\right) \\ \stackrel{\text{Lemma C.12}}{=}\sum_{\substack{\mathbf{k}\in \mathbb{Z}^{s}\\ -(2m - 1)\leq \mathbf{k}\leq 2m - 1}}a_{\mathbf{k}}^{m}\widehat{f^{*}} (\mathbf{k})\prod_{j = 1}^{s}\cos \left(\mathbf{k}_{j}x_{j}\right) \\ = \sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}2^{\| \mathbf{k}\|_{0}}a_{\mathbf{k}}^{m}\widehat{f^{*}} (\mathbf{k})\prod_{j = 1}^{s}\cos \left(\mathbf{k}_{j}x_{j}\right) \\ \end{array}
$$
$$
\| \mathbf {k} \| _ {0} := \# \left\{j \in \{1, \dots , s \}: \mathbf {k} _ {j} \neq 0 \right\}.
$$
In the last step we again used that cos is an even function and that
$$
\widehat {f ^ {*}} (\mathbf {k}) = \widehat {f ^ {*}} (\sigma \odot \mathbf {k})
$$
for all  $\sigma \in \{-1, 1\}^s$ , which also follows easily since  $f^*$  and  $V_m^s$  are even in every component. Letting
$$
\mathcal {V} _ {\mathbf {k}} ^ {m} (f) := 2 ^ {\| \mathbf {k} \| _ {0}} a _ {\mathbf {k}} ^ {m} \widehat {f ^ {*}} (\mathbf {k}),
$$
we have the desired form. The fact that  $\mathcal{V}_{\mathbf{k}}^{m}$  is a continuous linear functional on  $C_{2\pi}^{k}([-1,1]^{s};\mathbb{C})$  follows directly since  $f\mapsto \widehat{f^{*}} (\mathbf{k})$  is a continuous linear functional for every  $\mathbf{k}$ . If  $f$  is real-valued, so is  $\widehat{f^*} (\mathbf{k})$  for every  $\mathbf{k}\in \mathbb{N}_0^s$  with  $\mathbf{k}\leq 2m - 1$ , since  $f^{*}$  is real-valued and even in every component.
Our main approximation result involves linear combinations of Chebyshev polynomials where the coefficients in this linear combination are given as  $\mathcal{V}_{\mathbf{k}}^{m}(f)$ . It is therefore important to be able to bound the sum of the absolute values  $|\mathcal{V}_{\mathbf{k}}^{m}(f)|$ .
Lemma C.14. Let  $s \in \mathbb{N}$ . There exists a constant  $c = c(s) > 0$ , such that the inequality
$$
\sum_{\substack{\boldsymbol {k}\in \mathbb{N}_{0}^{s}\\ \boldsymbol {k}\leq 2m - 1}}|\mathcal{V}_{\boldsymbol{k}}^{m}(f)|\leq c\cdot m^{s / 2}\cdot \| f\|_{L^{\infty}([-1,1]^{s};\mathbb{C})}
$$
holds for all  $m \in \mathbb{N}$  and  $f \in C([-1, 1]^s; \mathbb{C})$ , where  $\mathcal{V}_k^m$  is as in Proposition C.13.
In fact, we have  $c(s) \leq \exp(C \cdot s)$  with an absolute constant  $C > 0$ .
Proof. Let  $f \in C([-1,1]^s; \mathbb{C})$  and  $m \in \mathbb{N}$ . For any multi-index  $\ell \in \mathbb{N}_0^s$ , it follows from Proposition C.13 that
$$
\widehat{v_{m}(f^{*})} (\ell) = \sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}\mathcal{V}_{\mathbf{k}}^{m}(f)\widehat{g}_{\mathbf{k}}(\ell),
$$
$$
g _ {\mathbf {k}}: \quad \mathbb {R} ^ {s} \to \mathbb {R}, \quad (x _ {1}, \dots , x _ {s}) \mapsto \prod_ {j = 1} ^ {s} \cos \left(\mathbf {k} _ {j} x _ {j}\right).
$$
Now, a calculation using Fubini's theorem and using
$$
g_{\mathbf{k}} = \prod_{j = 1}^{s}\frac{1}{2}\left(e_{\mathbf{k}_{j}} + e_{-\mathbf{k}_{j}}\right) = \prod_{\substack{1\leq j\leq s\\ \mathbf{k}_{j}\neq 0}}\frac{1}{2}\left(e_{\mathbf{k}_{j}} + e_{-\mathbf{k}_{j}}\right)
$$
for any number  $k\in \mathbb{N}_0$  shows
$$
\widehat {g _ {\mathbf {k}}} (\boldsymbol {\ell}) = \left\{ \begin{array}{l l} \frac {1}{2 ^ {| | \mathbf {k} | | _ {0}}}, & \mathbf {k} = \boldsymbol {\ell}, \\ 0, & \text {o t h e r w i s e} \end{array} \right. \quad \text {f o r} \mathbf {k}, \boldsymbol {\ell} \in \mathbb {N} _ {0} ^ {s}.
$$
Therefore, we have the bound  $|\mathcal{V}_{\ell}^{m}(f)| \leq 2^{s} \cdot \left| \widehat{v_{m}(f^{*})}(\ell) \right|$  for  $\ell \in \mathbb{N}_0^s$  with  $|\ell| \leq 2m - 1$ . Using the Cauchy-Schwarz and the Parseval inequality, we therefore see
$$
\sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}|\mathcal{V}_{\mathbf{k}}^{m}(f)|\leq 2^{s}\cdot \sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}\left|\widehat{v_{m}(f^{*})} (\mathbf{k})\right| \overset {\mathrm{CS}}{\leq}2^{s}\cdot (2m)^{s / 2}\cdot \left(\sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}\left|\widehat{v_{m}(f^{*})} (\mathbf{k})\right|^{2}\right)^{1 / 2}
$$
$$
\begin{array}{l} \stackrel {\text {P a r s e v a l}} {\leq} 2 ^ {s} \cdot 2 ^ {s / 2} \cdot m ^ {s / 2} \cdot \| v _ {m} (f ^ {*}) \| _ {L ^ {2} ([ - \pi , \pi ] ^ {s}; \mathbb {C})} \\ \leq \underbrace {2 ^ {s} \cdot 2 ^ {s / 2}} _ {=: c _ {1} (s)} \cdot m ^ {s / 2} \cdot \| v _ {m} \left(f ^ {*}\right) \| _ {L ^ {\infty} \left([ - \pi , \pi ] ^ {s}; \mathbb {C}\right)}. \\ \end{array}
$$
Using Proposition C.7, we get a constant  $c_{2}(s) \leq \exp(C_{0} \cdot s)$  such that
$$
\sum_{\substack{\mathbf{k}\in \mathbb{N}_{0}^{s}\\ \mathbf{k}\leq 2m - 1}}|\mathcal{V}_{\mathbf{k}}^{m}(f)|\leq c_{1}(s)\cdot c_{2}(s)\cdot m^{s / 2}\cdot \| f^{*}\|_{L^{\infty}([-\pi ,\pi ]^{s};\mathbb{C})} = c(s)\cdot m^{s / 2}\cdot \| f\|_{L^{\infty}([-1,1]^{s};\mathbb{C})},
$$
For any natural number  $\ell \in \mathbb{N}_0$ , we denote by  $T_{\ell}$  the  $\ell$ -th Chebyshev polynomial, satisfying
$$
T _ {\ell} \left(\cos (x)\right) = \cos (\ell x), \quad x \in \mathbb {R}.
$$
One can show that  $T_{\ell}$  is in fact a polynomial of degree  $\ell$ . For a multi-index  $\mathbf{k} \in \mathbb{N}_0^s$ , we define
$$
T _ {\mathbf {k}} (x) := \prod_ {j = 1} ^ {s} T _ {\mathbf {k} _ {j}} \left(x _ {j}\right), \quad x \in \mathbb {R} ^ {s}.
$$
We then get the following approximation result about approximating (non-periodic)  $C^k$ -functions by linear combinations of Chebyshev polynomials.
Theorem C.15. Let  $k, s, m \in \mathbb{N}$ . Then there exists a constant  $c = c(s, k) > 0$  with the following property: For any  $f \in C^k([-1, 1]^s; \mathbb{R})$  the polynomial  $P$  defined as
$$
P(x):= \sum_{\substack{\boldsymbol {k}\in \mathbb{N}_{0}^{s}\\ k\leq 2m - 1}}\mathcal{V}_{\boldsymbol{k}}^{m}(f)\cdot T_{\boldsymbol{k}}(x),
$$
with  $\mathcal{V}_k^m$  as in Proposition C.13, satisfies
$$
\left\| f - P \right\| _ {L ^ {\infty} ([ - 1, 1 ] ^ {s}; \mathbb {R})} \leq \frac {c}{m ^ {k}} \cdot \left\| f \right\| _ {C ^ {k} ([ - 1, 1 ] ^ {s}; \mathbb {R})}.
$$
Here, the maps
$$
C \left(\left[ - 1, 1 \right] ^ {s}; \mathbb {R}\right)\rightarrow \mathbb {R}, \quad f \mapsto \mathcal {V} _ {k} ^ {m} (f)
$$
are continuous and linear functionals with respect to the  $L^{\infty}$ -norm. Furthermore, there exists a constant  $\tilde{c} = \tilde{c}(s) > 0$ , such that the inequality
$$
\sum_{\substack{\boldsymbol {k}\in \mathbb{N}_{0}^{s}\\ \boldsymbol {k}\leq 2m - 1}}|\mathcal{V}_{\boldsymbol{k}}^{m}(f)|\leq \tilde{c}\cdot m^{s / 2}\cdot \| f\|_{L^{\infty}([-1,1]^{s};\mathbb{R})}
$$
holds for all  $f \in C([-1, 1]^s; \mathbb{R})$ .
Moreover, we have  $c(s,k) \leq \exp(C \cdot ks) \cdot k^{2k}$  and  $\tilde{c}(s) \leq \exp(C \cdot s)$  with an absolute constant  $C > 0$ .
Proof. We choose the constant  $c_0 = c_0(s, k)$  according to Corollary C.10. Let  $f \in C^k([-1, 1]^s; \mathbb{R})$  be arbitrary. Then we define the corresponding function  $f^* \in C_{2\pi}^k(\mathbb{R}^s; \mathbb{R})$  as above. Let  $P$  be defined as in the statement of the theorem. Then it follows from the definition of the Chebyshev polynomials  $T_k$ , the definition of  $P$ , and the formula for  $v_m(f^*)$  from Proposition C.13 that
$$
P ^ {*} (x) = v _ {m} \left(f ^ {*}\right) (x)
$$
is satisfied, where  $P^{*}$  is the corresponding function to  $P$  defined similarly to  $f^{*}$ . Overall, we get the bound
$$
\| f - P \| _ {L ^ {\infty} ([ - 1, 1 ] ^ {s}; \mathbb {R})} = \| f ^ {*} - P ^ {*} \| _ {L ^ {\infty} ([ - \pi , \pi ] ^ {s}; \mathbb {R})} \stackrel {{\text {C o r . C . 1 0}}} {\leq} \frac {c _ {0}}{m ^ {k}} \cdot \| f ^ {*} \| _ {C ^ {k} ([ - \pi , \pi ] ^ {s}; \mathbb {R})}.
$$
The first claim then follows using the continuity of the map  $f \mapsto f^*$  as proven in Lemma C.11. The second part of the theorem has already been proven in Lemma C.14.
C.2 Details on bounding the constant  $c$  in Theorem 3.2
In this appendix we provide details on the bound of the constant  $c$  that appears in the formulation of Theorem 3.2. Specifically, we perform a careful investigation of several results from [29] to get an upper bound for the constant appearing in Proposition C.8. Moreover, we analyze the operator norm of the operator
$$
C ^ {k} \left(\left[ - 1, 1 \right] ^ {s}; \mathbb {C}\right)\rightarrow C _ {2 \pi} ^ {k} \left(\mathbb {R} ^ {s}; \mathbb {C}\right), \quad f \mapsto f ^ {*} \quad \text {w i t h} \quad f ^ {*} (x) := f \left(\cos \left(x _ {1}\right), \dots , \cos \left(x _ {s}\right)\right)
$$
appearing in Lemma C.11 and show that it is bounded from above by  $k^k$ .
We start with the analysis of some bounds in [29, Chapter 4.3]. Here, a generalization of Jackson's kernel is defined for any  $m, r \in \mathbb{N}$  as
$$
L _ {m, r} (t) := \lambda_ {m, r} ^ {- 1} \cdot \left(\frac {\sin (m t / 2)}{\sin (t / 2)}\right) ^ {2 r}, \quad t \in \mathbb {R},
$$
where  $\lambda_{m,r}$  is chosen such that
$$
\int_ {[ - \pi , \pi ]} L _ {m, r} (t) d t = 1.
$$
The first two important bounds are provided in the following proposition.
Proposition C.16. Let  $m, r \in \mathbb{N}$ . Then it holds
$$
\lambda_ {m, r} ^ {- 1} \leq \exp (C \cdot r) \cdot m ^ {1 - 2 r} \quad a n d \quad \int_ {[ 0, \pi ]} t ^ {k} L _ {m, r} (t) d t \leq \exp (C \cdot r) \cdot m ^ {- k}
$$
for any  $k \leq 2r - 2$ , with an absolute constant  $C > 0$ .
Proof. Since  $L_{m,r} \geq 0$  and since  $\sin(t/2) \leq t/2$  for  $t \in [0, \pi]$ , we get
$$
\begin{array}{l} \lambda_ {m, r} \geq \int_ {[ 0, \pi ]} \left(\frac {\sin (m t / 2)}{t / 2}\right) ^ {2 r} d t = 2 ^ {2 r} \cdot \int_ {[ 0, \pi ]} \left(\frac {\sin (m t / 2)}{t}\right) ^ {2 r} d t \\ = 2 ^ {2 r} \cdot \int_ {[ 0, \pi m / 2 ]} \left(\frac {\sin (u)}{(2 u) / m}\right) ^ {2 r} d u \cdot \frac {2}{m} \geq m ^ {2 r - 1} \cdot \int_ {[ 0, \pi m / 2 ]} \left(\frac {\sin (u)}{u}\right) ^ {2 r} d u \\ \geq m ^ {2 r - 1} \cdot \int_ {[ 0, \pi / 2 ]} \left(\frac {\sin (u)}{u}\right) ^ {2 r} d u \geq m ^ {2 r - 1} \cdot \int_ {[ 0, \pi / 2 ]} \left(\frac {2 u}{\pi \cdot u}\right) ^ {2 r} d u \geq \left(\frac {2}{\pi}\right) ^ {2 r} \cdot m ^ {2 r - 1}. \\ \end{array}
$$
Here, we employed the inequality  $\sin (u)\geq \frac{2}{\pi} u$  for  $u\in [0,\pi /2]$  in the penultimate step. This shows the first part of the claim.
To see that this inequality holds, note that  $\sin^{\prime \prime}(u) = -\sin (u)\leq 0$  for  $u\in [0,\pi /2]$ , so that sin is concave on that interval, and hence  $\sin (u) = \sin ((1 - \frac{2u}{\pi})\cdot 0 + \frac{2u}{\pi}\cdot \frac{\pi}{2})\geq \frac{2u}{\pi}\sin (\frac{\pi}{2}) = \frac{2u}{\pi}$ .
For the second part, we again use the estimate  $\sin (u)\geq \frac{2}{\pi} u$  for  $u\in [0,\pi /2]$  to compute
$$
\begin{array}{l} \int_ {[ 0, \pi ]} t ^ {k} L _ {m, r} (t) d t = \lambda_ {m, r} ^ {- 1} \cdot \int_ {[ 0, \pi ]} t ^ {k} \left(\frac {\sin (m t / 2)}{\sin (t / 2)}\right) ^ {2 r} d t \leq \lambda_ {m, r} ^ {- 1} \cdot \int_ {[ 0, \pi ]} t ^ {k} \left(\frac {\sin (m t / 2)}{t / \pi}\right) ^ {2 r} d t \\ = \lambda_ {m, r} ^ {- 1} \cdot \pi^ {2 r} \cdot \int_ {[ 0, \pi ]} t ^ {k - 2 r} \sin (m t / 2) ^ {2 r} d t \\ = \lambda_ {m, r} ^ {- 1} \cdot \pi^ {2 r} \cdot \int_ {[ 0, \pi m / 2 ]} \left(\frac {2 u}{m}\right) ^ {k - 2 r} \sin (u) ^ {2 r} d u \cdot \frac {2}{m} \\ \leq \lambda_ {m, r} ^ {- 1} \cdot \pi^ {2 r} \cdot m ^ {2 r - 1 - k} \int_ {[ 0, \pi m / 2 ]} u ^ {k - 2 r} \sin (u) ^ {2 r} d u \\ \leq \exp (C _ {1} \cdot r) \cdot \int_ {[ 0, \infty)} u ^ {k - 2 r} \cdot \sin (u) ^ {2 r} d u \cdot m ^ {- k} \\ \end{array}
$$
with an absolute constant  $C_1 > 0$ . Here, we employed the first part of this proposition. It remains to bound the integral. This is done via
$$
\begin{array}{l} \int_ {[ 0, \infty)} u ^ {k - 2 r} \cdot \sin (u) ^ {2 r} d u = \int_ {[ 0, 1 ]} u ^ {k - 2 r} \cdot \sin (u) ^ {2 r} d u + \int_ {[ 1, \infty)} u ^ {k - 2 r} \cdot \sin (u) ^ {2 r} d u \\ \leq \int_ {[ 0, 1 ]} \underbrace {u ^ {k} \cdot \left(\frac {\sin (u)}{u}\right) ^ {2 r}} _ {\leq 1} d u + \int_ {[ 1, \infty)} u ^ {- 2} d u \leq C _ {2} \\ \end{array}
$$
with an absolute constant  $C_2 > 0$ . This proves the claim.
The proof in [29] proceeds by defining
$$
K _ {m, r} (t) := L _ {m ^ {\prime}, r} (t), \quad m ^ {\prime} = \left\lfloor \frac {m}{r} \right\rfloor + 1.
$$
Proposition C.16 shows for  $k \leq 2r - 2$  that
$$
\int_ {[ 0, \pi ]} t ^ {k} K _ {m, r} (t) d t \leq \exp (C \cdot r) \cdot \left(m ^ {\prime}\right) ^ {- k}.
$$
Since  $m^{\prime}\geq \frac{m}{r}$  we infer
$$
\int_ {[ 0, \pi ]} t ^ {k} K _ {m, r} (t) d t \leq \exp (C \cdot r) \cdot \left(\frac {r}{m}\right) ^ {k} \leq \exp (C \cdot r) \cdot r ^ {k} \cdot m ^ {- k} \tag {C.6}
$$
with an absolute constant  $C > 0$
We can now quantify the constant appearing in [29, Theorem 4.3].
Theorem C.17 (cf. [29, Theorem 4.3]). Let  $k, m \in \mathbb{N}$  and  $f \in C_{2\pi}^{k}(\mathbb{R}; \mathbb{R})$ . Let
$$
\omega (f ^ {(k)}, 1 / m) := \max _ {x \in \mathbb {R}, | t | \leq 1 / m} | f ^ {(k)} (x + t) - f ^ {(k)} (x) |.
$$
$$
E _ {m} ^ {1} (f) \leq \left(\exp (C \cdot k) \cdot k ^ {k}\right) \cdot m ^ {- k} \cdot \omega \left(f ^ {(k)}, 1 / m\right).
$$
Here, we recall that  $E_m^1(f)$  denotes the best possible approximation error when approximating  $f$  using trigonometric polynomials of degree  $m$ ; see Equation (C.1).
Proof. We follow the proof of [29, Theorem 4.3]. Take  $r = k + 1$  and define
$$
I_{m}(x):= -\int_{[-\pi ,\pi ]}K_{m,r}(t)\sum_{\ell = 1}^{k + 1}(-1)^{\ell}\binom {k + 1}{\ell}f(x + \ell t)  dt.
$$
Then it is shown in the proof of [29, Theorem 4.3] that  $I_{m}$  is a trigonometric polynomial of degree at most  $m$  and that
$$
| f (x) - I _ {m} (x) | \leq 2 \cdot \omega_ {k + 1} (f, 1 / m) \cdot \int_ {[ 0, \pi ]} (m t + 1) ^ {k + 1} K _ {m, r} (t) d t.
$$
Here,  $\omega_{k + 1}(f,1 / m)$  denotes the modulus of smoothness of  $f$  as defined on [29, p. 47]. The integral can be bounded via
$$
\begin{array}{l} \int_ {[ 0, \pi ]} (m t + 1) ^ {k + 1} K _ {m, r} (t) d t \\ = \int_ {[ 0, 1 / m ]} (\underbrace {m t + 1} _ {\leq 2}) ^ {k + 1} K _ {m, r} (t) d t + \int_ {[ 1 / m, \pi ]} (\underbrace {m t + 1} _ {\leq 2 m t}) ^ {k + 1} K _ {m, r} (t) d t \\ \leq 2 ^ {k + 1} \cdot \underbrace {\int_ {[ - \pi , \pi ]} K _ {m , r} (t)   d t} _ {= 1} + 2 ^ {k + 1} m ^ {k + 1} \cdot \int_ {[ 0, \pi ]} t ^ {k + 1} K _ {m, r} (t)   d t \\ \stackrel {(C. 6)} {\leq} 2 ^ {k + 1} + 2 ^ {k + 1} m ^ {k + 1} \exp (C _ {1} \cdot r) \cdot (k + 1) ^ {k + 1} \cdot m ^ {- (k + 1)} \stackrel {r \leq 2 k} {\leq} \exp (C \cdot k) \cdot k ^ {k} \\ \end{array}
$$
with absolute constants  $C, C_1 > 0$ . Since  $\omega_{k+1}(f, 1/m) \leq m^{-k} \cdot \omega(f^{(k)}, 1/m)$  follows from [29, Equation 3.6(5)], the claim is shown.
Therefore, we can bound the constant appearing in [29, Theorem 4.3] by  $\exp(C \cdot k) \cdot k^k$ . It remains to deal with the approximation of multivariate periodic functions by multivariate trigonometric polynomials which is contained in [29, Theorem 6.6].
Theorem C.18 (cf. [29, Theorem 6.6]). Let  $s, k \in \mathbb{N}$  and  $f \in C_{2\pi}^{k}(\mathbb{R}^{s};\mathbb{R})$ . Let  $\omega_{j}$  denote the modulus of continuity of  $\frac{\partial^k f}{\partial x_j^k}$  for  $j = 1, \dots, s$ . Then, with  $E_m^s$  as introduced in Equation (C.1), it holds
$$
E _ {m} ^ {s} (f) \leq \exp (C \cdot k s) \cdot k ^ {k} \cdot m ^ {- k} \sum_ {j = 1} ^ {s} \omega_ {j} (1 / m),
$$
Proof. We follow the proof of [29, Theorem 6.6] with  $p_j = k$  and  $n_j = m$  for every index  $j = 1, \dots, s$ . For  $j = 1, \dots, s + 1$  define the set  $\mathcal{T}_j$  consisting of all functions  $g \in C_{2\pi}(\mathbb{R}^s; \mathbb{R})$  that are a trigonometric polynomial in  $x_\ell$  of degree at most  $m$  for  $\ell < j$ ; in  $x_\ell$  for  $\ell \geq j$  they should have continuous partial derivatives  $\frac{\partial^p g}{\partial x_\ell^p}$  for  $0 \leq p \leq k$ ; the modulus of continuity of  $\frac{\partial^k g}{\partial x_\ell^k}$  should not exceed  $2^{K_j} \omega_j$ , where
$$
K _ {j} = (j - 1) (k + 1) \leq 2 k s \quad \text {f o r} j > 1 \text {a n d} K _ {1} = 1 \leq 2 k s.
$$
Then it is shown that if  $j \in \{1, \dots, s\}$  and  $f_j \in \mathcal{T}_j$  there exists a function  $f_{j+1} \in \mathcal{T}_{j+1}$  for which
$\| f_{j} - f_{j + 1}\|_{L^{\infty}(\mathbb{R}^{s};\mathbb{R})}\leq \exp (C_{1}\cdot k)\cdot k^{k}\cdot m^{-k}\cdot 2^{2ks}\cdot \omega_{j}(1 / m)\leq \exp (C_{2}\cdot ks)\cdot k^{k}\cdot m^{-k}\cdot \omega_{j}(1 / m)$  for absolute constants  $C_1,C_2 > 0$  . This is an application of Theorem C.17. Hence, defining  $f_{1}\coloneqq f$  we see
$$
\| f - f _ {s + 1} \| _ {L ^ {\infty} \left(\mathbb {R} ^ {s}; \mathbb {R}\right)} \leq \sum_ {j = 1} ^ {s} \| f _ {j} - f _ {j + 1} \| _ {L ^ {\infty} \left(\mathbb {R} ^ {s}; \mathbb {R}\right)} \leq \exp \left(C _ {2} \cdot k s\right) \cdot k ^ {k} \cdot m ^ {- k} \cdot \sum_ {j = 1} ^ {s} \omega_ {j} (1 / m).
$$
Therefore, we have shown that the constant appearing in [29, Theorem 6.6] can be bounded from above by  $\exp (C\cdot ks)\cdot k^{k}$
In the rest of this section, we discuss the operator norm of the operator defined in Lemma C.11. In Lemma C.11 the closed graph theorem is used to show that the operator is bounded. However, the closed graph theorem does not provide any bound on the norm of the operator. Therefore, in order to quantify the operator norm, we need to apply a different technique, which is Faa di Bruno's formula. This formula is a generalization of the chain rule to higher order derivatives.
Theorem C.19. Let  $s, k \in \mathbb{N}$ . We define the operator
$$
T: \quad C ^ {k} ([ - 1, 1 ] ^ {s}; \mathbb {C}) \to C _ {2 \pi} ^ {k} ([ - \pi , \pi ] ^ {s}; \mathbb {C}), \quad (T f) (x _ {1}, \dots , x _ {s}) := f (\cos (x _ {1}), \dots , \cos (x _ {s})).
$$
Let  $\alpha \in \mathbb{N}_0^s$  with  $|\alpha| \leq k$ . Then, for any  $f \in C^k([-1, 1]^s; \mathbb{C})$ , we have
$$
\left\| \partial^ {\boldsymbol {\alpha}} (T f) \right\| _ {L ^ {\infty} ([ \pi , \pi ] ^ {s}; \mathbb {C})} \leq \prod_ {j = 1} ^ {s} \boldsymbol {\alpha} _ {j} ^ {\boldsymbol {\alpha} _ {j}} \cdot \| f \| _ {C ^ {| \boldsymbol {\alpha} |} ([ - 1, 1 ] ^ {s})}.
$$