**5. Proofs**

The following three lemmas will be useful in the sequel.

**Lemma 1.** *Let f* ∈ *DT and Pm* ∈ P*<sup>m</sup>*, *m* ≥ 2*. Then*

$$\int\_0^{\frac{1}{m}} \frac{\omega\_\wp(f - P\_{\mathfrak{m}}, t)}{t} dt \le \mathcal{C} \left( \| (f - P\_{\mathfrak{m}}) \|\_{\infty} + \int\_0^{\frac{1}{m}} \frac{\omega\_\wp^r(f, t)}{t} dt \right), t$$

*where r* ∈ N *with r* < *m and* 0 < C = C(*<sup>m</sup>*, *f*).

**Proof.** Taking into account that *ωϕ*(*f* , *t*) is a non-decreasing function of *t*, we have

$$\int\_0^{\frac{1}{m}} \frac{\omega\_\Psi(f - P\_{m,t}t)}{t} dt \quad = \sum\_{j=m}^\infty \int\_{\frac{1}{j+1}}^{\frac{1}{j}} \frac{\omega\_\Psi(f - P\_{m,t}t)}{t} dt \\ \leq \mathcal{C} \sum\_{j=m}^\infty \frac{\omega\_\Psi\left(f - P\_{m,\frac{t}{j}}\frac{1}{j}\right)}{j}.$$

Then, by applying the following Stechkin type inequality ([22], Theorem 7.2.4)

$$
\omega\_{\varphi}(f, t) \le \mathcal{C}t \sum\_{i=0}^{\lfloor \frac{1}{2} \rfloor} E\_i(f), \quad 0 < \mathcal{C} \ne \mathcal{C}(f, t).
$$

we ge<sup>t</sup>

$$\begin{split} \int\_{0}^{\frac{1}{m}} \frac{\omega\_{g}(f - P\_{m}\prime)}{t} dt &\leq \begin{aligned} \mathscr{C} &\sum\_{j=m}^{\infty} \frac{1}{j^{2}} \sum\_{i=0}^{j} E\_{i}(f - P\_{m}) \\ &= \ \mathscr{C} \sum\_{j=m}^{\infty} \frac{1}{j^{2}} \left[ \sum\_{i=0}^{m-1} E\_{i}(f - P\_{m}) + \sum\_{i=m}^{j} E\_{i}(f - P\_{m}) \right] \\ &\leq \ \mathscr{C} \|(f - P\_{m})\|\_{\infty} \left( \sum\_{j=m}^{\infty} \frac{m}{j^{2}} \right) + \mathscr{C} \sum\_{j=m}^{\infty} \frac{1}{j^{2}} \sum\_{i=m}^{j} E\_{i}(f) . \end{aligned}$$

and taking into account that ∑∞*j*=*<sup>n</sup>* 1*j*2 ≤ C*n* holds for all *n* ∈ N, with C = C(*n*), we obtain

$$\begin{split} \int\_{0}^{\frac{1}{m}} \frac{\omega\_{\boldsymbol{\theta}}(f - P\_{\mathrm{W}}, t)}{t} dt &\leq \|\mathcal{C}\| (f - P\_{\mathrm{W}}) \|\_{\infty} + \mathcal{C} \sum\_{i=m}^{\infty} E\_{i}(f) \sum\_{j=i}^{\infty} \frac{1}{j^{2}} \\ &\leq \|\mathcal{C}\| (f - P\_{\mathrm{W}}) \|\_{\infty} + \mathcal{C} \sum\_{i=m}^{\infty} \frac{E\_{i}(f)}{i}. \end{split}$$

Finally, by applying the Jackson type inequality ([22], Theorem 7.2.1) (see also [31], Section 2.5.2),

$$E\_m(f) \le \mathcal{C}\omega\_\varphi^r\left(f, \frac{1}{m}\right), \quad r < m, \quad \mathcal{C} \ne \mathcal{C}(m, f),$$

and recalling that ([22], (4.1.3))

$$
\omega\_{\varPhi}(\mathbf{g}, \mathfrak{a}t) \le \mathcal{C}a\omega\_{\varPhi}(\mathbf{g}, t), \qquad \forall a \ge 1, \quad \mathcal{C} \ne \mathcal{C}(\mathbf{g}, t, \mathfrak{a}), \tag{56}
$$

we deduce

$$\begin{split} \sum\_{i=m}^{\infty} \frac{\mathbb{E}\_{i}(f)}{i} &\leq \quad \mathscr{C} \sum\_{i=m}^{\infty} \frac{\omega\_{\varphi}^{r} \left(f, \frac{1}{i}\right)}{i} = \sum\_{i=m}^{\infty} \omega\_{\varphi}^{r} \left(f, \frac{1}{i}\right) (i-1) \int\_{\frac{1}{i}}^{\frac{1}{i-1}} du \\ &\leq \quad \mathscr{C} \sum\_{i=m}^{\infty} \int\_{\frac{1}{i}}^{\frac{1}{i-1}} \frac{\omega\_{\varphi}^{r} \left(f, u\right)}{u} du = \mathscr{C} \int\_{0}^{\frac{1}{m-1}} \frac{\omega\_{\varphi}^{r} \left(f, u\right)}{u} du \\ &= \quad \mathscr{C} \int\_{0}^{\frac{1}{m}} \frac{\omega\_{\varphi}^{r} \left(f, \frac{m}{m-1} t\right)}{t} dt \leq \mathscr{C} \int\_{0}^{\frac{1}{m}} \frac{\omega\_{\varphi}^{r} \left(f, t\right)}{t} dt, \end{split}$$

which completes the proof.

**Lemma 2.** *For any* −1 < *t* ≤ −12 *, and for any f s.t. f* ∈ *DT, we have*

$$\oint\_{-1}^{2t+1} \frac{f(\mathbf{x})}{(\mathbf{x}-t)^2} d\mathbf{x} \le \mathcal{C} \left( \int\_0^1 \frac{\omega\_\phi(f', \sigma)}{\sigma} d\sigma + \frac{||f||}{1+t} \right) \cdot \mathbf{1}$$

*where* C = C(*f* , *t*).

**Proof.** Since .−2*t*+<sup>1</sup> −1 *dx x*−*t* = 0, we write

$$\begin{aligned} \underset{\mathbf{x}}{\overset{2t+1}{\underset{\mathbf{x}}{\leftarrow}}} \frac{f(\mathbf{x})}{(\mathbf{x}-t)^{2}} d\mathbf{x} &=& \underset{\mathbf{x}}{\overset{2t+1}{\underset{\mathbf{x}}{\leftarrow}}} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x}-t)}{(\mathbf{x}-t)^{2}} d\mathbf{x} + f(t) \underset{\mathbf{x}}{\overset{2t+1}{\underset{\mathbf{x}}{\leftarrow}}} \frac{d\mathbf{x}}{(\mathbf{x}-t)^{2}} \\ &=& A\_{1}(t) + A\_{2}(t). \end{aligned} \tag{57}$$

Concerning *A*1, by reasoning as done in proving Proposition 3 we have that *f* ∈ *DT* implies

$$A\_1(t) = \int\_{-1}^{2t+1} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x}$$

and using

$$f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t) = \int\_{t}^{\mathbf{x}} [f'(\tau) - f'(t)] d\tau,\tag{58}$$

we obtain the form

$$A\_1(t) = \int\_{-1}^{t} \left[ \int\_x^t [f'(t) - f'(\tau)] d\tau \right] \frac{d\mathbf{x}}{(\mathbf{x} - t)^2} + \int\_t^{2t + 1} \left[ \int\_t^{\mathbf{x}} [f'(\tau) - f'(t)] d\tau \right] \frac{d\mathbf{x}}{(\mathbf{x} - t)^2}.$$

Hence, changing the variables *x* = *t* − *σ*2 √1 − *t*2, *τ* = *t* − *h*2√<sup>1</sup> − *t*2 in the first addendum and *x* = *t* + *σ*2 √1 − *t*2, *τ* = *t* + *h*2√<sup>1</sup> − *t*2 in the second one, we ge<sup>t</sup>

$$\begin{split} A\_{1}(t) &= \int\_{0}^{2\sqrt{\frac{1+t}{1-t}}} \left[ \int\_{0}^{\sigma} \left[ f'\left(t + \frac{h}{2}\sqrt{1-t^{2}}\right) - f'\left(t - \frac{h}{2}\sqrt{1-t^{2}}\right) \right] dh \right] \frac{d\sigma}{\sigma^{2}} \\ &= \int\_{0}^{2\sqrt{\frac{1+t}{1-t}}} \left( \int\_{0}^{\sigma} \Delta\_{h\eta(t)} f'(t) dh \right) \frac{d\sigma}{\sigma^{2}}. \end{split}$$

Consequently, for any −1 < *t* ≤ −12we obtain

$$\begin{split} |A\_{1}(t)| &\leq \int\_{0}^{2\sqrt{\frac{1+t}{1-t}}} \left( \int\_{0}^{\sigma} ||\Delta\_{l\eta}f'||dh| \right) \frac{d\sigma}{\sigma^{2}} \leq \int\_{0}^{2\sqrt{\frac{1+t}{1-t}}} \sup\_{h\leq\sigma} ||\Delta\_{h\eta}f'||\frac{d\sigma}{\sigma} \\ &= \int\_{0}^{2\sqrt{\frac{1+t}{1-t}}} \frac{\omega\_{\varrho}(f',\sigma)}{\sigma} d\sigma \leq \int\_{0}^{\frac{2}{\sqrt{\delta}}} \frac{\omega\_{\varrho}(f',\sigma)}{\sigma} d\sigma, \end{split}$$

and using (56), we conclude that

$$|A\_1(t)| \le \int\_0^{\frac{2}{\sqrt{3}}} \frac{\omega\_\varrho(f', \sigma)}{\sigma} d\sigma = \int\_0^1 \omega\_\varrho\left(f', \frac{2}{\sqrt{3}}\mu\right) \frac{du}{u} \le \mathcal{C} \int\_0^1 \frac{\omega\_\varrho(f', u)}{u} du. \tag{59}$$

Finally, since

$$\int\_{-1}^{2t+1} \frac{dx}{(x-t)^2} = \lim\_{\varepsilon \to 0^+} \left[ \int\_{-1}^{t-\varepsilon} \frac{1}{(x-t)^2} dx + \int\_{t+\varepsilon}^{2t+1} \frac{1}{(x-t)^2} dx - \frac{2}{\varepsilon} \right] = -\frac{2}{1+t'} $$

we have

$$|A\_2(t)| = \frac{2}{1+t}|f(t)| \le 2\frac{||f||}{1+t'} $$

and the statement follows by collecting this last inequality, (59) and (57).

Similarly, we can prove the following

**Lemma 3.** *For any* 1 2≤ *t* < 1*, and for any f s.t. f* ∈ *DT, we have*

$$\int\_{-2t-1}^{1} \frac{f(x)}{(x-t)^2} dx \le \mathcal{C} \left( \int\_{0}^{1} \frac{\omega\_{\rho}(f', \sigma)}{\sigma} d\sigma + \frac{||f||}{1-t} \right) \lambda$$

*where* C = C(*f* , *t*).

**Proof of Theorem 4.** Assume first that −1 < *t* ≤ −1 2. In this case, *ϕ*<sup>2</sup>(*t*) ∼ (1 + *t*) and we have

$$\left|\varphi^{2}(t)\left|\mathcal{H}^{1}(f,t)\right|\sim(1+t)\left|\int\_{-1}^{2t+1}\frac{f(\mathbf{x})}{(\mathbf{x}-t)^{2}}d\mathbf{x}+\int\_{2t+1}^{1}\frac{f(\mathbf{x})}{(\mathbf{x}-t)^{2}}d\mathbf{x}\right|.\tag{60}$$

Since

$$(1+t)\left|\int\_{2t+1}^{1} \frac{f(x)}{(x-t)^2} dx\right| \le \mathcal{C} \|f\|\_{\mathcal{H}}$$

the statement follows from Lemma 2 for any −1 < *t* ≤ −1 2.

Assume now 1 2≤ *t* < 1, so that *ϕ*<sup>2</sup>(*t*) ∼ (1 − *t*). By using the decomposition

$$
\log^2(t) \left| \mathcal{H}^1(f, t) \right| \sim (1 - t) \left| \int\_{-1}^{2t - 1} \frac{f(\mathbf{x})}{(\mathbf{x} - t)^2} d\mathbf{x} + \oint\_{2t - 1}^1 \frac{f(\mathbf{x})}{(\mathbf{x} - t)^2} d\mathbf{x} \right|, \tag{61}
$$

and taking into account that

$$(1-t)\left|\int\_{-1}^{2t-1} \frac{f(x)}{(x-t)^2} dx\right| \le \mathcal{C} \|f\|\_{\ell'}$$

the statement follows from Lemma 3 for any 1 2≤ *t* < 1.

Finally, suppose |*t*| < 1 2 and fix 1 4 < *a* < 1 2 . In this case, *ϕ*(*t*) ∼ 1 and we consider the following decomposition

$$\begin{split} \left| q^2(t) \left| \mathcal{H}^1(f, t) \right| \right| &\sim \left| \int\_{|\mathbf{x} - t| \ge a} \frac{f(\mathbf{x})}{(\mathbf{x} - t)^2} d\mathbf{x} + \oint\_{t - a}^{t + a} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} + \mathbf{1} \right| \\ &+ f(t) \oint\_{t - a}^{t + a} \frac{d\mathbf{x}}{(\mathbf{x} - t)^2} \Big| \,. \end{split} \tag{62}$$

For the first term at the right-hand side of (62) we ge<sup>t</sup>

$$\left| \int\_{|\mathbf{x}-t| \ge a} \frac{f(\mathbf{x})}{(\mathbf{x}-t)^2} d\mathbf{x} \right| \le \mathcal{C} \|f\|.$$

Concerning the second addendum of (62), we proceed analogously to the estimate of *<sup>A</sup>*1(*t*) in Lemma 2. More precisely, by using *f* ∈ *DT* and (58) we obtain

$$\begin{aligned} \int\_{t-a}^{t+a} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} &= \quad \int\_{t-a}^{t+a} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} \\ &= \quad \left( \int\_{t-a}^{t} + \int\_{t}^{t+a} \right) \left( \int\_{t}^{\mathbf{x}} \left[ f'(\mathbf{r}) - f'(t) \right] d\mathbf{r} \right) \frac{d\mathbf{x}}{(\mathbf{x} - t)^2} \end{aligned}$$

and by changing the variables *x* = *t* ± *σ* 2 *ϕ*(*t*) and *τ* = *t* ± *h* 2 *ϕ*(*t*), we ge<sup>t</sup>

$$\left| \int\_{t-a}^{t+a} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} \right| \le \int\_0^{\frac{2a}{\rho(t)}} \int\_0^{\sigma} \left| \Delta\_{h\rho(t)} f'(t) \right| d\mathbf{h} \frac{d\sigma}{\sigma^2} \le \mathcal{C} \int\_0^1 \frac{\omega\_\rho(f', \mathbf{u})}{\mathbf{u}} d\mathbf{u}.$$

Finally, as regards the third term at the right-hand side of (62), since . =*t*+*<sup>a</sup> t*−*a dx* (*<sup>x</sup>*−*<sup>t</sup>*)<sup>2</sup> = −2 *a* , we have

$$\left| f(t) \neq \int\_{t-a}^{t+a} \frac{dx}{(x-t)^2} \right| \le \frac{2}{a} \|f\|\_{\varkappa}$$

and the theorem is completely proven.

**Proof of Proposition 1.** Start from the standard decomposition

$$\mathcal{H}(f,t) = \int\_{-1}^{1} \frac{f(\mathbf{x}) - f(t)}{\mathbf{x} - t} d\mathbf{x} + f(t)\mathcal{H}(\mathbf{1}, t), \tag{63}$$

and taking into account

$$\mathcal{H}(\mathbf{1},t) := \int\_{-1}^{1} \frac{d\mathbf{x}}{\mathbf{x} - t} = \log\left(\frac{1 - t}{1 + t}\right) \, dt$$

we must prove that the principal value integral in (63) is indeed an improper integral. To this aim, let us first prove that

$$\int\_{t}^{1} \frac{f(\mathbf{x}) - f(t)}{\mathbf{x} - t} d\mathbf{x} = \lim\_{\mathbf{c} \to 0^{+}} \int\_{t + \mathbf{c}}^{1} \frac{f(\mathbf{x}) - f(t)}{\mathbf{x} - t} d\mathbf{x} < \infty. \tag{64}$$

Please note that for any *-* > 0,

$$\int\_{t+\mathcal{c}}^{1} \frac{f(\mathbf{x}) - f(t)}{\mathbf{x} - t} d\mathbf{x} = \int\_{\mathcal{c}}^{1-t} \frac{f(\mathbf{u} + t) - f(t)}{\mathbf{u}} d\mathbf{u}.$$

Moreover, for any *g* ∈ AC, we note that

$$\begin{aligned} f(u+t) - f(t) &=& f(u+t) - g(u+t) - f(t) + g(t) + g(u+t) - g(t) \\ &\le& 2\|f - \mathbf{g}\| + \int\_{t}^{u+t} \mathbf{g}'(\sigma)d\sigma \\ &\le& 2\|f - \mathbf{g}\| + \|\mathbf{g}'\mathbf{g}\| \int\_{t}^{u+t} \frac{d\sigma}{\mathbf{g}(\sigma)} \\ &=& 2\|f - \mathbf{g}\| + u\|\mathbf{g}'\mathbf{g}\| \left[\frac{\arcsin(u+t) - \arcsin(t)}{u}\right]. \end{aligned}$$

On the other hand, recalling that

$$\arcsin y = y + \frac{y^3}{6} + \frac{3}{40}y^5 + \frac{5}{112}y^7 + \frac{35}{1152}y^9 + \dots, \qquad |y| < 1,$$

we easily ge<sup>t</sup>

$$\frac{\arcsin(u+t) - \arcsin(t)}{u} \le \mathcal{C} \ne \mathcal{C}(t, u), \qquad |t| < 1, \quad 0 < u \le 2\pi$$

and therefore, the previous estimate and (3) yield

$$\mathcal{I}\left(f(u+t) - f(t) \le \mathcal{C}\left(\inf\_{\mathcal{g}\in\mathcal{AC}}\{\|f-\mathcal{g}\|+\mathfrak{u}\|\mathcal{g}'\mathcal{g}\|\right)\right) = \mathcal{C}K\_{1,\mathfrak{p}}(f,\mathfrak{u}) \sim \omega\_{\mathfrak{P}}(f,\mathfrak{u}).$$

Hence, for all |*t*| < 1 it follows

$$\lim\_{\mathfrak{c}\to 0^{+}} \int\_{\mathfrak{c}}^{1-t} \frac{f(u+t) - f(t)}{u} du \le \mathcal{C} \lim\_{\mathfrak{c}\to 0^{+}} \int\_{\mathfrak{c}}^{2} \frac{\omega\_{\mathbb{P}}(f, u)}{u} du, \qquad \mathcal{C} \ne \mathcal{C}(t),$$

i.e., under the assumption *f* ∈ *DT*, (64) holds.

Similarly proceeding, we can prove that

$$\int\_{-1}^{t} \frac{f(\mathbf{x}) - f(t)}{\mathbf{x} - t} d\mathbf{x} = \lim\_{\varepsilon \to 0^{+}} \int\_{\varepsilon}^{1 + t} \frac{f(t) - f(t - u)}{u} du < \infty$$

and the statement follows.

**Proof of Proposition 2.** For 1 ≤ *k* ≤ *m* − 1, by using the recurrence relation (15) and taking into account that . 1−1 *pm*,*<sup>h</sup>*(*x*)*dx* = 2 *m*+1 holds ∀*h* ∈ <sup>N</sup>*<sup>m</sup>*0 , we ge<sup>t</sup>

$$\begin{split} q\_{m,k}(t) &= \quad \frac{1}{2} \int\_{-1}^{1} \frac{(1-x)p\_{m-1,k}(x) - (1-t)p\_{m-1,k}(t)}{x-t} dx \\ &\quad + \frac{1}{2} \int\_{-1}^{1} \frac{(1+x)p\_{m-1,k-1}(x) - (1+t)p\_{m-1,k-1}(t)}{x-t} dx \\ &= \quad \frac{1}{2} \left( q\_{m-1,k}(t) - \int\_{-1}^{1} \frac{xp\_{m-1,k}(x) - tp\_{m-1,k}(t)}{x-t} dx \right) \\ &\quad + \frac{1}{2} \left( q\_{m-1,k-1}(t) + \int\_{-1}^{1} \frac{xp\_{m-1,k-1}(x) - tp\_{m-1,k-1}(t)}{x-t} dx \right) \\ &= \quad \frac{1}{2} \left( q\_{m-1,k}(t) - \frac{2}{m} - tq\_{m-1,k}(t) \right) + \frac{1}{2} \left( q\_{m-1,k-1}(t) + \frac{2}{m} + tq\_{m-1,k-1}(t) \right) \\ &\quad = \quad \frac{(1-t)}{2} q\_{m-1,k}(t) + \frac{(1+t)}{2} q\_{m-1,k-1}(t). \end{split}$$

For *k* = 0, we have

$$\begin{aligned} q\_{m,0}(t) &= \frac{1}{2} \int\_{-1}^{1} \frac{(1-x)p\_{m-1,0}(x) - (1-t)p\_{m-1,0}(t)}{x-t} dx = \frac{1}{2} \left( q\_{m-1,k}(t) - \frac{2}{m} - tq\_{m-1,k}(t) \right) \\ &= \frac{(1-t)}{2} q\_{m-1,k}(t) - \frac{1}{m} .\end{aligned}$$

For *k* = *m* we proceed analogously.

**Proof of Theorem 5.** Set *Rm*,*<sup>s</sup> f* = *f* − *Bm*,*<sup>s</sup> f* , we have

$$\mathcal{E}\_{m,s}(f,t) = \mathcal{H}(R\_{m,s}f,t), \qquad \text{and} \qquad \Phi\_{m,s}(f,t) = \mathcal{F}(R\_{m,s}f,t).$$

Applying Theorem 3, E*<sup>m</sup>*,*<sup>s</sup>*(*f* , *t*) can be estimated as follows

$$|\mathcal{E}\_{m,s}(f,t)| \le \mathcal{C} \log\left(\frac{\varepsilon}{1-t^2}\right) \left[\|R\_{m,s}f\| + \int\_0^1 \frac{\omega\_\Psi(R\_{m,s}f,u)}{u} du\right], \qquad \mathcal{C} \ne \mathcal{C}(m,f,t), \tag{65}$$

and by Theorem 1 we further obtain

$$\|R\_{m,s}f\| \le \mathcal{C}\left[\omega\_{\varphi}^{2s}\left(f,\frac{1}{\sqrt{m}}\right) + \frac{\|f\|}{m^{\varepsilon}}\right], \qquad \mathcal{C} \ne \mathcal{C}(m,f). \tag{66}$$

Moreover, by Lemma 1 we ge<sup>t</sup>

$$\begin{aligned} \int\_0^1 \frac{\omega\_\theta(R\_{m,s}f, u)}{u} du &= \int\_0^{\frac{1}{m}} \frac{\omega\_\theta(R\_{m,s}f, u)}{u} du + \int\_{\frac{1}{m}}^1 \frac{\omega\_\theta(R\_{m,s}f, u)}{u} du \\ &\leq \quad \mathcal{C} \left( \int\_0^{\frac{1}{m}} \frac{\omega\_\theta^I(f, u)}{u} du + \|R\_{m,s}f\| \log m \right) \end{aligned}$$

and (42) follows from this last estimate, (65) and (66).

Regarding the quadrature error <sup>Φ</sup>*<sup>m</sup>*,*<sup>s</sup>*(*f* , *t*), we observe that

$$\Phi\_{\mathfrak{m},\mathfrak{s}}(f,t) = \mathcal{F}(R\_{\mathfrak{m},\mathfrak{s}}f,t) = \mathcal{H}(R\_{\mathfrak{m},\mathfrak{s}}f,t) - \log\left(\frac{1-t}{1+t}\right)R\_{\mathfrak{m},\mathfrak{s}}f(t),$$

which leads to

$$\begin{split} \left| \log^{-1} \left( \frac{\varepsilon}{1 - t^2} \right) |\Phi\_{m,s}(f, t)| \right| &\leq \left| \log^{-1} \left( \frac{\varepsilon}{1 - t^2} \right) |\mathcal{H}(R\_{m,s}f, t)| + \mathcal{C} |R\_{m,s}f(t)| \right| \\ &\leq \left| \mathcal{C} \log^{-1} \left( \frac{\varepsilon}{1 - t^2} \right) |\mathcal{E}\_{m,s}(f, t)| + \mathcal{C} \|R\_{m,t}f\|. \end{split}$$

Hence, in the case that *f* ∈ *DT*, the estimate (42) holds for <sup>Φ</sup>*<sup>m</sup>*,*<sup>s</sup>*(*f* , *t*) as well. Finally, if *f* ∈ *C*<sup>1</sup> then, by applying the mean value theorem, we ge<sup>t</sup>

$$|\Phi\_{m,s}(f,t)| = \left| \int\_{-1}^{1} \frac{R\_{m,s}f(x) - R\_{m,s}f(t)}{x - t} dx \right| \le \mathcal{C} \|(f - B\_{m,s}f)'\|\_{2}$$

and (43) follows from Theorem 2.

**Proof of Proposition 3.** We start from the standard decomposition

$$\mathcal{H}^1(f,t) = \oint\_{-1}^1 \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x}-t)}{(\mathbf{x}-t)^2} d\mathbf{x} + \oint\_{-1}^1 \frac{f(t) + f'(t)(\mathbf{x}-t)}{(\mathbf{x}-t)^2} d\mathbf{x},\tag{67}$$

and recalling the definitions

$$\begin{array}{rcl}\displaystyle\displaystyle\int\_{-1}^{1}\frac{\operatorname{g}(\mathbf{x})}{(\mathbf{x}-t)^{2}}d\mathbf{x} &=& \displaystyle\lim\_{\varepsilon\to 0^{+}}\left[\int\_{-1}^{t-\varepsilon}\frac{\operatorname{g}(\mathbf{x})}{(\mathbf{x}-t)^{2}}d\mathbf{x}+\int\_{t+\varepsilon}^{1}\frac{\operatorname{g}(\mathbf{x})}{(\mathbf{x}-t)^{2}}d\mathbf{x}-\frac{2\operatorname{g}(t)}{\varepsilon}\right],\\ \displaystyle\int\_{-1}^{1}\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}-t}d\mathbf{x} &=& \displaystyle\lim\_{\varepsilon\to 0^{+}}\left[\int\_{-1}^{t-\varepsilon}\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}-t}d\mathbf{x}+\int\_{t+\varepsilon}^{1}\frac{\operatorname{g}(\mathbf{x})}{\mathbf{x}-t}d\mathbf{x}\right],\end{array}$$

we note that

$$\int\_{-1}^{1} \frac{d\mathbf{x}}{(\mathbf{x} - t)^2} = -\frac{2}{1 - t^2}, \qquad \int\_{-1}^{1} \frac{(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} = \int\_{-1}^{1} \frac{d\mathbf{x}}{(\mathbf{x} - t)} = \log\left(\frac{1 - t}{1 + t}\right).$$

Moreover, taking into account that

$$f(\mathbf{x}) - f(t) = f'(\mathbf{j}\_{\mathbf{x},t})(\mathbf{x} - t), \qquad \min\{\mathbf{x}, t\} < \mathbf{j}\_{\mathbf{x},t} < \max\{\mathbf{x}, t\}, \mathbf{j}\_{\mathbf{x},t}$$

we have

$$\oint\_{-1}^{1} \frac{f(\mathbf{x}) - f(t) - f'(t)(\mathbf{x} - t)}{(\mathbf{x} - t)^2} d\mathbf{x} = \oint\_{-1}^{1} \frac{f'(\mathbf{\tilde{y}}\_{\mathbf{x},t}) - f'(t)}{(\mathbf{x} - t)} d\mathbf{x} dt$$

Hence to complete the proof, we have to prove that this last principal value integral is indeed an improper integral if *f* ∈ *DT*.

We are going to prove that

$$\int\_{t}^{1} \frac{f'(\xi\_{\mathbf{x},t}) - f'(t)}{(\mathbf{x} - t)} d\mathbf{x} = \lim\_{\varepsilon \to 0^{+}} \int\_{t + \varepsilon}^{1} \frac{f'(\xi\_{\mathbf{x},t}) - f'(t)}{(\mathbf{x} - t)} d\mathbf{x} < \infty,\tag{68}$$

being the proof of

$$\int\_{-1}^{t} \frac{f'(\xi\_{x,t}) - f'(t)}{(x - t)} d\mathbf{x} = \lim\_{\varepsilon \to 0^{+}} \int\_{-1}^{t - \varepsilon} \frac{f'(\xi\_{x,t}) - f'(t)}{(x - t)} d\mathbf{x} < \infty$$

analogous.

Set *ξ<sup>x</sup>*,*<sup>t</sup>* = (*x* − *t*)*θ* + *t*, with 0 < *θ* < 1, for any *-* > 0, we have

$$\int\_{t+\varepsilon}^{1} \frac{f'(\xi\_{\varepsilon}t) - f'(t)}{(\mathbf{x} - t)} d\mathbf{x} = \int\_{t+\varepsilon}^{1} \frac{f'((\mathbf{x} - t)\theta + t) - f'(t)}{(\mathbf{x} - t)} d\mathbf{x} = \int\_{\varepsilon}^{1-t} \frac{f'(u\theta + t) - f'(t)}{u} du.$$

On the other hand, for any *g* ∈ AC, |*t*| < 1, 0 < *θ* < 1 and 0 < *u* ≤ 2, similarly to the proof of Proposition 1, we have

$$\begin{split} f'(u\theta + t) - f'(t) &= \left. f'(u\theta + t) - g(u\theta + t) - f'(t) + g(t) + g(u\theta + t) - g(t) \right| \\ &\leq \left. 2\| f' - \mathbf{g} \| + \int\_{t}^{u\theta + t} \mathbf{g}'(\sigma) d\sigma \\ &\leq \left. 2\| f' - \mathbf{g} \| + u\| \| \mathbf{g}' \mathbf{g} \| \right| \left[ \frac{\arcsin(u\theta + t) - \arcsin(t)}{u} \right] \\ &\leq \mathcal{C} \left( \| f' - \mathbf{g} \| + u \| \| \mathbf{g}' \mathbf{g} \| \right), \qquad \mathcal{C} \neq \mathcal{C}(\mathcal{g}, u, \theta, t). \end{split}$$

Hence, by means of (3), we ge<sup>t</sup>

$$\lim\_{\varepsilon \to 0^{+}} \int\_{\varepsilon}^{1-t} \frac{f'(u\theta + t) - f'(t)}{u} du \le \mathcal{C} \lim\_{\varepsilon \to 0^{+}} \int\_{\varepsilon}^{2} \frac{\omega\_{\varphi}(f', u)}{u} du$$

and under the assumption *f* ∈ *DT*, (68) follows.

**Proof of Theorem 6.** We start from

$$\mathcal{E}\_{m,s}^{1}(f,t) = \mathcal{H}^{1}(R\_{m,s}f,t), \qquad R\_{m,s}f(t) = f(t) - B\_{m,s}f(t).$$

By Theorem 4, we have

$$|(1 - t^2)|\mathcal{H}^1(R\_{m, \mathfrak{s}}f, t)| \le \mathcal{C} \left( ||R\_{m, \mathfrak{s}}f|| + \int\_0^1 \frac{\omega\_\varrho((R\_{m, \mathfrak{s}}f)', \tau)}{\tau} d\tau \right) . \ t$$

Since

$$\begin{split} \int\_{0}^{1} \frac{\omega\_{\varphi}((R\_{m,\sf s}f)',\tau)}{\tau} d\tau &= \left\{ \int\_{0}^{\frac{1}{m}} + \int\_{\frac{1}{m}}^{1} \right\} \frac{\omega\_{\varphi}((R\_{m,\sf s}f)',\tau)}{\tau} d\tau \\ &\leq \int\_{0}^{\frac{1}{m}} \frac{\omega\_{\varphi}((R\_{m,\sf s}f)',\tau)}{\tau} d\tau + 2||(R\_{m,\sf s}f)'||\int\_{\frac{1}{m}}^{1} \frac{d\tau}{\tau} \\ &= \int\_{0}^{\frac{1}{m}} \frac{\omega\_{\varphi}((R\_{m,\sf s}f)',\tau)}{\tau} d\tau + 2||(R\_{m,\sf s}f)'||\log m\_{\sf s} \end{split}$$

by Lemma 1 we ge<sup>t</sup>

$$|(1 - t^2)|\mathcal{H}^1(\mathbb{R}\_{m, \mathfrak{s}} f, t)| \le \mathcal{C} \left( \| |\mathcal{R}\_{m, \mathfrak{s}} f| \| + \| (\mathcal{R}\_{m, \mathfrak{s}} f)' \| \log m + \int\_0^{\frac{1}{m}} \frac{\omega\_\varrho^r(f', \tau)}{\tau} d\tau \right),$$

i.e., (53) holds.

> The same estimate (53) also holds for <sup>Φ</sup>1*m*,*s*, since by (44) we have

$$\Phi\_{m,s}^1(f,t) = \mathcal{H}^1(R\_{m,s}f,t) - \log\left(\frac{1-t}{1+t}\right)(R\_{m,s}f)'(t) + \frac{2}{1-t^2}R\_{m,s}f(t),$$

and we note that

$$\begin{aligned} \left| (1 - t^2) \left| \log \left( \frac{1 - t}{1 + t} \right) (R\_{m, \mathcal{f}} f)'(t) \right| \right| &\leq \mathcal{C} \| (R\_{m, \mathcal{f}} f)' \|\_{\prime} \qquad \mathcal{C} \neq \mathcal{C} (t, f, m), \\\ \| (1 - t^2) \left| \frac{2}{1 - t^2} R\_{m, \mathcal{f}} f(t) \right| &\leq 2 \| R\_{m, \mathcal{f}} f \|. \end{aligned}$$

Finally, (54) follows from the Peano form of the Taylor's remainder term, namely

$$g(\mathbf{x}) = g(t) + g'(t)(\mathbf{x} - t) + g''(\xi)\frac{(\mathbf{x} - t)^2}{2}, \qquad \min\{\mathbf{x}, t\} \le \xi \le \max\{\mathbf{x}, t\}, \xi$$

which for *g* = *Rm*,*<sup>s</sup> f* , yields

$$\begin{aligned} \left| \Phi\_{\mathfrak{m},\mathfrak{s}}^{1}(f,t) \right| &= \left| \mathcal{F}^{1}(\mathbb{R}\_{\mathfrak{m},\mathfrak{s}}f,t) \right| \\ &\leq \int\_{-1}^{1} \frac{\left| R\_{\mathfrak{m},\mathfrak{s}}f(\mathbf{x}) - R\_{\mathfrak{m},\mathfrak{s}}f(t) - (R\_{\mathfrak{m},\mathfrak{s}}f)'(t)(\mathbf{x}-t) \right|}{(\mathbf{x}-t)^{2}} d\mathbf{x} \\ &\leq \left| \left| (R\_{\mathfrak{m},\mathfrak{s}}f)'' \right| \right|. \end{aligned}$$

**Author Contributions:** All authors equally contributed to the paper. Conceptualization, F.F., D.O. and W.T.; methodology, F.F., D.O. and W.T.; software, F.F., D.O. and W.T.; validation, F.F., D.O. and W.T.; analysis, F.F., D.O. and W.T.; investigation, F.F., D.O. and W.T.; resources, F.F., D.O. and W.T.; data curation, F.F., D.O. and W.T.; writing–original draft preparation, writing–review and editing, F.F., D.O. and W.T.; visualization, F.F., D.O. and W.T.; supervision F.F., D.O. and W.T. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was partially supported by INdAM - GNCS Project 2019 "Discretizzazione di misure, approssimazione di operatori integrali ed applicazioni". The research of the first author was partially supported by the Helmholtz Association under the project Ptychography4.0.

**Acknowledgments:** The authors thank the anonymous referees for their suggestions and remarks, which allowed to improve the paper. This research has been accomplished within the RITA "Research ITalian network on Approximation".

**Conflicts of Interest:** The authors declare no conflict of interest.
