In this case the desired estimate follows from the inequalities

\[ \sum _ { n = 0 } ^ { n - 1 } ( k + 1 ) ^ { r + 1 } \rho ^ { k } \leq C ( 1 - \rho ) ^ { - r - 2 } \, , \quad ( 1 - \rho ) ^ { r + 1 } \rho ^ { n } \leq C ^ { \prime } ( n + 1 ) ^ { - r - 1 } \, , \]

for 0 \( < \rho < \) 1.

(\( \Leftarrow \)=) We begin with (i) and assume first that \( p \) = 2. Use (2.6) to write for 0 \( < \rho < \) 1,\( \lambda \in \mathbb { T } \),

\[ \frac { ( n + 1 ) ( n + 2 ) } { 2 } M _ { n } ^ { ( 2 ) } ( \lambda T ) = \int _ { 0 } ^ { 2 \pi } ( I - \rho \lambda e ^ { i t } T ) ^ { - 1 } \sum _ { j = 0 } ^ { n } ( n - j + 1 ) e ^ { - i j t } \rho ^ { - j } \frac { \mathrm { d } t } { 2 \pi } \, . \]

so that

\[ \begin{array} { r l } & { \left\| \frac { ( n + 1 ) ( n + 2 ) } { 2 } M _ { n } ^ { ( 2 ) } ( \lambda T ) \right\| \le \frac { C } { ( 1 - \rho ) ^ { r + 1 } } \rho ^ { - n } \int _ { 0 } ^ { 2 \pi } \frac { 1 } { | 1 - \rho e ^ { - i t } | ^ { 2 } } \frac { \mathrm { d } t } { 2 \pi } } \\ & { \qquad \qquad \qquad \qquad = \frac { C } { ( 1 - \rho ) ^ { r + 1 } ( 1 - \rho ^ { 2 } ) } \rho ^ { - n } } \\ & { \qquad \qquad \qquad \qquad \le \frac { C } { ( 1 - \rho ) ^ { r + 2 } } \rho ^ { - n } \, . } \end{array} \]

If we now choose \( \textstyle \rho = 1 - { \frac { 1 } { n } } \) we obtain the estimate for \( M _ { n } ^ { ( 2 ) } ( T ) \) in (i) for all \( r \ge \) 0. For \( p > \) 2, this follows immediately from the first equality in (2.6).

Let us turn to the case when \( p \) = 1. For \( n \geq \) 1, \( \mu \in \mathbb { T } \) and \( \lambda \in \mathbb { C } \) with \|λ\| > 1 we have by the Abel summation formula

\[ \begin{array} { c } { { \displaystyle M _ { n } ( \mu T ) = \frac { 1 } { n + 1 } \sum _ { k = 0 } ^ { n } ( \mu T ) ^ { k } = \frac { 1 } { n + 1 } \sum _ { k = 0 } ^ { n } \lambda ^ { k + 1 } \lambda ^ { - k - 1 } ( \mu T ) ^ { k } } } \\ { { = \frac { 1 } { n + 1 } \left( \lambda ^ { n + 1 } \sum _ { k = 0 } ^ { n } \lambda ^ { - k - 1 } ( \mu T ) ^ { k } + ( 1 - \lambda ) \sum _ { k = 0 } ^ { n - 1 } \lambda ^ { k + 1 } \sum _ { j = 0 } ^ { k } \lambda ^ { - j - 1 } ( \mu T ) ^ { j } \right) } } \\ { { = \frac { 1 } { n + 1 } \left( \lambda ^ { n + 1 } \overline { { { \mu } } } \sum _ { k = 0 } ^ { n } ( \lambda \overline { { { \mu } } } ) ^ { - k - 1 } T ^ { k } + ( 1 - \lambda ) \sum _ { k = 0 } ^ { n - 1 } \lambda ^ { k + 1 } \overline { { { \mu } } } \sum _ { j = 0 } ^ { k } ( \lambda \overline { { { \mu } } } ) ^ { - j - 1 } T ^ { j } \right) . } } \end{array} \]

From the assumption

\[ \operatorname* { s u p } _ { \stackrel { n \in \mathbb { N } } { | \lambda | > 1 } } \frac { ( | \lambda | - 1 ) ^ { r + 1 } } { | \lambda | ^ { r } } \left\| \sum _ { k = 0 } ^ { n } \lambda ^ { - k - 1 } T ^ { k } \right\| < \infty \, , \]

we obtain

\[ \| M _ { n } ( \mu T ) \| \leq C \frac { | \lambda | ^ { r + 1 } } { ( | \lambda | - 1 ) ^ { r + 1 } ( n + 1 ) } \left( | \lambda | ^ { n } + \frac { | 1 - \lambda | ( | \lambda | ^ { n } - 1 ) } { | \lambda | - 1 } \right) \, , \]

and if \( \textstyle \lambda = 1 + { \frac { 1 } { n } } \),

\[ \| M _ { n } ( \mu T ) \| \le 2 ^ { r } ( 2 e - 1 ) C n ^ { r } \]

for \( n \geq \) 1 and \( \mu \in \mathbb { T } \), which concludes the proof.

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