satisfying the condition (4.29). Then there exists a martingale $f\in H_{p}$
such that

$\sup_{n\in\mathbb{N}}\left\|\frac{t_{n}f}{\Phi_{n}}\right\|_{weak-
L_{p}}=\infty.$

###### Corollary 4.17

Let $\left\\{\Phi_{n}:n\in\mathbb{N}\right\\}$ be any non-decreasing sequence
satisfying the condition (4.29). Then the maximal operator

$\sup_{n\in\mathbb{N}}\frac{\left|t_{n}f\right|}{\Phi_{n}}$

is not bounded from the Hardy space $H_{p}$ to the space $weak-L_{p}.$

We now formulate our final result in this section.

###### Theorem 4.18

Let $f\in H_{1/(1+\alpha)},$ where $0<\alpha\leq 1$ and
$\\{q_{k}:k\in\mathbb{N}\\}$ be a sequence of non-increasing numbers
satisfying the conditions (1.39) and (1.40). Then there exists an absolute
constant $c_{\alpha}$ depending only on $\alpha$ such that the maximal
operator

$\overset{\sim}{t}_{\alpha}^{\ast}:=\frac{\left|t_{n}f\right|}{\log^{1+\alpha}\left(n+1\right)}$

is bounded from the martingale Hardy space $H_{1/(1+\alpha)}$ to the Lebesgue
space $L_{1/(1+\alpha)}.$

b) Let $\left\\{\Phi_{n}:n\in\mathbb{N}_{+}\right\\}$ be any non-decreasing
sequence satisfying the condition

$\overline{\lim_{n\rightarrow\infty}}\frac{\log^{1+\alpha}\left(n+1\right)}{\Phi_{n}}=+\infty.$
(4.35)

Then there exists Nörlund means with non-increasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}$ satisfying the conditions (4.22) and (4.23) such
that

$\sup_{k\in\mathbb{N}}\frac{\left\|\sup_{n}\left|\frac{t_{n}f_{k}}{\Phi_{n}}\right|\right\|_{1/(1+\alpha)}}{\left\|f\right\|_{H_{1/(1+\alpha)}}}=\infty.$

###### Remark 4.19

Part a) of this result can be found in the paper Bhahota and Tephnadze [7].
Part b) has not been stated before for such a general case.

Proof: According to Lemma 1.39 the proof of part a) will be complete if we
show that

$\int_{\overline{I_{N}}}\left|\overset{\sim}{t}_{\alpha}^{\ast}a\right|^{1/\left(1+\alpha\right)}d\mu<\infty$

for every $1/\left(1+\alpha\right)$-atom $a.$ We may assume that $a$ is an
arbitrary $1/\left(1+\alpha\right)$-atom with support$\ I,$
$\mu\left(I\right)=M_{N}^{-1}$ and $I=I_{N}.$ It is easy to see that
$t_{m}\left(a\right)=0,$ when $m\leq M_{N}.$ Therefore we can suppose that
$m>M_{N}.$

Let $x\in I_{N}.$ Since $t_{m}$ is bounded from $L_{\infty}$ to $L_{\infty}$
(the boundedness follows from Corollary 1.33) and
$\left\|a\right\|_{\infty}\leq M_{N}^{1/\left(1+\alpha\right)}$ we obtain that

$\left|t_{m}a\left(x\right)\right|\leq\int_{I_{N}}\left|a\left(t\right)\right|\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq\left\|a\left(x\right)\right\|_{\infty}\int_{I_{N}}\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq
M_{N}^{1+\alpha}\int_{I_{N}}\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right).$

Let $x\in I_{N}^{k,l},\,0\leq k<l<N.$ From Lemma 1.34 we get that

$\left|t_{m}a\left(x\right)\right|\leq\frac{c_{\alpha}M_{k}M_{l}^{\alpha}M_{N}^{\alpha}}{m^{\alpha}}.$
(4.36)

Let $x\in I_{N}^{k,N},\,0\leq k<N.$ In the view of Lemma 1.34 we have that

$\left|t_{m}a\left(x\right)\right|\leq c_{\alpha}M_{k}M_{N}^{\alpha}.$ (4.37)

Let $x\in I_{N}^{k,l},\,0\leq k<l\leq N.$ Since $n>M_{N}$ we can conclude that

$\frac{\left|t_{n}a\left(x\right)\right|}{\log^{1+\alpha}n}\leq\frac{c_{\alpha}M_{k}M_{l}^{\alpha}}{N^{1+\alpha}}.$
(4.38)

The expression on the right-hand side of (4.38) does not depend on $n.$ Hence,

$\left|\overset{\sim}{t_{\alpha}}^{\ast}a\left(x\right)\right|\leq\frac{c_{\alpha}M_{k}M_{N}^{\alpha}}{N^{1+\alpha}},$
(4.39)

for $x\in I_{N}^{k,l},\,0\leq k<l\leq N.$

According to (1.1) and (4.39) we obtain that

$\int_{\overline{I_{N}}}\left|\overset{\sim}{t_{\alpha}}^{\ast}a\right|^{1/\left(1+\alpha\right)}d\mu$
$=\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\sum\limits_{x_{j}=0,j\in\\{l+1,\dots,N-1\\}}^{m_{j}-1}\int_{I_{N}^{k,l}}\left|\overset{\sim}{t_{\alpha}}^{\ast}a\right|^{1/\left(1+\alpha\right)}d\mu$
$+\overset{N-1}{\underset{k=0}{\sum}}\int_{I_{N}^{k,N}}\left|\overset{\sim}{t_{\alpha}}^{\ast}a\right|^{1/\left(1+\alpha\right)}d\mu$
$\leq\frac{c_{\alpha}}{N}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{1}{M_{l}}M_{l}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}$
$+\frac{c_{\alpha}}{N}\overset{N-1}{\underset{k=0}{\sum}}\frac{1}{M_{N}}M_{N}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}$
$\leq\frac{c_{\alpha}}{N}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{M_{l}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}}{M_{l}}$
$+\frac{c_{\alpha}}{N}\overset{N-1}{\underset{k=0}{\sum}}\frac{M_{k}^{1/\left(1+\alpha\right)}}{M_{N}^{1/\left(1+\alpha\right)}}\leq
c_{\alpha}<\infty.$

The proof of part a) is complete.

Under condition (4.35) there exists a positive integers $m_{k}^{,}$ such that
$M_{2m_{k}^{,}+1}\leq\lambda_{k}<2M_{2m_{k}^{,}+1}.$ Since $\Phi_{n}$ is a
non-decreasing function we have that

$\overline{\underset{k\rightarrow\infty}{\lim}}\frac{\left(m_{k}^{{}^{\prime}}\right)^{1+\alpha}}{\Phi_{M_{2m_{k}^{,}+1}}}\geq
c\lim_{k\rightarrow\infty}\frac{\log^{1+\alpha}\left(\lambda_{k}\right)}{\Phi_{M_{2\lambda_{k}+1}}}=\infty.$

Let $\
\left\\{n_{k}:k\in\mathbb{N}_{+}\right\\}\subset\left\\{m_{k}^{\prime}:k\in\mathbb{N}_{+}\right\\}$
be a sequence of positive numbers such that

$\lim_{k\rightarrow\infty}\frac{n_{k}^{1+\alpha}}{\Phi_{M_{2n_{k}+1}}}=\infty.$

To prove part b) we use the $1/\left(1+\alpha\right)$-atoms defined in Example
1.44. If we apply (1.11) in Lemma 1.2 with (1.52) and (1.53) and invoke Abel
transformation we get that

$\frac{\left|t_{M_{2n_{k}}+M_{2s}}f_{k}\right|}{\Phi_{M_{2n_{k}}+M_{2s}}}=\frac{1}{\Phi_{M_{2n_{k}}+M_{2s}}Q_{M_{2n_{k}}+M_{2s}}}\left|\sum_{j=M_{2n_{k}}+1}^{M_{2n_{k}}+M_{2s}}q_{M_{2n_{k}}+M_{2s}-j}\left(D_{j}-D_{M_{2n_{k}}}\right)\right|$
$=\frac{1}{\Phi_{M_{2n_{k}}+M_{2s}}Q_{M_{2n_{k}}+M_{2s}}}\left|\sum_{j=1}^{M_{2s}}q_{M_{2s}-j}\left(D_{j+M_{2n_{k}}}-D_{M_{2n_{k}}}\right)\right|$
$=\frac{1}{\Phi_{M_{2n_{k}}+M_{2s}}Q_{M_{2n_{k}}+M_{2s}}}\left|\psi_{M_{2n_{k}}}\sum_{j=1}^{M_{2s}}q_{M_{2s}-j}D_{j}\right|$
$=\frac{1}{\Phi_{M_{2n_{k}}+M_{2s}}Q_{M_{2n_{k}}+M_{2s}}}\left|\sum_{j=1}^{M_{2s}}q_{M_{2s}-j}j\right|.$

Let $x\in I_{2s}/I_{2s+1},$ $s=\left[n_{k}/2\right],...,n_{k}.$ If we again
use abel transformation, then under the conditions (4.22) and (4.23) we find
that

$\frac{\left|t_{M_{2n_{k}}+M_{2s}}f_{k}\right|}{\Phi_{M_{2n_{k}}+M_{2s}}}\geq\frac{c}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\left|\sum_{j=1}^{M_{2s}}\left(q_{M_{2s}-j}-q_{M_{2s}-j-1}\right)j^{2}\right|$
$\geq\frac{c}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\sum_{j=\left[M_{2s}\right]/2}^{M_{2s}}\left|q_{M_{2s}-j}-q_{M_{2s}-j-1}\right|j^{2}$
$\geq\frac{cM_{2s}^{2}}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\sum_{j=\left[M_{2s}\right]/2}^{M_{2s}}\left|q_{M_{2s}-j}-q_{M_{2s}-j-1}\right|$
$\geq\frac{cM_{2s}^{2}}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\sum_{j=1}^{\left[M_{2s}\right]/2}\left|q_{j}-q_{j+1}\right|$
$\geq\frac{cM_{2s}^{2}}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\sum_{j=1}^{\left[M_{2s}\right]/2}j^{\alpha-2}\geq\frac{cM_{2s}^{2}M_{2s}^{\alpha-1}}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}$
$\geq\frac{cM_{2s}^{\alpha+1}}{\Phi_{M_{2n_{k}}+M_{2s}}M_{2n_{k}}^{\alpha}}\geq\frac{cM_{2s}^{\alpha+1}}{\Phi_{M_{2n_{k}+1}}M_{2n_{k}}^{\alpha}}.$

Hence,

$\int_{G_{m}}\left(\sup_{n}\left|\frac{t_{n}f_{k}}{\Phi_{n}}\right|\right)^{1/\left(1+\alpha\right)}d\mu$
$\geq\text{
}\overset{n_{k}-1}{\underset{s=\left[n_{k}/2\right]}{\sum}}\int_{I_{2s}/I_{2s+1}}\left|\frac{t_{M_{2n_{k}}+M_{2s}}f_{k}}{\Phi_{M_{2n_{k}}+M_{2s}}}\right|^{1/\left(1+\alpha\right)}d\mu$
$\geq\frac{c}{\Phi_{M_{2n_{k}+1}}^{1/\left(1+\alpha\right)}}\text{
}\overset{n_{k}-1}{\underset{s=\left[n_{k}/2\right]}{\sum}}\int_{I_{2s}/I_{2s+1}}\frac{M_{2s}}{M_{2n_{k}}^{\alpha/\left(1+\alpha\right)}}d\mu$
$\geq\frac{c}{\Phi_{M_{2n_{k}+1}}^{1/\left(1+\alpha\right)}}\text{
}\overset{n_{k}-1}{\underset{s=\left[n_{k}/2\right]}{\sum}}\frac{M_{2s}^{\alpha+1}}{M_{2n_{k}}^{\alpha/\left(1+\alpha\right)}}\frac{1}{M_{2s}^{\alpha+1}}$
$\geq\frac{c}{\Phi_{M_{2n_{k}+1}}^{1/\left(1+\alpha\right)}}\text{
}\overset{n_{k}-1}{\underset{s=\left[n_{k}/2\right]}{\sum}}\frac{1}{M_{2n_{k}}^{\alpha/\left(1+\alpha\right)}}\geq\frac{cn_{k}}{M_{2n_{k}}^{\alpha/\left(1+\alpha\right)}\Phi_{M_{2n_{k}+1}}^{1/\left(1+\alpha\right)}}.$

Therefore, by also using (1.54) for $p=1/\left(1+\alpha\right)$ we have that

$\frac{\left(\int_{G_{m}}\left(\sup_{n}\left|\frac{t_{n}f_{k}}{\Phi_{n}}\right|\right)^{1/\left(1+\alpha\right)}d\mu\right)^{1+\alpha}}{\left\|f_{k}\right\|_{H_{1/\left(1+\alpha\right)}}}$
$\geq\frac{cn_{k}^{1+\alpha}}{M_{{}_{2n_{k}}}^{\alpha}\Phi_{M_{2n_{k}+1}}}M_{{}_{2n_{k}}}^{\alpha}\geq\frac{cn_{k}^{1+\alpha}}{\Phi_{M_{2n_{k}+1}}}\rightarrow\infty,\text{when
\qquad}k\rightarrow\infty.$

The proof is complete.

###### Corollary 4.20

Let $\left\\{\Phi_{n}:n\in\mathbb{N}_{+}\right\\}$ be any non-decreasing
sequence satisfying the condition (4.35). Then the following maximal operator

$\sup_{n}\left|\frac{t_{n}f}{\Phi_{n}}\right|$

is not bounded from the Hardy space $H_{1/\left(1+\alpha\right)}$ to the space
$L_{1/\left(1+\alpha\right)}.$

###  4.3 Strong convergence of Nörlund means on martingale Hardy spaces

The first result in this section is due to Persson, Tephnadze and Wall [51].

###### Theorem 4.21

Let $0<p<1/2$, $f\in H_{p}$ and the sequence
$\left\\{q_{k}:k\in\mathbb{N}\right\\}$ be non-decreasing. Then there exists
an absolute constant $c_{p}$ depending only on $p$ such that

$\overset{\infty}{\underset{k=1}{\sum}}\frac{\left\|t_{k}f\right\|_{p}^{p}}{k^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}}^{p}.$

###### Remark 4.22

Since the Fejér means are examples of Nörlund means with non-decreasing
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}$ we immediately obtain from
part b) of Theorem 3.25 that the asymptotic behaviour of the sequence of
weights

$\left\\{1/k^{2-2p}:k\in\mathbb{N}\right\\}$

in Nörlund means in Theorem 4.21 can not be improved.

Proof: The proof is similar to that of part a) of Theorem 3.25, but since this
case is more general we give the details.

By Lemma 1.38 the proof is complete if we show that

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|t_{m}a\right\|_{p}^{p}}{m^{2-2p}}\leq
c_{p}$ (4.40)

for every $p$-atom $a$ with support$\ I$, $\mu\left(I\right)=M_{N}^{-1}.$ We
may assume that $I=I_{N}.$ It is easy to see that
$S_{n}\left(a\right)=t_{n}\left(a\right)=0,$ when $n\leq M_{N}$. Therefore, we
can suppose that $n>M_{N}$.

Let $x\in I_{N}.$ Since Nörlund means $t_{n}$ with non-decreasing sequence
$\left\\{q_{k}:k\in\mathbb{N}\right\\}$ are bounded from $L_{\infty}$ to
$L_{\infty}$ (the boundedness follows from Corollary 1.23) and
$\left\|a\right\|_{\infty}\leq M_{N}^{1/p}$ we obtain that

$\int_{I_{N}}\left|t_{m}a\right|^{p}d\mu$
$\leq\frac{\left\|a\right\|_{\infty}^{p}}{M_{N}}\leq 1,\text{ \ }0<p\leq 1/2.$

Hence,

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\int_{I_{N}}\left|t_{m}a\right|^{p}d\mu}{m^{2-2p}}$
(4.41) $\leq\overset{\infty}{\underset{k=1}{\sum}}\frac{1}{m^{2-2p}}\leq
c<\infty.$

It is easy to see that

$\left|t_{n}a\left(x\right)\right|$
$=\left|\int_{I_{N}}a\left(t\right)\frac{1}{Q_{n}}\overset{n}{\underset{k=M_{N}}{\sum}}q_{n-k}D_{k}\left(x-t\right)d\mu\left(t\right)\right|$
$\leq\int_{I_{N}}\left|a\left(t\right)\right|\left|\frac{1}{Q_{n}}\overset{n}{\underset{k=M_{N}}{\sum}}q_{n-k}D_{k}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq\left\|a\right\|_{\infty}\int_{I_{N}}\left|\frac{1}{Q_{n}}\overset{n}{\underset{k=M_{N}}{\sum}}q_{n-k}D_{k}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq
M_{N}^{1/p}\int_{I_{N}}\left|\frac{1}{Q_{n}}\overset{n}{\underset{k=M_{N}}{\sum}}q_{n-k}D_{k}\left(x-t\right)\right|d\mu\left(t\right).$

Let $x\in I_{N}^{k,l},\,0\leq k<l\leq N.$ Then, in the view of Lemma 1.27, we
get that

$\left|t_{m}a\left(x\right)\right|\leq c_{p}M_{l}M_{k}M_{N}^{1/p-2},\text{ for
}0<p<1/2.$ (4.42)

According to (1.1) with (4.42) we find that

$\int_{\overline{I_{N}}}\left|t_{m}a\right|^{p}d\mu$ (4.43)
$=\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\sum\limits_{x_{j}=0,\text{
}j\in\\{l+1,\dots,N-1\\}}^{m_{j-1}}\int_{I_{N}^{k,l}}\left|t_{m}a\right|^{p}d\mu$
$+\overset{N-1}{\underset{k=0}{\sum}}\int_{I_{N}^{k,N}}\left|t_{m}a\right|^{p}d\mu$
$\leq
c_{p}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{m_{l+1}\dotsm
m_{N-1}}{M_{N}}M_{l}^{p}M_{k}^{p}M_{N}^{1-2p}$
$+c_{p}\overset{N-1}{\underset{k=0}{\sum}}\frac{1}{M_{N}}M_{k}^{p}M_{N}^{1-p}$
$\leq
c_{p}M_{N}^{1-2p}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{M_{l}^{p}M_{k}^{p}}{M_{l}}$
$+c_{p}\overset{N-1}{\underset{k=0}{\sum}}\frac{M_{k}^{p}}{M_{N}^{p}}\leq
cM_{N}^{1-2p}.$

Moreover, according to (4.43), we get that

$\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\right|^{p}d\mu}{m^{2-2p}}$
$\leq
c_{p}\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{M_{N}^{1-2p}}{m^{2-2p}}<c<\infty,\text{
\ }\left(0<p<1/2\right).$

Now, by combining this estimate with (4.41) we obtain (4.40) so the proof is
complete.

Also the next theorem is proved in Persson, Tephnadze and Wall [51].

###### Theorem 4.23

Let $f\in H_{1/2}$ and the sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}$ be
non-decreasing satisfying condition (1.28). Then there exists an absolute
constant $c,$ such that

$\frac{1}{\log
n}\overset{n}{\underset{k=1}{\sum}}\frac{\left\|t_{k}f\right\|_{1/2}^{1/2}}{k}\leq
c\left\|f\right\|_{H_{1/2}}^{1/2}.$

Proof: The proof is similar to that of part a) of Theorem 3.28, but since this
case is more general we present the details.

According to Lemma 1.38 it suffices to show that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|t_{m}a\right\|_{1/2}^{1/2}}{m}\leq
c$

for every $p$-atom $a$ with support$\ I$, $\mu\left(I\right)=M_{N}^{-1}.$ We
may assume that $I=I_{N}.$ It is easy to see that
$S_{n}\left(a\right)=t_{n}\left(a\right)=0$ when $n\leq M_{N}$. Therefore we
can suppose that $n>M_{N}$.

Let $x\in I_{N}.$ Since $t_{n}$ is bounded from $L_{\infty}$ to $L_{\infty}$
(the boundedness follows from Corollary 1.23) and
$\left\|a\right\|_{\infty}\leq M_{N}^{2}$ we obtain that

$\int_{I_{N}}\left|t_{m}a\right|^{1/2}d\mu\leq\frac{\left\|a\right\|_{\infty}^{1/2}}{M_{N}}\leq
c<\infty.$

Hence,

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\int_{I_{N}}\left|t_{m}a\right|^{1/2}d\mu}{m}$
(4.44) $\leq\frac{1}{\log n}\overset{n}{\underset{m=1}{\sum}}\frac{1}{m}\leq
c<\infty,\text{ \ }n\in\mathbb{N}\text{.}$

It is easy to see that

$\left|t_{m}a\left(x\right)\right|\leq\int_{I_{N}}\left|a\left(t\right)\right|\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq\left\|a\right\|_{\infty}\int_{I_{N}}\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right)$
$\leq
M_{N}^{2}\int_{I_{N}}\left|F_{m}\left(x-t\right)\right|d\mu\left(t\right).$

Let $x\in I_{N}^{k,l},\,0\leq k<l<N.$ Then, in the view of Lemma 1.25, we get
that

$\left|t_{m}a\left(x\right)\right|\leq\frac{cM_{l}M_{k}M_{N}}{m}.$ (4.45)

Let $x\in I_{N}^{k,N}.$ Then, according to Lemma 1.25, we find that

$\left|t_{m}a\left(x\right)\right|\leq cM_{k}M_{N}.$ (4.46)

By combining (4.45) and (4.46) with (1.1) we can conclude that

$\int_{\overline{I_{N}}}\left|t_{m}a\left(x\right)\right|^{1/2}d\mu\left(x\right)$
$\leq
c\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{m_{l+1}\dotsm
m_{N-1}}{M_{N}}\frac{M_{l}^{1/2}M_{k}^{1/2}M_{N}^{1/2}}{m^{1/2}}$
$+c\overset{N-1}{\underset{k=0}{\sum}}\frac{1}{M_{N}}M_{k}^{1/2}M_{N}^{1/2}$
$\leq
cM_{N}^{1/2}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{M_{l}^{1/2}M_{k}^{1/2}}{m^{1/2}M_{l}}+c\overset{N-1}{\underset{k=0}{\sum}}\frac{M_{k}^{1/2}}{M_{N}^{1/2}}$
$\leq\frac{cM_{N}^{1/2}N}{m^{1/2}}+c.$

It follows that

$\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\left(x\right)\right|^{1/2}d\mu\left(x\right)}{m}$
(4.47) $\leq\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\left(\frac{cM_{N}^{1/2}N}{m^{3/2}}+\frac{c}{m}\right)<c<\infty.$

The proof is complete by just combining (4.44) and (4.47).

Next, we investigate Nörlund means with non-increasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}.$ At first we consider the case $0$
$<p\,<1/\left(1+\alpha\right)$ where $0<\alpha<1.$ For details see the paper
of Blahota and Tephnadze [6].

###### Theorem 4.24

Let $f\in H_{p},$ where $0$ $<p\,<1/\left(1+\alpha\right),$ $0<\alpha\leq 1$
and $\\{q_{k}:k\in\mathbb{N}\\},$ be a sequence of non-increasing numbers
satisfying the conditions (1.39) and (1.40). Then there exists an absolute
constant $c_{\alpha,p},$ depending only on $\alpha$ and $p$ such that

$\overset{\infty}{\underset{k=1}{\sum}}\frac{\left\|t_{k}f\right\|_{H_{p}}^{p}}{k^{2-\left(1+\alpha\right)p}}\leq
c_{\alpha,p}\left\|f\right\|_{H_{p}}^{p}.$

Proof: By Lemma 1.38 the it suffices to show that

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|t_{m}a\right\|_{p}^{p}}{m^{2-\left(1+\alpha\right)p}}\leq
c_{\alpha,p}<\infty$

for every $p$-atom $a.$ Analogously to the proofs of the previous theorems we
may assume that $a$ be an arbitrary $p$-atom with support $I,\
\mu\left(I\right)=M_{N}^{-1}$ and $I=I_{N}$ and $m>M_{N}.$

Let $x\in I_{N}.$ Since $t_{m}$ is bounded from $L_{\infty}$ to $L_{\infty}$
(the boundedness follows from Corollary 1.33) and
$\left\|a\right\|_{\infty}\leq M_{N}^{1/p}$ we obtain that

$\int_{I_{N}}\left|t_{m}a\right|^{p}d\mu\leq\left\|a\right\|_{\infty}^{p}M_{N}^{-1}\leq
1.$

Hence

$\overset{\infty}{\underset{m=M_{N}}{\sum}}\frac{\int_{I_{N}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu}{m^{2-\left(1+\alpha\right)p}}$
$\leq\overset{\infty}{\underset{m=1}{\sum}}\frac{1}{m^{2-\left(1+\alpha\right)p}}\leq
c_{\alpha,p}<\infty.$

According to (1.1) and (4.30)-(4.31) we can conclude that

$\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\right|^{p}d\mu}{m^{2-\left(1+\alpha\right)p}}$
$=\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\sum\limits_{x_{j}=0,j\in\\{l+1,\dots,N-1\\}}^{m_{j}-1}\frac{\int_{I_{N}^{k,l}}\left|t_{m}a\right|^{p}d\mu}{m^{2-\left(1+\alpha\right)p}}$
$+\overset{n}{\underset{m=M_{N}+1}{\sum}}\overset{N-1}{\underset{k=0}{\sum}}\frac{\int_{I_{N}^{k,N}}\left|t_{m}a\right|^{p}d\mu}{m^{2-\left(1+\alpha\right)p}}$
$\leq
c_{\alpha,p}\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\left(\frac{M_{N}^{1-p}}{m^{2-p}}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{M_{l}^{p\alpha}M_{k}^{p}}{M_{l}}+\frac{M_{N}^{1-p}}{m^{2-\left(1+\alpha\right)p}}\overset{N-1}{\underset{k=0}{\sum}}\frac{M_{k}^{p}}{M_{N}}\right).$

Since

$\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{M_{l}^{p\alpha}M_{k}^{p}}{M_{l}}\leq\overset{N-2}{\underset{k=0}{\sum}}M_{k}^{p}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{1}{M_{l}^{1-p\alpha}}$
$\leq\overset{N-2}{\underset{k=0}{\sum}}\frac{1}{M_{k}^{1-p\alpha}}M_{k}^{p}\leq\overset{N-2}{\underset{k=0}{\sum}}\frac{1}{M_{k}^{1-p\left(\alpha+1\right)}}<c<\infty$

and

$\overset{N-1}{\underset{k=0}{\sum}}\frac{M_{k}^{p}}{M_{N}}\leq
M_{N}^{p-1}<\infty$

we obtain that

$\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\right|^{p}d\mu}{m^{2-\left(1+\alpha\right)p}}$
$<c_{\alpha,p}M_{N}^{1-p}\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{1}{m^{2-p}}+c_{\alpha,p}\overset{\infty}{\underset{m=M_{N}+1}{\sum}}\frac{1}{m^{2-\left(1+\alpha\right)p}}\leq
c_{\alpha,p}<\infty.$

The proof is complete.

The final result in this section is due to Blahota, Persson and Tephnadze [7].

###### Theorem 4.25

Let $f\in H_{1/(1+\alpha)}$ where $0<\alpha\leq 1$ and
$\\{q_{k}:k\in\mathbb{N}\\}$ be a sequence of non-increasing numbers
satisfying the conditions (1.39) and (1.40). Then there exists an absolute
constant $c_{\alpha}$ depending only on $\alpha$ such that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|t_{m}f\right\|_{H_{1/\left(1+\alpha\right)}}^{1/\left(1+\alpha\right)}}{m}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}^{1/\left(1+\alpha\right)}.$

Proof:  By Lemma 1.38 the proof is complete if we show that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|t_{m}a\right\|_{1/\left(1+\alpha\right)}^{1/\left(1+\alpha\right)}}{m}\leq
c_{\alpha}<\infty$

for every $1/\left(1+\alpha\right)$-atom $a.$ Analogously to the proofs of
previous resalts we may assume that $a$ is an arbitrary
$1/\left(1+\alpha\right)$-atom with support$\ I,$
$\mu\left(I\right)=M_{N}^{-1}$ and $I=I_{N}$ and $m>M_{N}.$

Let $x\in I_{N}.$ Since $t_{m}$ is bounded from $L_{\infty}$ to $L_{\infty}$
(the boundedness follows from Corollary 1.33) and
$\left\|a\right\|_{\infty}\leq M_{N}^{1+\alpha}$ we obtain that

$\int_{I_{N}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu$
$\leq\left\|a\right\|_{\infty}^{1/\left(1+\alpha\right)}/M_{N}\leq 1.$

Hence

$\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}}{\sum}}\frac{\int_{I_{N}}\left|t_{m}a\left(x\right)\right|^{1/\left(1+\alpha\right)}d\mu}{m}$
$\leq\frac{1}{\log n}\overset{n}{\underset{m=1}{\sum}}\frac{1}{m}\leq
c_{\alpha}<\infty.$

According to (1.1) together with (4.36)-(4.37) we can conclude that

$\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu}{m}$
$=\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}\sum\limits_{x_{j}=0,\text{
}j\in\\{l+1,\dots,N-1\\}}^{m_{j-1}}\frac{\int_{I_{N}^{k,l}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu}{m}$
$+\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\overset{N-1}{\underset{k=0}{\sum}}\frac{\int_{I_{N}^{k,N}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu}{m}$
$\leq\frac{c_{\alpha}}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{M_{N}^{\alpha/\left(1+\alpha\right)}}{m^{\alpha/\left(1+\alpha\right)+1}}\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}M_{l}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}\frac{m_{l+1}\dotsm
m_{N-1}}{M_{N}}$ $+\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{1}{mM_{N}}\overset{N-1}{\underset{k=0}{\sum}}M_{k}^{1/\left(1+\alpha\right)}M_{N}^{\alpha/\left(1+\alpha\right)}.$

Since

$\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}M_{l}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}\frac{m_{l+1}\dotsm
m_{N-1}}{M_{N}}$
$=\overset{N-2}{\underset{k=0}{\sum}}\overset{N-1}{\underset{l=k+1}{\sum}}M_{l}^{\alpha/\left(1+\alpha\right)}M_{k}^{1/\left(1+\alpha\right)}\frac{1}{M_{l}}$
$=\overset{N-2}{\underset{k=0}{\sum}}M_{k}^{1/\left(1+\alpha\right)}\overset{N-1}{\underset{l=k+1}{\sum}}\frac{1}{M_{l}^{1/\left(1+\alpha\right)}}$
$\leq\overset{N-2}{\underset{k=0}{\sum}}M_{k}^{1/\left(1+\alpha\right)}\frac{1}{M_{k}^{1/\left(1+\alpha\right)}}\leq\overset{N-2}{\underset{k=0}{\sum}}1\leq
N$

and

$\overset{N-1}{\underset{k=0}{\sum}}M_{k}^{1/\left(1+\alpha\right)}M_{N}^{\alpha/\left(1+\alpha\right)}$
$\leq M_{N}^{1/\left(1+\alpha\right)}M_{N}^{\alpha/\left(1+\alpha\right)}\leq
M_{N}$

we obtain that

$\frac{1}{\log
n}\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{\int_{\overline{I_{N}}}\left|t_{m}a\right|^{1/\left(1+\alpha\right)}d\mu}{m}$
$\leq\frac{c_{\alpha}}{\log
n}\left(\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{NM_{N}^{\alpha/\left(1+\alpha\right)}}{m^{\alpha/\left(1+\alpha\right)+1}}+\overset{n}{\underset{m=M_{N}+1}{\sum}}\frac{1}{m}\right)<c_{\alpha}<\infty.$

The proof is complete.

###  4.4 Maximal operators of Riesz and Nörlund logarithmic means on
martingale Hardy spaces

In our previous sections we investigated Nörlund means with non-increasing
sequences $\left\\{q_{k}:k\in\mathbb{N}\right\\}$, but the case when
$q_{k}=1/k$ was excluded, since this sequence does not satisfies the condition
(1.39) for any $0<\alpha\leq 1$. On the other hand, Riesz logarithmic means
are not examples of Nörlund means. In this subsection we fill up this gap
simultaneously for both cases.

Both theorems in this section are due to Tephnadze [75].

###### Theorem 4.26

a)

The maximal operator of Riesz logarithmic means $R^{\ast}$ is bounded from the
Hardy space $H_{1/2}$ to the space $weak-L_{1/2}.$

b) Let $0<p\leq 1/2$. Then there exists a martingale $f\in H_{p}$ such that

$\left\|R^{\ast}f\right\|_{p}=+\infty.$

Proof: By using Abel transformation we obtain that

$R_{n}f=\frac{1}{l_{n}}\overset{n-1}{\underset{j=1}{\sum}}\frac{\sigma_{j}f}{j+1}+\frac{\sigma_{n}f}{l_{n}}.$

Consequently,

$R^{\ast}f\leq c\sigma^{\ast}f.$ (4.48)

Since $\sigma^{\ast}$ is bounded from the martingale Hardy space $H_{1/2}$ to
the space $weak-L_{1/2}$ by using (4.48) we can conclude that

$\left\|R^{\ast}f\right\|_{weak-L_{1/2}}\leq c\left\|f\right\|_{H_{1/2}}$

and the proof of part a) is complete.

Let $f=\left(f^{\left(n\right)}:n\in\mathbb{N}\right)$ be martingale defined
in Example 1.49.

Set $q_{n}^{s}=M_{2n}+M_{2s}-1,$ $n>s.$ Then we can write that

$R_{q_{\alpha_{k}}^{s}}f=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=1}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\frac{S_{j}f}{j}$
$=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=1}{\overset{M_{2\alpha_{k}}}{\sum}}\frac{S_{j}f}{j}+\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=M_{2\alpha_{k}}+1}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\frac{S_{j}f}{j}:=I+II.$

According to (1.76) we have that

$\left|I\right|\leq\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=1}{\overset{M_{2\alpha_{k}}-1}{\sum}}\frac{\left|S_{j}f\left(x\right)\right|}{j}$
$\leq\frac{1}{\alpha_{k}}\frac{2\lambda
M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}\sum_{j=1}^{M_{2\alpha_{k}}-1}\frac{1}{j}$
$\leq\frac{2\lambda
M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}\leq\frac{2\lambda
M_{\alpha_{k}}^{1/p}}{\alpha_{k}^{3/2}}.$

Let $M_{2\alpha_{k}}\leq j\leq q_{\alpha_{k}}^{s}.$ According to the second
inequality of (1.75) in the case when $l=k,$ we deduce that

$S_{j}f=S_{M_{{}_{2\alpha_{k}}}}f+\frac{M_{2\alpha_{k}}^{1/p-1}\psi_{{}_{M_{2\alpha_{k}}}}D_{j-M_{{}_{2\alpha_{k}}}}}{\alpha_{k}^{1/2}}.$
(4.49)

Hence, we can rewrite $II$ as

$II=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=M_{2\alpha_{k}}}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\
\frac{S_{M_{{}_{2\alpha_{k}}}}f}{j}$ (4.50)
$+\frac{1}{l_{q_{\alpha_{k}}^{s}}}\frac{M_{2\alpha_{k}}^{1/p-1}\psi_{{}_{M_{2\alpha_{k}}}}}{\alpha_{k}^{1/2}}\sum_{j=M_{2\alpha_{k}}}^{q_{\alpha_{k}}^{s}}\frac{D_{j-M_{{}_{2\alpha_{k}}}}}{j}$
$:=II_{1}+II_{2}.$

In view of (1.76) we find that

$\left|II_{1}\right|\leq\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=M_{2\alpha_{k}}}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\
\frac{1}{j}\left|S_{M_{{}_{2\alpha_{k}}}}f\right|$ (4.51)
$\leq\left|S_{M_{{}_{2\alpha_{k}}}}f\right|\leq\frac{2\lambda
M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}.$

Let $0<p\leq 1/2,$ $x\in I_{2s}\backslash I_{2s+1}$ for
$s=\left[2\alpha_{k}/3\right],\ldots,\alpha_{k}.$ Since

$\sum_{j=0}^{M_{2s}-1}\frac{D_{{}_{j}}\left(x\right)}{{}_{j+M_{2\alpha_{k}}}}\geq\sum_{j=0}^{M_{2s}-1}\frac{j}{{}_{j+M_{2\alpha_{k}}}}$
$\geq\sum_{j=0}^{M_{2s}-1}\frac{j}{{}_{2M_{2\alpha_{k}}}}\geq\frac{cM_{2s}^{2}}{M_{2\alpha_{k}}}$

we obtain that

$\left|II_{2}\right|=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\frac{M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{1/2}}\left|\psi_{{}_{M_{2\alpha_{k}}}}\sum_{j=0}^{M_{2s}-1}\frac{D_{j}}{{}_{j+M_{2\alpha_{k}}}}\right|$
(4.52)
$\geq\frac{c}{\alpha_{k}}\frac{M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{1/2}}\frac{M_{2s}^{2}}{M_{2\alpha_{k}}}\geq\frac{cM_{2\alpha_{k}}^{1/p-2}M_{2s}^{2}}{\alpha_{k}^{3/2}}.$

By combining (1.71)-(1.73) with (4.50)-(4.52) we get that

$\left|R_{q_{\alpha_{k}}^{s}}f\right|=\left|II_{2}-I-II_{1}\right|$
$\geq\left|II_{2}\right|-\left|I\right|-\left|II_{1}\right|\geq\left|II_{2}\right|-\frac{2\lambda
M_{\alpha_{k}}^{1/p}}{\alpha_{k}^{3/2}}$
$\geq\frac{cM_{2\alpha_{k}}^{1/p-2}M_{2s}^{2}}{\alpha_{k}^{3/2}}-\frac{cM_{\alpha_{k}}^{1/p}}{\alpha_{k}^{3/2}}$
$\geq\frac{cM_{2\alpha_{k}}^{1/p-2}M_{2s}^{2}}{\alpha_{k}^{3/2}}.$

Hence,

$\int_{G_{m}}\left|R^{\ast}f\left(x\right)\right|^{p}d\mu\left(x\right)$ $\geq
c_{p}\sum_{s=\left[2\alpha_{k}/3\right]}^{\alpha_{k}}\underset{I_{2s}\backslash
I_{2s+1}}{\int}\left|R_{q_{\alpha_{k}}^{s}}f\left(x\right)\right|^{p}d\mu\left(x\right)$
$\geq
c_{p}\overset{\alpha_{k}}{\underset{s=\left[2\alpha_{k}/3\right]}{\sum}}\underset{I_{2s}\backslash
I_{2s+1}}{\int}\frac{cM_{2\alpha_{k}}^{1-2p}M_{2s}^{2p}}{\alpha_{k}^{3p/2}}d\mu\left(x\right)$
$\geq
c_{p}\sum_{s=\left[2\alpha_{k}/3\right]}^{\alpha_{k}}\frac{M_{2\alpha_{k}}^{1-2p}M_{2s}^{2p-1}}{\alpha_{k}^{3p/2}}$
$\geq\left\\{\begin{array}[]{ll}\frac{c_{p}2^{\alpha_{k}(1-2p)/3}}{\alpha_{k}^{3p/2}},&\text{\thinspace
when \ }0<p<1/2,\\\ c\alpha_{k}^{1/4},&\text{when \
}p=1/2,\end{array}\right.\rightarrow\infty\text{, when \ }k\rightarrow\infty.$

The proof is complete.

###### Theorem 4.27

Let $0<p\leq 1$. Then there exists a martingale $f\in H_{p}$ such that

$\left\|L^{\ast}f\right\|_{p}=+\infty.$

Proof: We write that

$L_{q_{\alpha_{k}}^{s}}f=\frac{1}{l_{q_{\alpha_{k},s}}}\underset{j=1}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\frac{S_{j}f}{q_{\alpha_{k}}^{s}-j}$
(4.53)
$=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\underset{j=1}{\overset{M_{2\alpha_{k}}-1}{\sum}}\frac{S_{j}f}{q_{\alpha_{k}}^{s}-j}+\frac{1}{q_{\alpha_{k}}^{s}}\underset{j=M_{2\alpha_{k}}}{\overset{q_{\alpha_{k}}^{s}}{\sum}}\frac{S_{j}f}{q_{\alpha_{k}}^{s}-j}:=III+IV.$

In the view of (1.76) for $III$ we get the following estimate:

$\left|III\right|\leq\frac{1}{\alpha_{k}}\sum_{j=0}^{M_{2\alpha_{k-1}}}\frac{1}{q_{\alpha_{k}}^{s}-j}\frac{M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}\text{
}\leq\frac{M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}\leq\frac{M_{\alpha_{k}}^{1/p}}{\alpha_{k}^{3/2}}.$
(4.54)

Moreover, according to the second inequality of (1.75) in the case when $l=k,$
(see also (4.49)) we can rewrite $IV$ as

$IV=\frac{1}{l_{q_{\alpha_{k}}^{s}}}\sum_{j=M_{2\alpha_{k}}}^{q_{\alpha_{k}}^{s}}\frac{S_{M_{{}_{2\alpha_{k}}}}f}{q_{\alpha_{k},s}-j}$
(4.55)
$+\frac{1}{l_{q_{\alpha_{k}}^{s}}}\frac{M_{2\alpha_{k}}^{1/p-1}\psi_{{}_{M_{2\alpha_{k}}}}}{\alpha_{k}^{1/2}}\sum_{j=M_{2\alpha_{k}}}^{q_{\alpha_{k}}^{s}}\frac{D_{j-M_{{}_{2\alpha_{k}}}}}{q_{\alpha_{k}}^{s}-j}:=IV_{1}+IV_{2}.$

By now applying (1.76) again we have that

$\left|IV_{1}\right|\leq\frac{2\lambda
M_{2\alpha_{k-1}}^{1/p}}{\alpha_{k-1}^{1/2}}\leq\frac{2\lambda
M_{\alpha_{k}}^{1/p}}{\alpha_{k}^{3/2}}.$ (4.56)

Let $x\in I_{2s}\backslash I_{2s+1},$
$s=\left[2\alpha_{k}/3\right],...,\alpha_{k}.$ Since

$\sum_{j=0}^{M_{2s}-1}\frac{D_{j}}{{}_{M_{2s}}-j}=\sum_{j=0}^{M_{2s}-1}\frac{j}{{}_{M_{2s}}-j}=\sum_{j=0}^{M_{2s}-1}\frac{M_{2s}}{{}_{M_{2s}}-j}-\sum_{j=0}^{M_{2s}-1}\frac{M_{2s}-j}{{}_{M_{2s}}-j}$
$=c_{p}sM_{2s}-M_{2s}\geq c_{p}sM_{2s},$

we obtain that

$\left|IV_{2}\right|=\frac{1}{l_{q_{\alpha_{k},s}}}\frac{M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{1/2}}\left|\sum_{j=0}^{M_{2s}-1}\frac{D_{j}}{{}_{M_{2s}}-j}\right|$
(4.57)
$\geq\frac{c_{p}M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{3/2}}sM_{2s},\qquad x\in
I_{2s}/I_{2s+1}.$

By combining (1.73) with (4.53)-(4.57) for $x\in I_{2s}\backslash
I_{2s+1},s=\left[2\alpha_{k}/3\right],...,\alpha_{k}$ and $0<p\leq 1$ we get
that

$\left|L_{q_{\alpha_{k}}^{s}}f\left(x\right)\right|=\left|III+IV_{1}+IV_{2}\right|\geq\left|IV_{2}\right|-\left|III\right|-\left|IV_{1}\right|$
$\geq\frac{c_{p}M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{3/2}}sM_{2s}-\frac{2\lambda
M_{\alpha_{k}}}{\alpha_{k}^{3/2}}\geq\frac{c_{p}M_{2\alpha_{k}}^{1/p-1}}{\alpha_{k}^{3/2}}sM_{2s}.$

Consequently,

$\int_{G_{m}}\left|L^{\ast}f\left(x\right)\right|^{p}d\mu\left(x\right)\geq\sum_{s=\left[2\alpha_{k}/3\right]}^{\alpha_{k}}\int_{I_{2s}\backslash
I_{2s+1}}\left|L^{\ast}f\left(x\right)\right|^{p}d\mu\left(x\right)$
$\geq\sum_{s=\left[2\alpha_{k}/3\right]}^{\alpha_{k}}\int_{I_{2s}\backslash
I_{2s+1}}\left|L_{q_{\alpha_{k}}^{s}}f\left(x\right)\right|^{p}d\mu\left(x\right)$
$\geq
c_{p}\sum_{s=\left[2\alpha_{k}/3\right]}^{\alpha_{k}}\int_{I_{2s}\backslash
I_{2s+1}}\frac{M_{2\alpha_{k-1}}^{1-p}}{\alpha_{k}^{3p/2}}s^{p}M_{2s}^{p}d\mu\geq
c_{p}\underset{s=\left[2\alpha_{k}/3\right]}{\overset{\alpha_{k}}{\sum}}\frac{M_{2\alpha_{k-1}}^{1-p}}{\alpha_{k}^{p/2}}M_{2s}^{p-1}$
$\geq\left\\{\begin{array}[]{ll}\frac{c_{p}2^{\alpha_{k}(1-p)/3}}{\alpha_{k}^{p/2}},&\text{when
}0<p<1,\\\ c\alpha_{k}^{1/2},&\text{when \
}p=1,\end{array}\right.\rightarrow\infty\text{, when \ }k\rightarrow\infty.$

The proof is complete.

###  4.5 Applications

First we consider Nörlund means $t_{n}$ with monotone and bounded sequence
$\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ The results in our previous sections
in particular imply the following results:

###### Theorem 4.28

a) Let $f\in H_{1/2}$ and $t_{n}$ be Nörlund means with monotone and bounded
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ Then there exists an
absolute constant $c$ such that

$\left\|t^{\ast}f\right\|_{weak-L_{1/2}}\leq c\left\|f\right\|_{H_{1/2}}.$

b) There exists a martingale $f\in H_{1/2},$ such that

$\left\|t^{\ast}f\right\|_{1/2}=\infty.$

###### Theorem 4.29

a) Let $p>1/2$, $f\in H_{p}$ and $t_{n}$ be Nörlund means with monotone and
bounded sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ Then there exists an
absolute constant $c_{p}$ depending only on $p$ such that

$\left\|\overset{\sim}{t_{p,1}}^{\ast}f\right\|_{p}\leq
c_{p}\left\|f\right\|_{H_{p}}.$

b) Let $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a non-decreasing
function satisfying the condition (3.6). Then the following maximal operator

$\sup_{n\in\mathbb{N}}\frac{\left|t_{n}f\right|}{\Phi_{n}}$

is not bounded from the Hardy space $H_{p}$ to the space $weak-L_{p}.$

###### Theorem 4.30

a) Let $f\in H_{1/2}$ and $t_{n}$ be Nörlund means with monotone and bounded
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ Then there exists an
absolute constant $c$ such that

$\left\|\overset{\sim}{t_{1}}^{\ast}f\right\|_{1/2}\leq
c\left\|f\right\|_{H_{1/2}}.$

b) Let $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a non-decreasing
function satisfying the condition (3.11). Then the following maximal operator

$\sup_{n\in\mathbb{N}}\frac{\left|B_{n}f\right|}{\Phi_{n}}$

is not bounded from the Hardy space $H_{1/2}$ to the Lebesgue space $L_{1/2}.$

###### Theorem 4.31

Let $0<p<1/2$, $f\in H_{p}$ and $t_{n}$ be Nörlund means with monotone and
bounded sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ Then there exists an
absolute constant $c_{p}$ depending only on $p$ such that

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|t_{m}f\right\|_{p}^{p}}{m^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}}^{p}.$

###### Theorem 4.32

Let $f\in H_{1/2}$ and $t_{n}$ be Nörlund means with monotone and bounded
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}.$ Then there exists an
absolute constant $c$ such that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|t_{m}f\right\|_{1/2}^{1/2}}{m}\leq
c\left\|f\right\|_{H_{1/2}}^{1/2},\text{ \ }n=2,3,\ldots.$

Next we remark that $\sigma_{n}$ are Nörlund means with monotone and bounded
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}$ and
$\kappa_{n}^{\alpha,\beta}$ are concrete examples of Nörlund means with non-
decreasing and unbounded sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}$ but
they have similar boundedness properties. The results in our previous sections
in particular imply the following results:

###### Theorem 4.33

a) Let $f\in H_{1/2}.$ Then there exists absolute constant $c$ such that

$\left\|\sigma^{\ast}f\right\|_{weak-L_{1/2}}\leq c\left\|f\right\|_{H_{1/2}}$

and

$\left\|\kappa^{\alpha,\beta,\ast}f\right\|_{weak-L_{1/2}}\leq
c\left\|f\right\|_{H_{1/2}}.$

b) There exists a martingale $f\in H_{1/2}$ such that

$\left\|\sigma^{\ast}f\right\|_{1/2}=\infty$

and

$\left\|\kappa^{\alpha,\beta,\ast}f\right\|_{1/2}=\infty.$

###### Theorem 4.34

a) Let $0<p<1/2$ and $f\in H_{p}.$ Then there exists an absolute constant
$c_{p}$ depending only on $p$ such that

$\left\|\overset{\sim}{\sigma_{p}}^{\ast}f\right\|_{p}\leq
c_{p}\left\|f\right\|_{H_{p}}$

and

$\left\|\widetilde{\kappa}_{p}^{\alpha,\beta,\ast}f\right\|_{p}\leq
c_{p}\left\|f\right\|_{H_{p}}.$

b) Let $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a non-decreasing
function satisfying the condition (3.6). Then the following maximal operators

$\text{\ }\sup_{n\in\mathbb{N}}\frac{\left|\sigma_{n}f\right|}{\Phi_{n}}\text{
\ \ \ and \ \ \
}\sup_{n\in\mathbb{N}}\frac{\left|\kappa_{n}^{\alpha,\beta}\right|}{\Phi_{n}}$

are not bounded from the Hardy space $H_{p}$ to the space $weak-L_{p}.$

###### Theorem 4.35

a) Let $f\in H_{1/2}.$ Then there exists an absolute constant $c$ such that

$\left\|\overset{\sim}{\sigma}^{\ast}f\right\|_{1/2}\leq
c\left\|f\right\|_{H_{1/2}}$

and

$\left\|\widetilde{\kappa}^{\alpha,\beta,\ast}f\right\|_{1/2}\leq
c\left\|f\right\|_{H_{1/2}}.$

b) Let $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a nondecreasing
function satisfying the condition (3.11). Then the following maximal operators

$\text{\ }\sup_{n\in\mathbb{N}}\frac{\left|\sigma_{n}f\right|}{\Phi_{n}}\text{
\ \ \ and \ \ \
}\sup_{n\in\mathbb{N}}\frac{\left|\kappa_{n}^{\alpha,\beta}\right|}{\Phi_{n}}$

are not bounded from the Hardy space $H_{1/2}$ to the Lebesgue space
$L_{1/2}.$

###### Theorem 4.36

Let $0<p<1/2$ and $f\in H_{p}.$ Then there exists an absolute constant $c_{p}$
depending only on $p$ such that

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{p}^{p}}{m^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}}^{p}$

and

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|\kappa_{m}^{\alpha,\beta}f\right\|_{p}^{p}}{m^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}}^{p}.$

###### Theorem 4.37

Let $f\in H_{1/2}.$ Then there exists an absolute constant $c$ such that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{1/2}^{1/2}}{m}\leq
c\left\|f\right\|_{H_{1/2}}^{1/2},\text{ \ }n=2,3,\ldots,$

and

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\kappa_{m}^{\alpha,\beta}f\right\|_{1/2}^{1/2}}{m}\leq
c\left\|f\right\|_{H_{1/2}}^{1/2},\text{ }n=2,3,\ldots.$

Furthemore, we note that it is obvious that Cesàro $\left(C,\alpha\right)$ and
Riesz $\left(R,\alpha\right)$ means are examples of Nörlund means with non-
increasing sequence satisfying the conditions (1.39) and (1.40). It follows
that all results concerning such summability methods are true also for the
Cesàro $\left(C,\alpha\right)$ and Riesz $\left(R,\alpha\right)$ means. Hence,
the results in our previous sections in particular imply the following
results:

###### Theorem 4.38

a) Let $0<p<1/2.$ Then there exists a martingale $f\in H_{p}$ such that

$\sup_{n\in\mathbb{N}}\left\|\sigma_{n}^{\alpha}f\right\|_{weak-L_{p}}=\infty$

and

$\sup_{n\in\mathbb{N}}\left\|\beta_{n}^{\alpha}f\right\|_{weak-L_{p}}=\infty.$

###### Theorem 4.39

a) Let $0<\alpha\leq 1$ and $f\in H_{1/(1+\alpha)}.$ Then there exists
absolute constant $c_{\alpha}$ depending only on $\alpha$ such that

$\left\|\sigma^{\alpha,\ast}f\right\|_{weak-L_{1/\left(1+\alpha\right)}}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}$

and

$\left\|\beta^{\alpha,\ast}f\right\|_{weak-L_{1/\left(1+\alpha\right)}}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}.$

b) Let $0<\alpha\leq 1.$ Then there exists a martingale $f\in
H_{1/(1+\alpha)}$ such that

$\left\|\sigma^{\alpha,\ast}f\right\|_{1/\left(1+\alpha\right)}=\infty$

and

$\left\|\beta^{\alpha,\ast}f\right\|_{1/\left(1+\alpha\right)}=\infty.$

###### Theorem 4.40

a) Let $0<\alpha\leq 1$, $0<p<1/\left(1+\alpha\right)$ and $f\in H_{p}.$ Then
there exists an absolute constant $c_{\alpha,p}$ depending only on $\alpha$
and $p$ such that

$\left\|\overset{\sim}{\sigma_{p}}^{\alpha,\ast}f\right\|_{p}\leq
c_{\alpha,p}\left\|f\right\|_{H_{p}}$

and

$\left\|\overset{\sim}{\beta_{p}}^{\alpha,\ast}f\right\|_{p}\leq
c_{\alpha,p}\left\|f\right\|_{H_{p}}.$

b) Let $0<\alpha\leq 1$ and $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a
non-decreasing function satisfying the condition (4.29). Then the following
maximal operators

$\sup_{n\in\mathbb{N}}\frac{\left|\sigma_{n}^{\alpha}f\right|}{\Phi_{n}}\text{
\ \ \ and \ \ \
}\sup_{n\in\mathbb{N}}\frac{\left|\beta_{n}^{\alpha}f\right|}{\Phi_{n}}$

are not bounded from the Hardy space $H_{p}$ to the space $weak-L_{p}.$

###### Theorem 4.41

a) Let $0<\alpha\leq 1$ and $f\in H_{1/(1+\alpha)}.$ Then there exists an
absolute constant $c_{\alpha}$ depending only on $\alpha$ such that

$\left\|\overset{\sim}{\sigma}^{\alpha,\ast}f\right\|_{1/\left(1+\alpha\right)}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}$

and

$\left\|\overset{\sim}{\beta}^{\alpha,\ast}f\right\|_{1/\left(1+\alpha\right)}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}.$

b) Let $0<\alpha\leq 1$ and $\varphi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be a
nondecreasing function satisfying the condition (4.35). Then the following
maximal operators

$\sup_{n\in\mathbb{N}}\frac{\left|\sigma_{n}^{\alpha}f\right|}{\Phi_{n}}\text{
\ \ and \ \ \ \ \
}\sup_{n\in\mathbb{N}}\frac{\left|\beta_{n}^{\alpha}f\right|}{\Phi_{n}}$

are not bounded from the Hardy space $H_{1/\left(1+\alpha\right)}$ to the
Lebesgue space $L_{1/\left(1+\alpha\right)}.$

###### Theorem 4.42

Let $0<\alpha\leq 1$, $0<p<1/\left(1+\alpha\right)$ and $f\in H_{p}.$ Then
there exists an absolute constant $c_{\alpha,p}$ depending only on $\alpha$
and $p$ such that

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}^{\alpha}a\right\|_{p}^{p}}{m^{\left(1+\alpha\right)\left(1-p\right)}}\leq
c_{\alpha,p}\left\|f\right\|_{H_{p}}^{p}$

and

$\overset{\infty}{\underset{m=1}{\sum}}\frac{\left\|\beta_{m}^{\alpha}a\right\|_{p}^{p}}{m^{\left(1+\alpha\right)\left(1-p\right)}}\leq
c_{\alpha,p}\left\|f\right\|_{H_{p}}^{p}.$

###### Theorem 4.43

Let $0<\alpha\leq 1$ and $f\in H_{1/(1+\alpha)}.$ Then there exists an
absolute constant $c_{\alpha}$ depending only on $\alpha$ such that

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}^{\alpha}f\right\|_{1/\left(1+\alpha\right)}^{1/\left(1+\alpha\right)}}{m}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}^{1/\left(1+\alpha\right)}$

and

$\frac{1}{\log
n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\beta_{m}^{\alpha}f\right\|_{1/\left(1+\alpha\right)}^{1/\left(1+\alpha\right)}}{m}\leq
c_{\alpha}\left\|f\right\|_{H_{1/\left(1+\alpha\right)}}^{1/\left(1+\alpha\right)}.$

Finally, we present some applications concerning almost everywhere convergence
of some summability methods. To study almost everywhere convergence of some
summability methods is one of the most difficult topics in Fourier analysis.
They involve techniques from function theory and Hardy spaces.

In most applications the a.e. convergence of
$\left\\{T_{n}:n\in\mathbb{N}\right\\}$ can be established for $f$ in some
dense class of $L_{1}\left(G_{m}\right).$ In particular, the following result
play an important role for studying this type of questions (see e.g. the books
[35], [55] and [88]).

###### Lemma 4.44

Let $f\in L_{1}$ and $T_{n}:L_{1}\rightarrow L_{1}$ be some sub-linear
operators and

$T^{\ast}:=\sup_{n\in\mathbb{N}}\left|T_{n}\right|.$

If

$T_{n}f\rightarrow f\text{ a.e. \ \ for every }f\in S$

where the set $S$ is dense in the space $\ L_{1}$ and the maximal operator
$T^{\ast}$ is bounded from the space $L_{1}$ to the space $weak-L_{1},$ that
is

$\sup_{\lambda>0}\lambda\mu\left\\{x\in G_{m}:\text{
}\left|T^{\ast}f\left(x\right)\right|>\lambda\right\\}\leq\left\|f\right\|_{1},$

then

$T_{n}f\rightarrow f,\text{ a.e. \ \ for every }f\in L_{1}\left(G_{m}\right).$

###### Remark 4.45

Since the Vilenkin function $\psi_{m}$ is constant on $I_{n}(x)$ for every
$x\in G_{m}$ and $0\leq m<M_{n},$ it is clear that each Vilenkin function is a
complex-valued step function, that is, it is a finite linear combination of
the characteristic functions

$\chi\left(E\right)=\left\\{\begin{array}[]{ll}1,&x\in E,\\\ 0,&x\notin
E.\end{array}\right.$

On the other hand, notice that, by Corollary 1.5 (Paley lemma), it yields that

$\chi\left(I_{n}(t)\right)\left(x\right)=\frac{1}{M_{n}}\sum_{j=0}^{M_{n}-1}\psi_{j}\left(x-t\right),\text{
\ \ }x\in I_{n}(t),$

for each $x,t\in$ $G_{m}$ and $n\in\mathbb{N}$. Thus each step function is a
Vilenkin polynomial. Consequently, we obtain that the collection of step
functions coincides with a collection of Vilenkin polynomials $\mathscr{P}$.
Since the Lebesgue measure is regular it follows from Lusin theorem that given
$f\in L_{1}$ there exist Vilenkin polynomials $P_{1},P_{2}...,$ such that
$P_{n}\rightarrow f$ a.e. when $n\rightarrow\infty.$ This means that the
Vilenkin polynomials are dense in the space $L_{1}.$

By using Lemma 4.44, remarks 4.45 and the results in our previous sections we
in particular obtain the following a.e. convergence results.

###### Theorem 4.46

Let $f\in L_{1}$ and $t_{n}$ be the regular Nörlund means with non-decreasing
sequence $\\{q_{k}:k\in\mathbb{N}\\}$. Then

$t_{n}f\rightarrow f\text{ \ \ a.e., \ \ when \ }n\rightarrow\infty.\text{\ }$

Proof: Since

$S_{n}P=P,\text{ for every }P\in\mathscr{P}$

we obtain that

$t_{n}P\rightarrow P\ \text{a.e.,}\ \ \ \text{when \ \ \ }\
n\rightarrow\infty,$

for every regular Nörlund mean with non-decreasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}$, where $P\in\mathscr{P}$ is dense in the space
$L_{1}$ (see Remark 4.45).

On the other hand, by combining Lemma 1.40 and Theorem 4.1 we obtain that the
maximal operator of Nörlund means, with non-decreasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}$ satisfying conditions (1.38) is bounded from the
space $L_{1}$ to the space $weak-L_{1},$ that is,

$\sup_{\lambda>0}\lambda\mu\left\\{x\in G_{m}:\text{
}\left|t^{\ast}f\left(x\right)\right|>\lambda\right\\}\leq\left\|f\right\|_{1}.$

According to Lemma 4.44 we obtain that under condition (1.38) we have a.e.
convergence

$t_{n}f\rightarrow f\text{ \ \ a.e., \ \ when \ }n\rightarrow\infty.$

The proof is complete.

###### Corollary 4.47

Let $f\in L_{1}$. Then

$\sigma_{n}f\rightarrow f\text{ \ \ \ a.e., \ \ \ when \
}n\rightarrow\infty,\text{\ \ \ }$

and

$\kappa_{n}^{\alpha,\beta}f\rightarrow f\text{ \ \ \ a.e., \ \ when \
}n\rightarrow\infty.$

Proof: Since $\sigma_{n}$ and $\kappa_{n}$ are regular Nörlund means with non-
decreasing sequence $\\{q_{k}:k\in\mathbb{N}\\}$ (see (1.5) and (1.10)) the
proof is complete by just using Theorem 4.46.

###### Theorem 4.48

Let $f\in L_{1}$ and $t_{n}$ be Nörlund means with non-increasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}$ satisfying conditions (1.39) and (1.40). Then

$t_{n}f\rightarrow f\text{ \ \ a.e., \ \ when\ }n\rightarrow\infty.$

Proof: By using part b) of Theorem 1.1 we get that every Nörlund mean, with
non-increasing sequence $\\{q_{k}:k\in\mathbb{N}\\}$ is regular. Since

$S_{n}P=P,\text{ for every }P\in\mathscr{P}$

we obtain that $t_{n}P\rightarrow P,$ when $n\rightarrow\infty,$ where
$P\in\mathscr{P}$ is dense in the space $L_{1}$ (see Remark 4.45).

On the other hand, by combining Lemma 1.40 and part a) of Theorem 4.11 we
obtain that the maximal operator of Nörlund means with non-increasing sequence
$\\{q_{k}:k\in\mathbb{N}\\}$ satisfying conditions (1.39) and (1.40) is
bounded from the space $L_{1}$ to the space $weak-L_{1},$ that is

$\sup_{\lambda>0}\lambda\mu\left\\{x\in G_{m}:\text{
}\left|t^{\ast}f\left(x\right)\right|>\lambda\right\\}\leq\left\|f\right\|_{1}.$

According to Lemma 4.44 we obtain that under conditions (1.39) and (1.40) we
have a.e. convergence

$t_{n}f\rightarrow f\text{ \ \ a.e., \ \ when \ }n\rightarrow\infty.$

The proof is complete.

###### Corollary 4.49

Let $f\in L_{1}$. Then

$\text{\ \ \ }\sigma_{n}^{\alpha}f\rightarrow f\text{ \ \ \ a.e., \ \ when \
}n\rightarrow\infty,\text{ \ \ when \ }0<\alpha<1.$

and

$\beta_{n}^{\alpha}f\rightarrow f\text{ \ \ \ a.e., \ \ when \
}n\rightarrow\infty,\text{ \ \ \ \ when \ }0<\alpha<1.$

Proof: By combining (1.6)-(1.7) with (1.8)-(1.9) we obtain that the Nörlund
means $\sigma_{n}^{\alpha}$ and $\beta_{n}^{\alpha}$, with non-increasing
sequence $\\{q_{k}:k\in\mathbb{N}\\}$ satisfy conditions (1.39) and (1.40).

Hence, the proof is complete by just using Theorem 4.48.

###### Corollary 4.50

Let $f\in L_{1}$ and $t_{n}$ be Nörlund means with monotone and bounded
sequence $\left\\{q_{k}:k\in\mathbb{N}\right\\}$. Then

$\text{\ \ \ }t_{n}f\rightarrow f\text{ \ \ \ a.e., \ \ when \
}n\rightarrow\infty.$

Proof: The proof is analogously to the proofs of Theorems 4.46 and 4.48 so we
leave out the details.

###### Theorem 4.51

Let $f\in L_{1}$. Then

$\text{\ \ \ }R_{n}f\rightarrow f,\text{ \ \ \ a.e. \ \ \ \ when \
}n\rightarrow\infty.$

Proof: The proof is analogously to the proofs of Theorems 4.46 and 4.48 so we
leave out the details.

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