$\displaystyle\frac{\sigma_{{}_{2^{\left|\alpha_{k}\right|}}}f}{\Phi\left(\alpha_{k}\right)\alpha_{k}}+\frac{\left(\alpha_{k}-2^{\left|\alpha_{k}\right|}\right)S_{2^{\left|\alpha_{k}\right|}}f}{\Phi\left(\alpha_{k}\right)\alpha_{k}}$
$\displaystyle+$
$\displaystyle\frac{2^{\left|\alpha_{k}\right|\left(1/p-1\right)}}{\Phi\left(\alpha_{k}\right)u\left(\alpha_{k}\right)\alpha_{k}}\sum_{j=2^{{}_{\left|\alpha_{k}\right|}}}^{\alpha_{k}-1}\left(D_{{}_{j}}-D_{2^{\left|\alpha_{k}\right|}}\right)$
$\displaystyle=$ $\displaystyle IV_{1}+IV_{2}+IV_{3}.$

Let $\alpha_{k}\in\mathbb{N}$ and
$E_{\left[\alpha_{k}\right]}:=I_{{}_{\left[\alpha_{k}\right]+1}}\left(e_{\left[\alpha_{k}\right]-1}+e_{\left[\alpha_{k}\right]}\right).$
Since
$\left[\alpha_{k}-2^{\left|\alpha_{k}\right|}\right]=\left[\alpha_{k}\right],$
analogously to (3.3), if we apply Lemma 3.26 for $IV_{3}$ we have the
following estimation

$\displaystyle\left|IV_{3}\right|$ $\displaystyle=$
$\displaystyle\frac{2^{\left|\alpha_{k}\right|\left(1/p-1\right)}}{\Phi\left(\alpha_{k}\right)u\left(\alpha_{k}\right)\alpha_{k}}\left(\alpha_{k}-2^{\left|\alpha_{k}\right|}\right)\left|K_{\alpha_{k}-2^{\left|\alpha_{k}\right|}}\right|$
$\displaystyle=$
$\displaystyle\frac{2^{\left|\alpha_{k}\right|\left(1/p-1\right)}}{\Phi\left(\alpha_{k}\right)u\left(\alpha_{k}\right)\alpha_{k}}\left|2^{\left[\alpha_{k}\right]}K_{\left[\alpha_{k}\right]}\right|$
$\displaystyle\geq$
$\displaystyle\frac{2^{\left|\alpha_{k}\right|\left(1/p-2\right)}2^{2\left[\alpha_{k}\right]-4}}{\Phi\left(\alpha_{k}\right)u\left(\alpha_{k}\right)}$
$\displaystyle\geq$ $\displaystyle
2^{\left|\alpha_{k}\right|\left(1/p-2\right)/2}2^{2\left[\alpha_{k}\right]-4}/\Phi^{1/2}\left(\alpha_{k}\right).$

Hence,

$\displaystyle\left\|IV_{3}\right\|_{weak-L_{p}(G)}^{p}$ $\displaystyle\geq$
$\displaystyle\left(\frac{2^{\left|\alpha_{k}\right|\left(1/p-2\right)/2}2^{2\left[\alpha_{k}\right]-4}}{\Phi^{1/2}\left(\alpha_{k}\right)}\right)^{p}\mu\left\\{x\in
G:\text{
}\left|IV_{3}\right|\geq\frac{2^{\left|\alpha_{k}\right|\left(1/p-2\right)/2}2^{2\left[\alpha_{k}\right]-4}}{\Phi^{1/2}\left(\alpha_{k}\right)}\right\\}$
$\displaystyle\geq$ $\displaystyle
c_{p}\left(2^{2\left[\alpha_{k}\right]+\left|\alpha_{k}\right|\left(1/p-2\right)/2}/\Phi^{1/2}\left(\alpha_{k}\right)\right)^{p}\mu(E_{\left[\alpha_{k}\right]})$
$\displaystyle\geq$ $\displaystyle
c_{p}\left(2^{\left(\left|\alpha_{k}\right|-\left[\alpha_{k}\right]\right)\left(1/p-2\right)}/\Phi\left(\alpha_{k}\right)\right)^{p/2}$
$\displaystyle=$ $\displaystyle
c_{p}\left(2^{d\left(\alpha_{k}\right)\left(1/p-2\right)}/\Phi\left(\alpha_{k}\right)\right)^{p/2}\rightarrow\infty,\text{\
as }k\rightarrow\infty.$

By combining Corollary 2.13 and first part of Theorem 3.29 we find that

$\left\|IV_{1}\right\|_{weak-L_{p}(G)}\leq c_{p}<\infty,\text{ \ \
}\left\|IV_{2}\right\|_{weak-L_{p}(G)}\leq c_{p}<\infty.$

On the other hand, for sufficiently large $n$ we can conclude that

$\displaystyle\left\|\sigma_{\alpha_{k}}f\right\|_{weak-L_{p}(G)}^{p}$
$\displaystyle\geq$ $\displaystyle\left\|IV_{3}\right\|_{weak-
L_{p}(G)}^{p}-\left\|IV_{2}\right\|_{weak-
L_{p}(G)}^{p}-\left\|IV_{1}\right\|_{weak-L_{p}(G)}^{p}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\left\|IV_{3}\right\|_{weak-
L_{p}(G)}^{p}\rightarrow\infty,\text{ as }k\rightarrow\infty.$

Theorem 3.29 is proved.

The proofs of Corollaries 3.30-3.32 are similar to the proofs of Corollaries
2.13-2.15. So, we leave out the details of proofs and just present these
results:

###### Corollary 3.30

Let $p>0$ and $f\in H_{p}(G)$. Then

$\left\|\sigma_{2^{k}}f-f\right\|_{H_{p}(G)}\rightarrow 0,\text{ as
}k\rightarrow\infty.$

###### Corollary 3.31

Let $p>0$ and $f\in H_{p}(G)$. Then

$\left\|\sigma_{2^{k}+2^{k-1}}f-f\right\|_{H_{p}(G)}\rightarrow 0,\text{ as
}k\rightarrow\infty.$

###### Corollary 3.32

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

$\left\|\sigma_{2^{k}+1}f-f\right\|_{weak-L_{p}(G)}\nrightarrow 0,\text{ as
}k\rightarrow\infty.$

On the other hand, for any $f\in H_{1/2}(G)$ the following is true:

$\left\|\sigma_{2^{k}+1}f-f\right\|_{H_{1/2}(G)}\rightarrow 0,\text{ as
}k\rightarrow\infty.$

###  3.4 Modulus of continuity and convergence in norm of subsequences of
Fejér means with respect to the one-dimensional Walsh-Fourier series on the
martingale Hardy spaces

In this section we apply Theorem 3.28 and Theorem 3.29 to find necessary and
sufficient conditions for modulus of continuity of martingale $f\in H_{p}$,
for which subsequences of Fejér means with respect to the one-dimensional
Walsh-Fourier series converge in $H_{p}$-norm.

First, we prove the following result:

###### Theorem 3.33

a) Let $f\in H_{1/2}(G),\ \
\sup_{k\in\mathbb{N}_{+}}V\left(n_{k}\right)=\infty$ and

$\omega_{H_{p}(G)}\left(1/2^{\left|n_{k}\right|},f\right)=o\left(1/V^{2}\left(n_{k}\right)\right),\text{
\ as \ }k\rightarrow\infty.$ (3.27)

Then

$\left\|\sigma_{n_{k}}f-f\right\|_{H_{1/2}(G)}\rightarrow 0,\text{ as
}k\rightarrow\infty.$

b) Let $\sup_{k\in\mathbb{N}_{+}}V\left(n_{k}\right)=\infty.$ Then there
exists a martingale $f\in H_{1/2}(G),$ such that

$\omega_{H_{1/2}(G)}\left(1/2^{\left|n_{k}\right|},f\right)=O\left(1/V^{2}\left(n_{k}\right)\right),\text{
\ as \ }k\rightarrow\infty$ (3.28)

and

$\left\|\sigma_{n_{k}}f-f\right\|_{H_{1/2}(G)}\nrightarrow 0,\text{ \ as \
}k\rightarrow\infty.$ (3.29)

Proof: Let $f\in H_{1/2}(G)$ and $2^{k}<n\leq 2^{k+1}.$ Then

$\displaystyle\left\|\sigma_{n}f-f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle\leq\left\|\sigma_{n}f-\sigma_{n}S_{2^{k}}f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle+\left\|\sigma_{n}S_{2^{k}}f-S_{2^{k}}f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle+\left\|S_{2^{k}}f-f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle=\left\|\sigma_{n}\left(S_{2^{k}}f-f\right)\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle+\left\|S_{2^{k}}f-f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle+\left\|\sigma_{n}S_{2^{k}}f-S_{2^{k}}f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle\leq
c\left(V\left(n\right)+1\right)\omega_{H_{1/2}(G)}^{1/2}\left(1/2^{k},f\right)$
$\displaystyle+\left\|\sigma_{n}S_{2^{k}}f-S_{2^{k}}f\right\|_{H_{1/2}(G)}^{1/2}.$

It is evident that

$\displaystyle\sigma_{n}S_{2^{k}}f-S_{2^{k}}f$
$\displaystyle=\frac{2^{k}}{n}\left(S_{2^{k}}\sigma_{2^{k}}f-S_{2^{k}}f\right)$
$\displaystyle=\frac{2^{k}}{n}S_{2^{k}}\left(\sigma_{2^{k}}f-f\right).$

Let $p>0.$ By combining Corollaries 2.13 and 3.30 we can conclude that

$\displaystyle\left\|\sigma_{n}S_{2^{k}}f-S_{2^{k}}f\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle\leq\frac{2^{k/2}}{n^{1/2}}\left\|S_{2^{k}}\left(\sigma_{2^{k}}f-f\right)\right\|_{H_{1/2}(G)}^{1/2}$
$\displaystyle\leq\left\|\sigma_{2^{k}}f-f\right\|_{H_{1/2}(G)}^{1/2}\rightarrow
0,\text{\ as \ }k\rightarrow\infty.$

Now, we prove part b) of Theorem 3.33. Since
$\sup_{k\in\mathbb{N}_{+}}V(\alpha_{k})=\infty,$ then there exists a
martingale
$\\{\alpha_{k}:k\in\mathbb{N}_{+}\\}\subset\\{n_{k}:k\in\mathbb{N}_{+}\\}$
such that $V(\alpha_{k})\uparrow\infty,$ as $k\rightarrow\infty$ and

$V^{2}(\alpha_{k})\leq V(\alpha_{k+1}).$ (3.30)

Let $f$ be martingale from Example 2.6, where

$\lambda_{k}=V^{-2}(\alpha_{k}),$

If we apply (3.30) we get that condition (2.3) is fulfilled and it follows
that $f\in H_{p}(G).$ By using (2.4) we find that

$\widehat{f}(j)=\left\\{\begin{array}[]{ll}2^{{}_{\left|\alpha_{k}\right|}}/V^{2}(\alpha_{k}),&\text{\thinspace}j\in\left\\{2^{\left|\alpha_{k}\right|},...,2^{{}_{\left|\alpha_{k}\right|+1}}-1\right\\},\text{
}k\in\mathbb{N}_{+}\\\
0\,,&j\notin\bigcup\limits_{k=0}^{\infty}\left\\{2^{{}_{\left|\alpha_{k}\right|}},...,2^{{}_{\left|\alpha_{k}\right|+1}}-1\right\\}.\text{
}\end{array}\right.$ (3.31)

By combining (2.7) and (3.30) we can conclude that

$\displaystyle
w_{H_{1/2}(G)}(1/2^{n},f)=\left\|f-S_{2^{n}}f\right\|_{H_{1/2}(G)}$ (3.32)
$\displaystyle\leq\sum\limits_{i=n+1}^{\infty}1/V^{2}(\alpha_{i})=O\left(1/V^{2}(\alpha_{n})\right),\text{
\ as \ }n\rightarrow\infty.$

Let $2^{{}_{\left|\alpha_{k}\right|}}<j\leq\alpha_{k}.$ By using (2.6) we get
that

$S_{j}f=S_{2^{{}_{\left|\alpha_{k}\right|}}}f+\frac{2^{{}_{\left|\alpha_{k}\right|}}w_{2^{{}_{\left|\alpha_{k}\right|}}}D_{j-2^{{}_{\left|\alpha_{k}\right|}}}}{V^{2}(\alpha_{k})}.$

Hence,

$\displaystyle\sigma_{\alpha_{k}}f-f$ $\displaystyle=$
$\displaystyle\frac{2^{{}_{\left|\alpha_{k}\right|}}}{\alpha_{k}}\left(\sigma_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\right)$
$\displaystyle+$
$\displaystyle\frac{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}{\alpha_{k}}\left(S_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\right)$
$\displaystyle+$
$\displaystyle\frac{2^{{}_{\left|\alpha_{k}\right|}}w_{2^{{}_{\left|\alpha_{k}\right|}}}\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}}{\alpha_{k}V^{2}(\alpha_{k})}.$

According to (1.2), (1.12) and (3.4) we have that

$\displaystyle\|\sigma_{\alpha_{k}}f-f\|_{1/2}^{1/2}$ $\displaystyle\geq$
$\displaystyle\frac{c}{V(\alpha_{k})}\|\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\|_{1/2}^{1/2}$
$\displaystyle-$
$\displaystyle\left(\frac{2^{{}_{\left|\alpha_{k}\right|}}}{\alpha_{k}}\right)^{1/2}\|\sigma_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\|_{1/2}^{1/2}$
$\displaystyle-$
$\displaystyle\left(\frac{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}{\alpha_{k}}\right)^{1/2}\|S_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\|_{1/2}^{1/2}.$

Let

$\alpha_{k}=\sum_{i=1}^{r_{k}}\sum_{k=l_{i}^{k}}^{m_{i}^{k}}2^{k},$

where

$m_{1}^{k}\geq l_{1}^{k}>l_{1}^{k}-2\geq m_{2}^{k}\geq
l_{2}^{k}>l_{2}^{k}-2>...>m_{s}^{k}\geq l_{s}^{k}\geq 0$

and

$E_{l_{i}^{k}}:=I_{{}_{l_{i}^{k}+1}}\left(e_{l_{i}^{k}-1}+e_{l_{i}^{k}}\right).$

By using Lemma 3.26 we get that

$\displaystyle\int_{G}\left|\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\left(x\right)\right|^{1/2}d\mu$
$\displaystyle\geq$
$\displaystyle\frac{1}{16}\underset{i=2}{\overset{r_{k}-2}{\sum}}\int_{E_{l_{i}^{k}}}\left|\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\left(x\right)\right|^{1/2}d\mu\left(x\right)$
$\displaystyle\geq$
$\displaystyle\frac{1}{16}\underset{i=2}{\overset{r_{k}-2}{\sum}}\frac{1}{2^{l_{i}^{k}}}2^{l_{i}^{k}}\geq
cr_{k}\geq cV(\alpha_{k}).$

By combining estimations (3.4-3.4), Corollaries 2.13 and 3.30 we get that
(3.29) holds true and Theorem 3.33 is proved.

###### Theorem 3.34

a) Let $0<p<1/2,$ $f\in H_{p}(G)$, $\
\sup_{k\in\mathbb{N}_{+}}d\left(n_{k}\right)=\infty$ and

$\omega_{H_{p}(G)}\left(1/2^{\left|n_{k}\right|},f\right)=o\left(1/2^{d\left(n_{k}\right)\left(1/p-2\right)}\right),\text{
as \ }k\rightarrow\infty.$ (3.36)

Then

$\left\|\sigma_{n_{k}}f-f\right\|_{H_{p}(G)}\rightarrow 0,\text{\ as \
}k\rightarrow\infty.$ (3.37)

b) Let $\sup_{k\in\mathbb{N}_{+}}d\left(n_{k}\right)=\infty.$ Then there
exists a martingale $f\in H_{p}(G)$ $\left(0<p<1/2\right),$ such that

$\displaystyle\omega_{H_{p}(G)}\left(1/2^{\left|n_{k}\right|},f\right)=O\left(1/2^{d\left(n_{k}\right)\left(1/p-2\right)}\right),\text{
\ as \ }k\rightarrow\infty$ (3.38)

and

$\left\|\sigma_{n_{k}}f-f\right\|_{weak-L_{p}(G)}\nrightarrow 0,\text{\ as \
}k\rightarrow\infty.$ (3.39)

Proof:  Let $0<p<1/2.$ Then under condition (3.36) if we repeat steps of the
proof of Theorem 3.33, we immediately get that (3.37) holds.

Let prove part b) of Theorem 3.34. Since $\sup_{k}d\left(n_{k}\right)=\infty,$
there exists
$\\{\alpha_{k}:k\in\mathbb{N}_{+}\\}\subset\\{n_{k}:k\in\mathbb{N}_{+}\\}$
such that $\sup_{k\in\mathbb{N}_{+}}d\left(\alpha_{k}\right)=\infty$ and

$2^{2d\left(\alpha_{k}\right)\left(1/p-2\right)}\leq
2^{d\left(\alpha_{k+1}\right)\left(1/p-2\right)}.$ (3.40)

Let $f$ be a martingale from Lemma 2.6, where

$\lambda_{k}=2^{-\left(1/p-2\right)d\left(\alpha_{i}\right)}.$

If we use (3.40) we conclude that condition (2.3) is fulfilled and it follows
that $f\in H_{p}(G).$

According to (2.4) we get that

$\widehat{f}(j)=\left\\{\begin{array}[]{ll}2^{\left(1/p-2\right)\left[\alpha_{k}\right]},&\text{\thinspace\thinspace}j\in\left\\{2^{\left|\alpha_{k}\right|},...,2^{{}_{\left|\alpha_{k}\right|+1}}-1\right\\},\text{
}k\in\mathbb{N}_{+}\\\
0\,,&\text{\thinspace}j\notin\bigcup\limits_{n=0}^{\infty}\left\\{2^{{}_{\left|\alpha_{n}\right|}},...,2^{{}_{\left|\alpha_{n}\right|+1}}-1\right\\}.\text{
}\end{array}\right.$ (3.41)

By combining (2.7) and (3.40) we have that

$\displaystyle\omega_{H_{p}(G)}(1/2^{\left|\alpha_{k}\right|},f)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{i=k}^{\infty}1/2^{d\left(\alpha_{i}\right)\left(1/p-2\right)}$
$\displaystyle=$ $\displaystyle
O\left(1/2^{d\left(\alpha_{k}\right)\left(1/p-2\right)}\right)\text{ \ as \
}k\rightarrow\infty.$

Analogously to the proof of previous theorem, if we use also Corollaries 2.13
and 3.30, for the sufficiently large $k$ we can conclude that

$\displaystyle\|\sigma_{\alpha_{k}}f-f\|_{weak-L_{p}(G)}^{p}$
$\displaystyle\geq$ $\displaystyle
2^{\left(1-2p\right)\left[\alpha_{k}\right]}\|\left(\alpha_{k}-2^{\left|\alpha_{k}\right|}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\|_{weak-
L_{p}(G)}^{p}$ $\displaystyle-$
$\displaystyle\left(\frac{2^{\left|\alpha_{k}\right|}}{\alpha_{k}}\right)^{p}\|\sigma_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\|_{weak-
L_{p}(G)}^{p}$ $\displaystyle-$
$\displaystyle\left(\frac{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}{\alpha_{k}}\right)^{p}\|S_{2^{{}_{\left|\alpha_{k}\right|}}}f-f\|_{weak-
L_{p}(G)}^{p}$ $\displaystyle\geq$ $\displaystyle
2^{\left(1-2p\right)\left[\alpha_{k}\right]-1}\|\left(\alpha_{k}-2^{\left|\alpha_{k}\right|}\right)K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\|_{weak-
L_{p}(G)}^{p}$

Let $x\in E_{\left[\alpha_{k}\right]}.$ Lemma 3.26 follows that

$\displaystyle\mu\left(x\in
G:\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)\left|K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\right|\geq
2^{2\left[\alpha_{k}\right]-4}\right)$
$\displaystyle\geq\mu\left(E_{\left[\alpha_{k}\right]}\right)\geq
1/2^{\left[\alpha_{k}\right]-4}$

and

$\displaystyle 2^{2p\left[\alpha_{k}\right]-4}\mu\left(x\in
G:\left(\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}\right)\left|K_{\alpha_{k}-2^{{}_{\left|\alpha_{k}\right|}}}\right|\geq
2^{2\left[\alpha_{k}\right]-4}\right)$ (3.44) $\displaystyle\geq
2^{\left(2p-1\right)\left[\alpha_{k}\right]-4}.$

Hence, by combining (1.2), (1.12), (3.4) and (3.44) we get that

$\left\|\sigma_{n_{k}}f-f\right\|_{weak-L_{p}(G)}\nrightarrow
0,\,\,\,\text{as}\,\,\,k\rightarrow\infty.$

The proof of Theorem 3.34 is complete.

By using Theorem 3.34 we easily get an important result which was proved in
[60]:

###### Corollary 3.35

a) Let $f\in H_{1/2}(G)$ and

$\omega_{H_{1/2}(G)}\left(1/2^{k},f\right)=o\left(\frac{1}{k^{2}}\right),\text{
as \ }k\rightarrow\infty.$

Then

$\left\|\sigma_{k}f-f\right\|_{H_{1/2}(G)}\rightarrow 0,\,\,\,\text{ \ as \
}k\rightarrow\infty.$

b) There exists a martingale $f\in H_{1/2}(G),$ for which

$\omega_{H_{1/2}(G)}\left(1/2^{k},f\right)=O\left(\frac{1}{k^{2}}\right),\text{
\ as \ }k\rightarrow\infty$

and

$\left\|\sigma_{k}f-f\right\|_{1/2}\nrightarrow 0,\,\,\,\text{ \ as \
}k\rightarrow\infty.$

###### Corollary 3.36

a) Let $0<p<1/2,$ $f\in H_{p}(G)$ and

$\omega_{H_{p}(G)}\left(1/2^{k},f\right)=o\left(1/2^{k(1/p-2)}\right),\text{
as \ }k\rightarrow\infty.$

Then

$\left\|\sigma_{k}f-f\right\|_{H_{p}(G)}\rightarrow
0,\,\,\,\text{as\thinspace\thinspace\thinspace}k\rightarrow\infty.$

b) Then there exists a martingale $f\in H_{p}(G)$ $\left(0<p<1/2\right),$ for
which

$\omega_{H_{p}(G)}\left(1/2^{k},f\right)=O\left(1/2^{k\left(1/p-2\right)}\right),\text{
\ as \ }k\rightarrow\infty$

and

$\left\|\sigma_{k}f-f\right\|_{weak-L_{p}(G)}\nrightarrow 0,\,\,\,\text{ \ as
\ }k\rightarrow\infty.$

###  3.5 Strong convergence of Fejér means with respect to the one-dimensional
Walsh-Fourier series on the martingale Hardy spaces

In this section we consider strong convergence results of Fejér means with
respect to the one-dimensional Walsh-Fourier series in the martingale Hardy
spaces, when $0<p\leq 1/2$ (for details see [59]).

The following is true:

###### Theorem 3.37

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

$\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{H_{p}(G)}^{p}}{m^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}(G)}^{p}.$

b) Let $0<p<1/2,$ $\Phi:\mathbb{N}_{+}\rightarrow[1,\infty)$ be non-decreasing
function, such that $\Phi\left(n\right)\uparrow\infty$ and

$\overline{\underset{k\rightarrow\infty}{\lim}}\frac{k^{2-2p}}{\Phi\left({k}\right)}=\infty.$

Then there exists a martingale $f\in H_{p}(G),$ such that

$\underset{m=1}{\overset{\infty}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{weak-
L_{p}(G)}^{p}}{\Phi\left(m\right)}=\infty.$

Proof: Suppose that

$\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{p}^{p}}{m^{2-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}(G)}^{p}.$

By combining (1.7), (1.15) and Lemma 3.27 we can conclude that

$\displaystyle\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{H_{p}(G)}^{p}}{m^{2-2p}}$
(3.45)
$\displaystyle\leq\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}f\right\|_{p}^{p}}{m^{2-2p}}+\|\widetilde{\sigma}_{\\#}^{\ast}f\|_{H_{p}(G)}+\|\widetilde{S}_{\\#}^{\ast}f\|_{H_{p}(G)}$
$\displaystyle\leq c_{p}\left\|f\right\|_{H_{p}(G)}^{p}.$

According to Lemma 2.7 and (3.45) Theorem 3.37 will be proved if we show that

$\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\left\|\sigma_{m}a\right\|_{p}^{p}}{m^{2-2p}}\leq
c<\infty,\text{ \ \ \ }m=2,3,...$

for any $p$-atom $a$. We may assume that $a$ is $p$-atom, with support $I$,
$\mu\left(I\right)=2^{-M}$ and $I=I_{M}.$ It is evident that
$\sigma_{n}\left(a\right)=0,$ when $n\leq 2^{M}.$ Therefore, we may assume
that $n>2^{M}.$

Let $x\in I_{M}.$ Since $\sigma_{n}$ is bounded from $L_{\infty}(G)$ to
$L_{\infty}(G)$ (The boundedness follows fact that Fejér kernels are uniformly
bounded in the space $L_{1}(G)$, which is proved in Lemma 3.22) and
$\left\|a\right\|_{\infty}\leq 2^{M/p}$ we can conclude that

$\int_{I_{M}}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)\leq\left\|\sigma_{m}a\right\|_{\infty}^{p}/2^{M}$
$\leq\left\|a\right\|_{\infty}^{p}/2^{M}\leq c<\infty,\text{ }0<p\leq 1/2.$

Let $0<p\leq 1/2.$ Then

$\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{\int_{I_{M}}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)}{m^{2-2p}}$
$\leq\frac{c}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=1}{\sum}}\frac{1}{m^{2-2p}}\leq
c<\infty.$

It is evident that

$\left|\sigma_{m}a\left(x\right)\right|\leq\int_{I_{M}}\left|a\left(t\right)\right|\left|K_{m}\left(x+t\right)\right|d\mu\left(t\right)$
$\leq
2^{M/p}\int_{I_{M}}\left|K_{m}\left(x+t\right)\right|d\mu\left(t\right).$

Lemma 3.23 follows that

$\left|\sigma_{m}a\left(x\right)\right|\leq\frac{c2^{k+l}2^{M\left(1/p-1\right)}}{m},\text{
\ \ }x\in I_{l+1}\left(e_{k}+e_{l}\right),\,0\leq k<l<M$ (3.46)

and

$\left|\sigma_{m}a\left(x\right)\right|\leq
c2^{M\left(1/p-1\right)}2^{k},\text{ \ }x\in I_{M}\left(e_{k}\right),\,0\leq
k<M.$ (3.47)

If we use identity (2.1) and (3.46-3.47) we get that

$\displaystyle\int_{\overline{I_{M}}}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)$
$\displaystyle=$
$\displaystyle\overset{M-2}{\underset{k=0}{\sum}}\overset{M-1}{\underset{l=k+1}{\sum}}\int_{I_{l+1}\left(e_{k}+e_{l}\right)}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)$
$\displaystyle+\overset{M-1}{\underset{k=0}{\sum}}\int_{I_{M}\left(e_{k}\right)}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)$
$\displaystyle\leq$ $\displaystyle
c\overset{M-2}{\underset{k=0}{\sum}}\overset{M-1}{\underset{l=k+1}{\sum}}\frac{1}{2^{l}}\frac{2^{p\left(k+l\right)}2^{M\left(1-p\right)}}{m^{p}}+c\overset{M-1}{\underset{k=0}{\sum}}\frac{1}{2^{M}}2^{M\left(1-p\right)}2^{pk}$
$\displaystyle\leq$
$\displaystyle\frac{c2^{M\left(1-p\right)}}{m^{p}}\overset{M-2}{\underset{k=0}{\sum}}\overset{M-1}{\underset{l=k+1}{\sum}}\frac{2^{p\left(k+l\right)}}{2^{l}}+c\overset{M-1}{\underset{k=0}{\sum}}\frac{2^{pk}}{2^{pM}}$
$\displaystyle\leq$
$\displaystyle\frac{c2^{M\left(1-p\right)}M^{\left[1/2+p\right]}}{m^{p}}+c.$

Hence,

$\displaystyle\frac{1}{\log^{\left[1/2+p\right]}n}\overset{n}{\underset{m=2^{M}+1}{\sum}}\frac{\int_{\overline{I_{M}}}\left|\sigma_{m}a\left(x\right)\right|^{p}d\mu\left(x\right)}{m^{2-2p}}$
$\displaystyle\leq$
$\displaystyle\frac{1}{\log^{\left[1/2+p\right]}n}\left(\overset{n}{\underset{m=2^{M}+1}{\sum}}\frac{c2^{M\left(1-p\right)}M^{\left[1/2+p\right]}}{m^{2-p}}+\overset{n}{\underset{m=2^{M}+1}{\sum}}\frac{c}{m^{2-2p}}\right)<c<\infty.$

The proof of part a) of theorem 3.37 is complete.

Now, we prove part b) of Theorem 3.37. Let $\Phi\left(n\right)$ non-decreasing
function satisfying the condition

$\underset{k\rightarrow\infty}{\lim}\frac{2^{\left(\left|n_{k}\right|+1\right)\left(2-2p\right)}}{\Phi\left(2^{\left|n_{k}\right|+1}\right)}=\infty.$
(3.49)

According to (3.49), there exists an increasing sequence
$\left\\{\alpha_{k}:\text{
}k\in\mathbb{N}_{+}\right\\}\subset\left\\{n_{k}:\text{
}k\in\mathbb{N}_{+}\right\\}$ such that

$\left|\alpha_{k}\right|\geq 2,\text{ \ \ \ where \ }k\in\mathbb{N}_{+}$
(3.50)

and

$\displaystyle\sum_{\eta=0}^{\infty}\frac{\Phi^{1/2}\left(2^{\left|\alpha_{\eta}\right|+1}\right)}{2^{\left|\alpha_{\eta}\right|\left(1-p\right)}}$
$\displaystyle=$ $\displaystyle
2^{1-p}\sum_{\eta=0}^{\infty}\frac{\Phi^{1/2}\left(2^{\left|\alpha_{\eta}\right|+1}\right)}{2^{\left(\left|\alpha_{\eta}\right|+1\right)\left(1-p\right)}}<c<\infty.$

Let $f=\left(f_{n},\text{ }n\in\mathbb{N}_{+}\right)\in H_{p}(G)$ be a
martingale from the Example 2.6, where

$\lambda_{k}=\frac{\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{\left(\left|\alpha_{k}\right|\right)\left(1/p-1\right)}}$

By combining (2.3) and (3.5) we get that $f\in H_{p}(G).$ According to (2.4)
we have that

$\displaystyle\widehat{f}(j)=\left\\{\begin{array}[]{l}\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)\text{
\ if \
}j\in\left\\{2^{\left|\alpha_{k}\right|},...,2^{\left|\alpha_{k}\right|+1}-1\right\\},\text{
}k\in\mathbb{N}_{+},\\\ 0\text{ },\text{ \ if \
}j\notin\bigcup\limits_{k=0}^{\infty}\left\\{2^{\left|\alpha_{k}\right|},...,2^{\left|\alpha_{k}\right|+1}-1\right\\}.\text{
}\end{array}\right.$ (3.54)

Let $2^{\left|\alpha_{k}\right|}<n<2^{\left|\alpha_{k}\right|+1}.$ Then

$\sigma_{{}_{n}}f=\frac{1}{n}\sum_{j=1}^{2^{\left|\alpha_{k}\right|}}S_{j}f+\frac{1}{n}\sum_{j=2^{\left|\alpha_{k}\right|}+1}^{n}S_{j}f=III+IV.$
(3.55)

It is evident that

$S_{j}f=0,\,\ \text{ if \thinspace\thinspace}0\leq j\leq
2^{\left|\alpha_{1}\right|}$ (3.56)

Let $2^{\left|\alpha_{s}\right|}<j\leq 2^{\left|\alpha_{s}\right|+1},$ where
$s=1,2,...,k.$ If we apply (2.6) we get that

$\displaystyle S_{j}f$ $\displaystyle=$
$\displaystyle\sum_{\eta=0}^{s-1}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\left(D_{2^{\left|\alpha_{\eta}\right|+1}}-D_{2^{\left|\alpha_{\eta}\right|}}\right)$
$\displaystyle+\Phi^{1/2p}\left(2^{\left|\alpha_{s}\right|+1}\right)w_{2^{\left|\alpha_{s}\right|}}D_{j-2^{\left|\alpha_{s}\right|}}.$

Let $2^{\left|\alpha_{s}\right|+1}\leq j\leq 2^{\left|\alpha_{s+1}\right|},$
$s=0,1,...k-1.$ Then if we use (2.6) we can conclude that

$S_{j}f=\sum_{\eta=0}^{s}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\left(D_{2^{\left|\alpha_{\eta}\right|+1}}-D_{2^{\left|\alpha_{\eta}\right|}}\right).$
(3.58)

Let $x\in I_{2}\left(e_{0}+e_{1}\right).$ Since (see Lemmas 2.3 and 3.23)

$D_{2^{n}}\left(x\right)=K_{2^{n}}\left(x\right)=0,\text{ where }n\geq 2$
(3.59)

by combining (3.50) and (3.56-3.59) we get that

$III=\frac{1}{n}\sum_{\eta=0}^{k-1}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\sum_{v=2^{\left|\alpha_{\eta}\right|}+1}^{2^{\left|\alpha_{\eta}\right|+1}}D_{v}\left(x\right)$
(3.60)
$=\frac{1}{n}\sum_{\eta=0}^{k-1}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\left(2^{\left|\alpha_{\eta}\right|+1}K_{2^{\left|\alpha_{\eta}\right|+1}}\left(x\right)-2^{\left|\alpha_{\eta}\right|}K_{2^{\left|\alpha_{\eta}\right|}}\left(x\right)\right)=0.$

If we use (3.5) when $s=k$ for $IV$ we can write that

$\displaystyle IV$ $\displaystyle=$
$\displaystyle\frac{n-2^{\left|\alpha_{k}\right|}}{n}\sum_{\eta=0}^{k-1}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\left(D_{2^{\left|\alpha_{\eta}\right|+1}}-D_{2^{\left|\alpha_{\eta}\right|}}\right)$
$\displaystyle+$
$\displaystyle\frac{\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{n}\sum_{j=2^{{}_{\left|\alpha_{k}\right|}}+1}^{n}w_{2^{\left|\alpha_{k}\right|}}D_{j-2^{\left|\alpha_{k}\right|}}$
$\displaystyle=$ $\displaystyle IV_{1}+IV_{2}.$

By combining (3.50) and (3.59) we can conclude that

$IV_{1}=0,\text{ \ where \ }x\in I_{2}\left(e_{0}+e_{1}\right).$ (3.62)

Let $\alpha_{k}\in\mathbb{A}_{0,2},$
$2^{\left|\alpha_{k}\right|}<n<2^{\left|\alpha_{k}\right|+1}$ and $x\in
I_{2}\left(e_{0}+e_{1}\right)$. Since
$n-2^{{}_{\left|\alpha_{k}\right|}}\in\mathbb{A}_{0,2},$ Lemmas 2.2 and 3.22
and (3.59) follows that

$\displaystyle\left|IV_{2}\right|=\frac{\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{n}\left|\sum_{j=1}^{n-2^{\left|{}_{\alpha_{k}}\right|}}D_{{}_{j}}\left(x\right)\right|$
$\displaystyle=$
$\displaystyle\frac{\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{n}\left|\left(n-2^{{}_{\left|\alpha_{k}\right|}}\right)K_{n-2^{{}_{\left|\alpha_{k}\right|}}}\left(x\right)\right|$
$\displaystyle\geq$
$\displaystyle\frac{\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{\left|\alpha_{k}\right|+1}}.$

Let $0<p<1/2$ and $n\in\mathbb{A}_{0,2}.$ By combining (3.55-3.5) we get that

$\displaystyle\left\|\sigma_{n}f\right\|_{weak-L_{p}(G)}^{p}$
$\displaystyle\geq$
$\displaystyle\frac{c_{p}\Phi^{1/2}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{p\left(\left|\alpha_{k}\right|+1\right)}}\mu\left\\{x\in
I_{2}\left(e_{0}+e_{1}\right):\text{
}\left|\sigma_{n}f\right|\geq\frac{c_{p}\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{\left|\alpha_{k}\right|+1}}\right\\}$
$\displaystyle\geq$
$\displaystyle\frac{c_{p}\Phi^{1/2}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{p\left(\left|\alpha_{k}\right|+1\right)}}\mu\left\\{I_{2}\left(e_{0}+e_{1}\right)\right\\}$
$\displaystyle\geq$
$\displaystyle\frac{c_{p}\Phi^{1/2}\left(2^{\left|\alpha_{k}\right|+1}\right)}{2^{p\left(\left|\alpha_{k}\right|+1\right)}}.$

Hence,

$\displaystyle\underset{n=1}{\overset{\infty}{\sum}}\frac{\left\|\sigma_{n}f\right\|_{weak-
L_{p}(G)}^{p}}{\Phi\left(n\right)}$ $\displaystyle\geq$
$\displaystyle\underset{\left\\{n\in\mathbb{A}_{0,2}:\text{
}2^{\left|\alpha_{k}\right|}<n<2^{\left|\alpha_{k}\right|+1}\right\\}}{\sum}\frac{\left\|\sigma_{n}f\right\|_{weak-
L_{p}(G)}^{p}}{\Phi\left(n\right)}$ $\displaystyle\geq$
$\displaystyle\frac{1}{\Phi^{1/2}\left(2^{\left|\alpha_{k}\right|+1}\right)}\underset{\left\\{n\in\mathbb{A}_{0,2}:\text{
}2^{\left|\alpha_{k}\right|}<n<2^{\left|\alpha_{k}\right|+1}\right\\}}{\sum}\frac{1}{2^{p\left(\left|\alpha_{k}\right|+1\right)}}$
$\displaystyle\geq$
$\displaystyle\frac{c_{p}2^{\left(1-p\right)\left(\left|\alpha_{k}\right|+1\right)}}{\Phi^{1/2}\left(2^{\left|\alpha_{k}\right|+1}\right)}\rightarrow\infty,\text{
\qquad as \ \ }k\rightarrow\infty.$

The proof of Theorem 3.37 is complete.

###### Theorem 3.38

Let $f\in H_{1/2}(G).$ Then

$\underset{n\in\mathbf{\mathbb{N}}_{+}}{\sup}\underset{\left\|f\right\|_{H_{p}}\leq
1}{\sup}\frac{1}{n}\underset{m=1}{\overset{n}{\sum}}\left\|\sigma_{m}f\right\|_{1/2}^{1/2}=\infty.$

Proof: Let $0<p\leq 1$ and

$f_{k}({x}):=2^{k}\left(D_{2^{k+1}}(x)-D_{2^{k}}(x)\right)$

Since

$\text{supp}(f_{k})=I_{k},\text{ \ \ }\int_{I_{k}}a_{k}d\mu=0$

and

$\left\|f_{k}\right\|_{\infty}\leq 2^{2k}=(\text{supp}f_{k})^{-2},$

we conclude that $f_{k}$ is $1/2$-atom, for every $k\in\mathbb{N}$

Moreover, if we use orthogonality of Walsh functions we get that

$S_{2^{n}}\left(f_{k},x)\right)$ $=\left\\{\begin{array}[]{ll}0,&n=0,...,k,\\\
\left(D_{2^{k+1}}(x)-D_{2^{k}}(x)\right),&n\geq k+1,\end{array}\right.$

and

$\displaystyle\sup\limits_{n\in\mathbb{N}}\left|S_{2^{n}}\left(f_{k},x\right)\right|$
$\displaystyle=$
$\displaystyle\left|\left(D_{2^{k+1}}(x)-D_{2^{k}}(x)\right)\right|,$

where ${x}\in G$.

By combining first equality of Lemma 2.2 and Lemma 2.8 we obtain that

$\displaystyle\left\|a_{k}\right\|_{H_{p}(G)}$ $\displaystyle=$
$\displaystyle{2^{k}}\left\|\sup\limits_{n\in\mathbb{N}}\left|S_{2^{n}}\left(D_{2^{k+1}}(x)-D_{2^{k}}(x)\right)\right|\right\|_{1/2}$
$\displaystyle=$
$\displaystyle{2^{k}}\left\|\left(D_{2^{k+1}}(x)-D_{2^{k}}(x)\right)\right\|_{1/2}$
$\displaystyle=$ $\displaystyle{2^{k}}\left\|D_{2^{k}}(x)\right\|_{1/2}$
$\displaystyle\leq$ $\displaystyle{2^{k}}\cdot 2^{-k}\leq 1.$

It is easy to easy to show that

$\widehat{f}_{m}\left(i\right)=\left\\{\begin{array}[]{l}\text{ }2^{m},\text{
if }i=2^{m},...,2^{m+1}-1,\\\ \text{ }0,\text{otherwise}\end{array}\right.$
(3.64)

and

$S_{i}f_{m}=\left\\{\begin{array}[]{l}2^{m}\left(D_{i}-D_{2^{m}}\right),\text{
\ if }i=2^{m}+1,...,2^{m+1}-1,\\\ \text{ }f_{m},\text{ \ if }i\geq 2^{m+1},\\\
0,\text{ \ otherwise.}\end{array}\right.$ (3.65)

Let $0<n<2^{m}.$ By using first equality of Lemma 2.2 we have that

$\displaystyle\left|\sigma_{n+2^{m}}f_{m}\right|$ $\displaystyle=$
$\displaystyle\frac{1}{n+2^{m}}\left|\overset{n+2^{m}}{\underset{j=2^{m}+1}{\sum}}S_{j}f_{m}\right|$
$\displaystyle=$
$\displaystyle\frac{1}{n+2^{m}}\left|2^{m}\overset{n+2^{m}}{\underset{j=2^{m}+1}{\sum}}\left(D_{j}-D_{2^{m}}\right)\right|$
$\displaystyle=$
$\displaystyle\frac{1}{n+2^{m}}\left|2^{m}\overset{n}{\underset{j=1}{\sum}}\left(D_{j+2^{m}}-D_{2^{m}}\right)\right|$
$\displaystyle=$
$\displaystyle\frac{1}{n+2^{m}}\left|2^{m}\overset{n}{\underset{j=1}{\sum}}D_{j}\right|=\frac{2^{m}}{n+2^{m}}n\left|K_{n}\right|.$

Let

$n=\sum_{i=1}^{s}\sum_{k=l_{i}}^{m_{i}}2^{k},$

where

$0\leq l_{1}\leq m_{1}\leq l_{2}-2<l_{2}\leq m_{2}\leq...\leq
l_{s}-2<l_{s}\leq m_{s}.$

By applying Lemma 3.26 and (3.5) we find that

$\left|\sigma_{n+2^{m}}f_{m}\left(x\right)\right|\geq c2^{2l_{i}},\text{ \ \ \
where \ \ \ }x\in I_{l_{i}+1}\left(e_{l_{i}-1}+e_{l_{i}}\right).$

Hence,

$\displaystyle\int_{G}\left|\sigma_{n+2^{m}}f_{m}(x)\right|^{1/2}d\mu\left(x\right)$
$\displaystyle\geq$ $\displaystyle\text{
}\underset{i=0}{\overset{s}{\sum}}\int_{I_{l_{i}+1}\left(e_{l_{i}-1}+e_{l_{i}}\right)}\left|\sigma_{n+2^{m}}f_{m}(x)\right|^{1/2}d\mu\left(x\right)$
$\displaystyle\geq$ $\displaystyle
c\overset{s}{\underset{i=0}{\sum}}\frac{1}{2^{l_{i}}}2^{l_{i}}\geq cs\geq
cV\left(n\right).$

According to the second estimation of Lemma 2.4 we can conclude that

$\displaystyle\underset{n\in\mathbf{\mathbb{N}}_{+}}{\sup}\underset{\left\|f\right\|_{H_{p}}\leq
1}{\sup}\frac{1}{n}\underset{k=1}{\overset{n}{\sum}}\left\|\sigma_{k}f\right\|_{1/2}^{1/2}$
$\displaystyle\geq$
$\displaystyle\frac{1}{2^{m+1}}\overset{2^{m+1}-1}{\underset{k=2^{m}+1}{\sum}}\left\|\sigma_{k}f_{m}\right\|_{1/2}^{1/2}$
$\displaystyle\geq$
$\displaystyle\frac{c}{2^{m+1}}\underset{k=2^{m}+1}{\overset{2^{m+1}-1}{\sum}}V\left(k-2^{m}\right)$
$\displaystyle\geq$
$\displaystyle\frac{c}{2^{m+1}}\underset{k=1}{\overset{2^{m}-1}{\sum}}V\left(k\right)\geq
c\log m\rightarrow\infty,\text{ \ as \ }m\rightarrow\infty.$

The proof is complete.

##  4 Convergence and summability of partial sums with respect to the two-
dimensional Walsh-Fourier series on the martingale Hardy spaces

###  4.1 Basic notations

Let denote by $\overrightarrow{x}$ the two-dimensional vector
$\overrightarrow{x}:=\left(x^{1},x^{2}\right)$ and by $G^{2}$ the direct
product of two Walsh groups. Let $I_{n}^{2}:=I_{n}\left(0\right)\times
I_{n}\left(0\right)$ for any $n\in\mathbb{N}$ and
$\overline{I_{n}^{2}}:=G^{2}\backslash I_{n}^{2}$.

The norms (or quasi-norms) of the spaces of $L_{p}(G^{2})$ space is defined by

$\left\|f\right\|_{p}:=\left(\int_{G^{2}}\left|f(\overrightarrow{x})\right|^{p}d\mu(\overrightarrow{x})\right)^{1/p}\qquad\left(0<p<\infty\right).$

The space $weak-L_{p}(G^{2})$ consists of all functions $f$ for which

$\left\|f\right\|_{weak-
L_{p}(G^{2})}:=\underset{\lambda>0}{\sup}\lambda\mu\left(\overrightarrow{x}\in{G^{2}}:|f|>\lambda\right)^{1/p}<+\infty.$

Two-dimensional Walsh system is defined by

$w_{n_{1},n_{2}}\left(x^{1},x^{2}\right):=w_{n_{1}}\left(x^{1}\right)w_{n_{2}}\left(x^{2}\right).$

The two-dimensional Walsh system is orthonormal and complete
$L_{2}\left({G}^{2}\right)$ (see [47]).

For $f\in L_{1}\left(G^{2}\right)$ the following number

$\displaystyle\widehat{f}\left(n_{1},n_{2}\right):=\int\limits_{G^{2}}f({\overrightarrow{x}})w_{n_{1},n_{2}}({\overrightarrow{x}})d\mu({\overrightarrow{x}})$

is called $\left(n_{1},n_{2}\right)$-th Fourier coeficients of function $f$.

$\left({n_{1},n_{2}}\right)$-th rectangular partial sum $S_{n_{1},n_{2}}$ of
function $f$ is defined by:

$S_{n_{1},n_{2}}(f;{\overrightarrow{x}}):=\sum\limits_{i_{1}=0}^{n_{1}-1}\sum\limits_{i_{2}=0}^{n_{2}-1}\widehat{f}\left(i_{1},i_{2}\right)w_{i_{1},i_{2}}\left({\overrightarrow{x}}\right).$

$n$-th Marcinkiewicz (Marcinkiewicz-Fejér) means of the two-dimensional Walsh-
Fourier series of function $f$ is defined by

$\mathscr{M}_{n}(f;{\overrightarrow{x}}):=\frac{1}{n}\sum\limits_{k=0}^{n}S_{k,...,k}(f;{\overrightarrow{x}}).$

Dirichlet and Marcinkiewicz kernels of the two-dimensional Walsh-Fourier
series are defined by

$D_{n_{1},n_{2}}\left({\overrightarrow{x}}\right)=D_{n_{1}}\left(x^{1}\right)D_{n_{2}}\left(x^{2}\right).$

and

$K_{n}({\overrightarrow{x}}):=\frac{1}{n}\sum\limits_{k=0}^{n}D_{k,k}({\overrightarrow{x}}).$

For the partial sums of the two-dimensional Walsh-Fourier let as define

$S_{M}^{\left(1\right)}f\left(x^{1},x^{2}\right):=\int\limits_{G}f\left(s,x^{2}\right)D_{M}\left(x^{1}+s\right)d\mu\left(s\right)$

and

$S_{N}^{\left(2\right)}f\left(x^{1},x^{2}\right):=\int\limits_{G}f\left(x^{1},t\right)D_{N}\left(x^{2}+t\right)d\mu\left(t\right).$

For the partial sums of the two-dimensional Walsh-Fourier let define the
following maximal operators $S_{\\#}^{\ast}$ and $\widetilde{S}_{\\#}^{\ast}$
by

$S_{\\#}^{\ast}\left(f;x^{1},x^{2}\right):=\sup_{n\in\mathbb{N_{+}}}\left|S_{n,n}\left(f;x^{1},x^{2}\right)\right|$

and

$\widetilde{S}_{\\#}^{\ast}\left(f;x^{1},x^{2}\right):=\sup_{n\in\mathbb{N}}\left|S_{2^{n},2^{n}}\left(f;x_{1},x_{2}\right)\right|.$

We define the maximal operator $\mathscr{M}^{\ast}$ and restricted maximal
operator $\widetilde{\mathscr{M}}_{\\#}^{\ast}$ of Marcinkiewicz means by

$\mathscr{M}^{\ast}\left(f;x^{1},x^{2}\right):=\sup_{n\in\mathbb{N_{+}}}\left|\mathscr{M}_{n}\left(f;x_{1},x_{2}\right)\right|$

and

$\widetilde{\mathscr{M}}_{\\#}^{\ast}\left(f;x^{1},x^{2}\right):=\sup_{n\in\mathbb{N}}\left|\mathscr{M}_{2^{n}}\left(f;x_{1},x_{2}\right)\right|.$

For the partial sums of the two-dimensional Walsh-Fourier let define the
following weighted maximal operators

$\widetilde{\mathscr{M}}^{\ast}(f;x_{1},x_{2}):=\sup_{n\in\mathbb{N_{+}}}\frac{\left|\mathscr{M}_{n}(f;x_{1},x_{2})\right|}{\log^{3/2}\left(n+1\right)}$

and

$\widetilde{\mathscr{M}}^{\ast,p}\left(f;x_{1},x_{2}\right)=\sup_{n\in\mathbb{N_{+}}}\left|\frac{\mathscr{M}_{n}(f;x_{1},x_{2})}{(n+1)^{2/p-3}}\right|.\quad$

The $\sigma$-algebra generated by the two-dimensional cubes

$I_{n}^{2}\left(\overrightarrow{x}\right):=I_{n}\left(x^{1}\right)\times
I_{n}\left(x^{2}\right),$

is defined by $\digamma_{n}\left(n\in\mathbb{N}\right)$.

The conditional expectation operator with respect to
$\digamma_{n}\left(n\in\mathbb{N}\right)$ is denoted by $E_{n}$ and in our
concrete case we have the following explicit expression for it:

$\displaystyle
E_{n}f(\overrightarrow{x})=S_{2^{n},2^{n}}f\left(\overrightarrow{x}\right)$
$\displaystyle=\sum_{k_{1}=0}^{2^{n}-1}\sum_{k_{2}=0}^{2^{n}-1}\widehat{f}\left(k_{1},k_{2}\right)w_{k_{1},k_{2}}(\overrightarrow{x})$
$\displaystyle=\frac{1}{\left|I_{n}^{2}\left(\overrightarrow{x}\right)\right|}\int_{I_{n}^{2}\left(\overrightarrow{x}\right)}f(\overrightarrow{x})d\mu(\overrightarrow{x}),$

where $\left|I_{n}^{2}\left(\overrightarrow{x}\right)\right|=2^{-2n}$ denotes
measure of cube $I_{n}^{2}\left(\overrightarrow{x}\right)$.

Sequence $f=\left(f_{n},\text{ }n\in\mathbb{N}\right)$ of functions $f_{n}\in
L_{1}\left(G^{2}\right)$ is called dyadic martingales if (for details see
[47])

$\left(i\right)$ $f_{n}$ is measurable with respect to $\sigma-$algebra
$\digamma_{n}$, for any $n\in\mathbb{N}$,

$\left(ii\right)$ $E_{n}f_{m}=f_{n}$ for every $n\leq m$.

The maximal function of martingale $f$ is defined by

$f^{\ast}=\sup_{n\in\mathbb{N}}\left|f_{n}\right|.$

If $f\in L_{1}\left(G^{2}\right),$ then it is weell-known that the maximal
operator is defined by

$f^{\ast}\left(\overrightarrow{x}\right)=\sup\limits_{n\in\mathbb{N}}\frac{1}{\mu\left(I_{n}^{2}\left(\overrightarrow{x}\right)\right)}\left|\int\limits_{I_{n}^{2}\left(\overrightarrow{x}\right)}f\left(\overrightarrow{u}\right)d\mu\left(\overrightarrow{u}\right)\right|.$

For $0<p<\infty$ the one-parameter martingale Hardy space
$H_{p}\left(G^{2}\right)$ consist of all martingales for which

$\left\|f\right\|_{H_{p}\left(G^{2}\right)}:=\left\|f^{\ast}\right\|_{p}<\infty.$

Next, we define $p$-atoms, which are very important to characterize martingale
Hardy spaces.

A function $a$ is called a $p$-atom, if there exists an interval $I^{2}$, such
that

$\int_{I^{2}}ad\mu=0,\\\
\qquad\left\|a\right\|_{\infty}\leq\mu\left(I^{2}\right)^{-1/p},\\\
\qquad\text{supp}\left(a\right)\subset I^{2}.$

It is easy to check that for every martingale $f=(f_{n},n\in\mathbb{N})$ and
for every $({k_{1},k_{2})}\in{\mathbb{N}}^{2}$ the limit

$\widehat{f}\left({k_{1},k_{2}}\right):=\lim_{{{n_{1},n_{2}}}\rightarrow\infty}\int_{{G}^{2}}f_{{{n_{1},n_{2}}}}\left({\overrightarrow{x}}\right)w_{{k_{1},k_{2}}}\left({\overrightarrow{x}}\right)d\mu\left({\overrightarrow{x}}\right)$

exists and it is called $\left({k_{1},k_{2}}\right)$-th Walsh-Fourier
coefficients of $f$.

If $f\in L_{1}\left(G^{2}\right)$ and $(E_{n}f:n\in\mathbb{N})$ is regular
martingale, then

$\displaystyle\widehat{f}\left({k_{1},k_{2}}\right)=\int_{{G^{2}}}f\left({{\overrightarrow{x}}}\right)w_{{k_{1},k_{2}}}({\overrightarrow{x}})d\mu\left({{\overrightarrow{x}}}\right)=\widehat{f}\left({k_{1},k_{2}}\right),\text{
}{k_{1},k_{2}}\in\mathbb{N}.$

For the two-dimensional case modulus of continuity in $H_{p}(G^{2})$ spaces
can be defined as

$\omega_{H_{p}\left(G^{2}\right)}\left(\frac{1}{2^{n}},f\right):=\left\|f-S_{2^{n},2^{n}}f\right\|_{H_{p}\left(G^{2}\right)}.$

It is necessary to describe how can be understood difference
$f-S_{2^{n},2^{n}}f$ where $f$ is martingale and $S_{2^{n},2^{n}}f$ is
function. The following is true:

###### Remark 4.39

Let $0<p\leq 1.$ Since

$S_{2^{n},2^{n}}f=f^{\left(n\right)}\in L_{1}(G^{2}),\text{ \ where
}f=\left(f^{\left(n\right)}:n\in\mathbb{N}\right)\in H_{p}(G^{2})$

and

$\displaystyle\left(S_{2^{k},2^{k}}f^{\left(n\right)}:k\in\mathbb{N}\right)$
$\displaystyle=$
$\displaystyle\left(S_{2^{k},2^{k}}S_{2^{n},2^{n}},k\in\mathbb{N}\right)$
$\displaystyle=$
$\displaystyle\left(S_{2^{0},2^{0}}f,\ldots,S_{2^{n-1},2^{n-1}}f,S_{2^{n},2^{n}}f,S_{2^{n},2^{n}}f,\ldots\right)$
$\displaystyle=$
$\displaystyle\left(f^{\left(0\right)},\ldots,f^{\left(n-1\right)},f^{\left(n\right)},f^{\left(n\right)},\ldots\right)$

we obtain that $f-S_{2^{n},2^{n}}f$ is martingale for which

$\left(f-S_{2^{n},2^{n}}f\right)^{\left(k\right)}=\left\\{\begin{array}[]{ll}0,&k=0,.\ldots,n,\\\
f^{\left(k\right)}-f^{\left(n\right)},&k\geq n+1,\end{array}\right.$

and norm

$\left\|f-S_{2^{n},2^{n}}f\right\|_{H_{p}\left(G^{2}\right)}$

can be understood as $H_{p}(G^{2})$ norm of martingale

$f-S_{2^{n},2^{n}}f=(\left(f-S_{2^{n},2^{n}}f\right)^{\left(k\right)},\ \
k\in\mathbb{N})$

###  4.2 Auxillary lemmas

In the following lemmas we investigate estimations of Marcinkiewicz means of
the two-dimensional Walsh-Fourier series (see Lemma 4.40-Lemma 4.43).

Glukhov [19] proved that the following is true:

###### Lemma 4.40

There exists an absolute constant $c$, such that

$\sup_{n\in\mathbb{N}}\int_{G^{2}}|K_{n}(x^{1},x^{2})|d\mu(x^{1},x^{2})\leq
c.$

The following lemma is proved in [25]:

###### Lemma 4.41

Let $n\geq 2^{N}$, $(x^{1},x^{2})\in(I_{l^{1}}\backslash
I_{l^{1}+1})\times(I_{m^{2}}\backslash I_{m^{2}+1})$ and $0\leq l^{1}\leq
m^{2}<N$. Then

$\displaystyle\int_{I_{N}\times
I_{N}}|K_{n}(x^{1}+t^{1},x^{2}+t^{2})|d\mu(t^{1},t^{2})$ $\displaystyle\leq$
$\displaystyle\frac{c}{n2^{2N}}\left\\{2^{l^{1}-m^{2}}\sum_{r^{1}=l^{1}+1}^{m^{2}+1}2^{r^{1}}D_{2^{m^{2}+1}}(x^{1}+e_{l^{1}}+e_{r^{1}})\sum_{s=m^{2}+1}^{N}D_{2^{s}}(x^{2}+e_{m^{2}}+x_{m^{2}+1,s-1}^{1})\right.$
$\displaystyle\left.+2^{l^{1}+m^{2}}\sum_{s=l^{1}}^{m^{2}}\sum_{r^{1}=l^{1}+1}^{s}D_{2^{s}}(x^{1}+e_{l^{1}}+e_{r^{1}})\right\\},$

where

$x_{i,j}:=\sum_{s=i}^{j}x_{s}e_{s},\text{ \ \ }(x_{i,i-1}=0).$

For our further investigation we need the following lemma (for details see
[25]):

###### Lemma 4.42

Let $(x^{1},x^{2})\in I_{N}\times(I_{m^{2}}\backslash I_{m^{2}+1})$ and $0\leq
m^{2}<N.$ Then

$\int_{I_{N}\times I_{N}}|K_{n}(x^{1}+t^{1},x^{2}+t^{2})|d\mu(t^{1},t^{2})$

$\leq
c\frac{2^{m^{2}}}{n2^{N}}\sum_{s=m^{2}}^{N-1}D_{2^{s}}(x^{2}+e_{m^{2}}),\text{
\ when \ }n>2^{N}.$

We also need the following lemma proved by Goginava [20]:

###### Lemma 4.43

Let

$x^{1}\in
I_{4A}\left(0,...,0,x_{4m}^{1}=1,0,...,0,x_{4l}^{1}=1,x_{4l+1}^{1},...,x_{4A-1}^{1}\right)$

and

$x^{2}\in
I_{4A}\left(0,...,0,x_{4l}^{2}=1,x_{4l+1}^{1}...,x_{4q-1}^{1},1-x_{4q}^{1},x_{4q+1}^{2},...,x_{4A-1}^{2}\right).$

Then

$n_{A-1}\left|K_{n_{A-1}}\left(x^{1},x^{2}\right)\right|\geq 2^{4q+4l+4m-3},$

where

$n_{A}=2^{4A}+2^{4A-4}+...+2^{4}+2^{0}.$

Hardy martingale spaces $H_{p}\left(G^{2}\right)$ have atomic decomposition
for $0<p\leq 1$. The following is true (for details see [50], [78] and [79]):

###### Lemma 4.44

A martingale $f=\left(f_{n},\text{ }n\in\mathbb{N}\right)$ belongs to
$H_{p}(G^{2})\left(0<p\leq 1\right)$ if and only if there exist a sequence of
p-atoms $\left(a_{k},\text{ }k\in\mathbb{N}\right)$ and sequence of real
numbers $\left(\mu_{k},\text{ }k\in\mathbb{N}\right)$ such that for every
$n\in\mathbb{N}$

$\qquad\sum_{k=0}^{\infty}\mu_{k}S_{2^{n},2^{n}}a_{k}=f_{n}$ (4.1)

and

$\qquad\sum_{k=0}^{\infty}\left|\mu_{k}\right|^{p}<\infty.$

Moreover,

$\left\|f\right\|_{H_{p}(G^{2})}\backsim\inf\left(\sum_{k=0}^{\infty}\left|\mu_{k}\right|^{p}\right)^{1/p},$

where infimum is taken over all decomposition of $f$ of the form (4.1).

###### Lemma 4.45

Let $0<p\leq 1$ and $T$ be $\sigma$-sublinear operator, such that

$\int\limits_{G^{2}}\left|Ta(\overrightarrow{x})\right|^{p}d\mu(\overrightarrow{x})\leq
c_{p}<\infty,$

for any $p$-atom $a$. Then there exists an absolute constant $c_{p},$ such
that

$\left\|Tf\right\|_{p}\leq c_{p}\left\|f\right\|_{H_{p}(G)}.$ (4.2)

In addition, if $T$ is bounded from $L_{\infty}(G^{2})$ to
$L_{\infty}(G^{2})$, then we only have to prove that

$\int\limits_{\overset{-}{I^{2}}}\left|Ta\left({\overrightarrow{x}}\right)\right|^{p}d\mu\left({\overrightarrow{x}}\right)\leq
c_{p}<\infty,$

for every $p$-atom $a$, where $I^{2}$ denotes support of $a$ and
$\overline{I^{2}}:=G^{2}\backslash I^{2}$.

In special cases there exists simpler ways how to calculate
$H_{p}(G^{2})$-norm of martingale $f\in H_{p}(G^{2})$ (For details see e.g.
[50], [78] and [79]):

###### Lemma 4.46

Let $g\in L_{1}\left(G^{2}\right)$ and $f:=(E_{n}g:n\in\mathbb{N})$ be regular
martingale. Then $H_{p}\left(G^{2}\right)\text{ }(0<p\leq 1)$ norm is
calculated by

$\left\|f\right\|_{H_{p}(G^{2})}=\left\|\sup\limits_{n\in\mathbb{N}}|S_{2^{n},2^{n}}g|\right\|_{p}.$

Proofs of Lemma 4.47 and Lemma 4.48 are proved in [35], [36], [66].

###### Lemma 4.47

Let $0<p\leq 1$, $2^{k}\leq n<2^{k+1}$ and $S_{n,n}f$ be $(n,n)$-th partial
sum, where $f\in H_{p}(G^{2})$. Then for any fixed $n\in\mathbb{N},$ we have
the following estimation:

$\displaystyle\left\|S_{n,n}f\right\|_{H_{p}(G^{2})}$ $\displaystyle\leq$
$\displaystyle\left\|\sup_{0\leq l\leq
k}\left|S_{2^{l},2^{l}}f\right|\right\|_{p}+\left\|S_{n,n}f\right\|_{p}$
$\displaystyle\leq\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{p}+\left\|S_{n,n}f\right\|_{p}.$

Proof: Let consider the following martingale

$f_{\\#}:=\left(S_{2^{k},2^{k}}S_{n,n}f,\text{
}k\in\mathbb{N}_{+}\right)=\left(S_{2^{0},2^{0}},S_{2^{k},2^{k}}f,...,\ \ \
S_{n,n}f,...,S_{n,n}f,...\right).$

By using Lemma 4.46 we immediately get that

$\displaystyle\left\|S_{n,n}f\right\|_{H_{p}(G^{2})}^{p}$ $\displaystyle\leq$
$\displaystyle\left\|\sup_{0\leq l\leq
k}\left|S_{2^{l},2^{l}}f\right|\right\|_{p}^{p}+\left\|S_{n,n}f\right\|_{p}^{p}$
$\displaystyle\leq\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{p}^{p}+\left\|S_{n,n}f\right\|_{p}^{p}.$

The proof is complete.

###### Lemma 4.48

Let $0<p\leq 1$, $2^{k}\leq n<2^{k+1}$ and $\mathscr{M}_{n}f$ be $n$-th
Marcinkiewicz means, where $f\in H_{p}(G^{2})$. Then for any fixed
$n\in\mathbb{N},$ we get that

$\displaystyle\left\|\mathscr{M}_{n}f\right\|_{H_{p}(G^{2})}^{p}$
$\displaystyle\leq$ $\displaystyle\left\|\sup_{0\leq l\leq
k}\left|\mathscr{M}_{2^{l}}f\right|\right\|_{p}^{p}+\left\|\sup_{0\leq l\leq
k}\left|S_{2^{l}}f\right|\right\|_{p}^{p}+\left\|\mathscr{M}_{n}f\right\|_{p}^{p}$
$\displaystyle\leq$
$\displaystyle\left\|\widetilde{\mathscr{M}}_{\\#}^{\ast}f\right\|_{p}^{p}+\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{p}^{p}+\left\|\mathscr{M}_{n}f\right\|_{p}^{p}.$

Proof: Let consider the following martingale

$f_{\\#}=\left(S_{2^{k}}\mathscr{M}_{n}f,\text{ }k\in\mathbb{N}\right)$
$=\left(\frac{2^{0}\mathscr{M}_{2^{0}}}{n}+\frac{(n-2^{0})S_{2^{0}}f}{n},...,\frac{2^{k}\mathscr{M}_{2^{k}}f}{n}+\frac{(n-2^{k})S_{2^{k}}f}{n},\mathscr{M}_{n}f,...,\mathscr{M}_{n}f,...\right).$

According to Lemma 4.46 we immediately get

$\displaystyle\left\|\mathscr{M}_{n}f\right\|_{H_{p}(G^{2})}^{p}$
$\displaystyle\leq$ $\displaystyle\left\|\sup_{0\leq l\leq
k}\left|\mathscr{M}_{2^{l}}f\right|\right\|_{p}^{p}+\left\|\sup_{0\leq l\leq
k}\left|S_{2^{l},2^{l}}f\right|\right\|_{p}^{p}+\left\|S_{n,n}f\right\|_{p}^{p}$
$\displaystyle\leq$
$\displaystyle\left\|\widetilde{\mathscr{M}}_{\\#}^{\ast}f\right\|_{p}^{p}+\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{p}^{p}+\left\|\mathscr{M}_{n}f\right\|_{p}^{p}.$

The proof is complete.

###  4.3 Strong convergence of partial sums with respect to the two-
dimensional Walsh-Fourier series on the martingale Hardy spaces

In this section we investigate strong convergence of partial sums with respect
to the two-dimensional Walsh-Fourier series on the martingale Hardy spaces
when $0<p\leq 1$ (see [66]). The following theorem is true:

###### Theorem 4.49

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

$\sum\limits_{n=1}^{\infty}\frac{\left\|S_{n,n}f\right\|_{H_{p}(G^{2})}^{p}}{n^{3-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}(G^{2})}^{p}.$

b) Let $0<p<1$ and $\Phi:\mathbb{N}\rightarrow[1,$ $\infty)$ be non-decreasing
function, satisfying the condition

$\underset{n\rightarrow\infty}{\lim}\Phi\left(n\right)=+\infty.$ (4.3)

Then there exists a martingale $f\in H_{p}(G^{2})$, such that

$\underset{n=1}{\overset{\infty}{\sum}}\frac{\left\|S_{n,n}f\right\|_{weak-
L_{p}(G^{2})}^{p}\Phi\left(n\right)}{n^{3-2p}}=\infty.$

Proof: Suppose that

$\sum\limits_{n=1}^{\infty}\frac{\left\|S_{n,n}f\right\|_{p}^{p}}{n^{3-2p}}\leq
c_{p}\left\|f\right\|_{H_{p}(G^{2})}^{p}.$

By combining Lemma 2.9 and inequality (1.18) we can conclude that

$\displaystyle{\sum_{n=1}^{\infty}}\frac{\left\|S_{n,n}f\right\|_{H_{p}(G^{2})}^{p}}{n^{3-2p}}\leq{\sum_{n=1}^{\infty}}\frac{\left\|S_{n,n}f\right\|_{p}^{p}}{n^{3-2p}}+\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{p}^{p}\leq\left\|f\right\|_{H_{p}(G^{2})}^{p}.$

According to Lemma 4.45 we only have to prove that

$\sum\limits_{n=1}^{\infty}\frac{\left\|S_{n,n}a\right\|_{p}^{p}}{n^{3-2p}}\leq
c_{p}<\infty,$ (4.4)

for avery $p$-atom $a$.

Let $a$ be $p$-atom with support $I_{N}\left(z^{1}\right)\times
I_{N}\left(z^{2}\right)$, where
$\mu\left(I_{N}\right)=\mu\left(I_{N}\right)=2^{-N}$. Without loss the
generality we may assume that $z^{1}=z^{2}=0.$

Let $\left(x^{1},x^{2}\right)\in\overline{I}_{N}\times\overline{I}_{N}$. Then

$D_{2^{i}}\left(x^{1}+t^{1}\right)1_{I_{N}}\left(t^{1}\right)=0,\text{ \ when
\ }i\geq N$

and

$D_{2^{i}}\left(x^{2}+t^{2}\right)1_{I_{N}}\left(t^{2}\right)=0,\text{ \ when
\ }i\geq N.$

If we apply $w_{2^{j}}\left(x^{i}+t^{i}\right)=w_{2^{j}}\left(x^{i}\right),$
where $t^{i}\in I_{N},\ \ i=1\vee 2$ and $j<N$ according to both equality of
Lemma 2.3 we get that

$\displaystyle S_{n,n}a\left(x^{1},x^{2}\right)$ $\displaystyle=$
$\displaystyle\int\limits_{G\times
G}a\left(t^{1},t^{2}\right)D_{n}\left(x^{1}+t^{1}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{1},t^{2}\right)$
$\displaystyle=$ $\displaystyle\int\limits_{I_{N}\times
I_{N}}a\left(t^{1},t^{2}\right)D_{n}\left(x^{1}+t^{1}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{1},t^{2}\right)$
$\displaystyle=$ $\displaystyle\int\limits_{I_{N}\times
I_{N}}a\left(t^{1},t^{2}\right)w_{n}\left(x^{1}+t^{1}+x^{2}+t^{2}\right)\sum\limits_{i=0}^{N-1}n_{i}w_{2^{i}}\left(x^{1}+t^{1}\right)D_{2^{i}}\left(x^{1}+t^{1}\right)$
$\displaystyle\times\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{2}+t^{2}\right)D_{2^{j}}\left(x^{2}+t^{2}\right)d\mu\left(t^{1},t^{2}\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}\right)\sum\limits_{i=0}^{N-1}n_{i}w_{2^{i}}\left(x^{2}\right)D_{2^{i}}\left(x^{1}\right)w_{n}\left(x^{2}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{2}\right)D_{2^{j}}\left(x^{2}\right)$
$\displaystyle\times\int\limits_{I_{N}\times
I_{N}}a\left(t^{1},t^{2}\right)w_{n}\left(t^{1}+t^{2}\right)d\mu\left(t^{1},t^{2}\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}+x^{2}\right)\sum\limits_{i=0}^{N-1}n_{i}w_{2^{i}}\left(x^{1}\right)D_{2^{i}}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{2}\right)D_{2^{j}}\left(x^{2}\right)$
$\displaystyle\times\int\limits_{I_{N}}\left(\int\limits_{I_{N}}a\left(t^{2}+\tau,t^{2}\right)d\mu\left(t^{2}\right)\right)w_{n}\left(\tau\right)d\mu\left(\tau\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}+x^{2}\right)\sum\limits_{i=0}^{N-1}n_{i}w_{2^{i}}\left(x^{1}\right)D_{2^{i}}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{2}\right)D_{2^{j}}\left(x^{2}\right)\int\limits_{I_{N}}\Phi\left(\tau\right)w_{n}\left(\tau\right)d\mu\left(\tau\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}+x^{2}\right)\sum\limits_{i=0}^{N-1}n_{i}w_{2^{i}}\left(x^{1}\right)D_{2^{i}}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{2}\right)D_{2^{j}}\left(x^{2}\right)\widehat{\Phi}\left(n\right),$

where

$\Phi\left(\tau\right)=\int\limits_{I_{N}}a\left(t^{i}+\tau,t^{i}\right)d\mu\left(t^{i}\right)\text{
\ and \ }i=1\vee 2.$

Let $x\in I_{s}\backslash I_{s+1},$ where $i=1\vee 2.$ By using again Lemma
2.3 can conclude that

$\overset{N-1}{\underset{i=0}{\sum}}D_{2^{i}}\left(x\right)\leq c2^{s}.$

By using (2.1) we have that

$\displaystyle\int_{\overline{I}_{N}}\left(\overset{N-1}{\underset{i=0}{\sum}}D_{2^{i}}\left(x\right)\right)^{p}d\mu\left(x\right)$
$\displaystyle\leq$ $\displaystyle
c_{p}\underset{s=0}{\overset{N-1}{\sum}}\int_{I_{s}\backslash
I_{s+1}}2^{ps}d\mu\left(x\right)$ $\displaystyle\leq$ $\displaystyle
c_{p}\underset{s=0}{\overset{\infty}{\sum}}2^{\left(p-1\right)s}$
$\displaystyle<$ $\displaystyle c_{p}<\infty,\text{ \qquad}0<p<1.$

From (4.3) we get that

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{\overline{I}_{N}\times\overline{I}_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{n=1}^{\infty}\frac{\left|\widehat{\Phi}\left(n\right)\right|^{p}}{n^{3-2p}}\int\limits_{\overline{I}_{N}}\left(\sum\limits_{i=0}^{N-1}D_{2^{i}}\left(x^{1}\right)\right)^{p}d\mu\left(x^{1}\right)\int\limits_{\overline{I}_{N}}\left(\sum\limits_{i=0}^{N-1}D_{2^{i}}\left(x^{2}\right)\right)^{p}d\mu\left(x^{2}\right)$
$\displaystyle\leq$ $\displaystyle
c_{p}\sum\limits_{n=1}^{\infty}\frac{\left|\widehat{\Phi}\left(n\right)\right|^{p}}{n^{3-2p}}.$

Let $n<2^{N}$. Since $w_{n}\left(\tau\right)=1,$ where $\tau\in I_{N}$ we
obtain that

$\displaystyle\widehat{\Phi}\left(n\right)$ $\displaystyle=$
$\displaystyle\int\limits_{I_{N}}\Phi\left(\tau\right)w_{n}\left(\tau\right)d\mu\left(\tau\right)$
$\displaystyle=$
$\displaystyle\int\limits_{I_{N}}\left(\int\limits_{I_{N}}a\left(t^{2}+\tau,t^{2}\right)d\mu\left(t^{2}\right)\right)w_{n}\left(\tau\right)d\mu\left(\tau\right)$
$\displaystyle=$ $\displaystyle\int\limits_{I_{N}\times
I_{N}}a\left(t^{1},t^{2}\right)d\mu\left(t^{1},t^{2}\right)=0.$

So, we may assume that $n\geq 2^{N}.$ If we apply Holder’s inequality we get
that

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{\left|\widehat{\Phi}\left(n\right)\right|^{p}}{n^{3-2p}}$
$\displaystyle\leq$
$\displaystyle\left(\sum\limits_{n=2^{N}}^{\infty}\left|\widehat{\Phi}\left(n\right)\right|^{2}\right)^{p/2}\left(\sum\limits_{n=2^{N}}^{\infty}\frac{1}{n^{\left(3-2p\right)\cdot\left(2/\left(2-p\right)\right)}}\right)^{\left(2-p\right)/2}$
$\displaystyle\leq$
$\displaystyle\left(\frac{1}{2^{N\left(2\left(3-2p\right)/\left(2-p\right)-1\right)}}\right)^{\left(2-p\right)/2}\left(\int\limits_{G}\left|\Phi\left(\tau\right)\right|^{2}d\mu\left(\tau\right)\right)^{p/2}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(4-3p\right)/2}}\left(\int\limits_{I_{N}}\left|\int\limits_{I_{N}}a\left(t^{2}+\tau,t^{2}\right)d\mu\left(t\right)\right|^{2}d\mu\left(\tau\right)\right)^{p/2}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(4-3p\right)/2}}\left\|a\right\|_{\infty}^{p}\frac{1}{2^{Np/2}}\frac{1}{2^{Np}}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(4-3p\right)/2}}2^{2N}\frac{1}{2^{3pN/2}}<c_{p}<\infty.$

Let $\left(x^{1},x^{2}\right)\in\overline{I}_{N}\times I_{N}$. Then

$\displaystyle S_{n,n}a\left(x^{1},x^{2}\right)$ $\displaystyle=$
$\displaystyle
w_{n}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{1}\right)D_{2^{j}}\left(x^{1}\right)$
$\displaystyle\times\int\limits_{G\times
G}a\left(t^{1},t^{2}\right)w_{n}\left(t^{1}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{1},t^{2}\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{1}\right)D_{2^{j}}\left(x^{1}\right)\int\limits_{G}S_{n}^{\left(2\right)}a\left(t^{1},x^{2}\right)w_{n}\left(t^{1}\right)d\mu\left(t^{1}\right)$
$\displaystyle=$ $\displaystyle
w_{n}\left(x^{1}\right)\sum\limits_{j=0}^{N-1}n_{j}w_{2^{j}}\left(x^{1}\right)D_{2^{j}}\left(x^{1}\right)\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right).$

By applying (4.3) we get that

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{\overline{I}_{N}\times
I_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{\overline{I}_{N}\times
I_{N}}\left(\sum\limits_{j=0}^{N-1}D_{2^{j}}\left(x^{1}\right)\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|\right)^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int_{\overline{I}_{N}}\left(\overset{N-1}{\underset{i=0}{\sum}}D_{2^{i}}\left(x^{1}\right)\right)^{p}d\mu\left(x^{1}\right)\cdot\int\limits_{I_{N}}\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|^{p}d\mu\left(x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{I_{N}}\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|^{p}d\mu\left(x^{2}\right).$

Let $n<2^{N}$. By applyng definition of $p$-atom we get that

$\displaystyle\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)$
$\displaystyle=$
$\displaystyle\int\limits_{G}\left(\int\limits_{G}a\left(t^{1},t^{2}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{2}\right)\right)w_{n}\left(t^{1}\right)d\mu\left(t^{1}\right)$
$\displaystyle=$ $\displaystyle
D_{n}\left(x^{2}\right)\int\limits_{I_{N}\times
I_{N}}a\left(t^{1},t^{2}\right)d\mu\left(t^{1},t^{2}\right)=0.$

Therefore, we can suppose that $n\geq 2^{N}$. If follows that

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{\overline{I}_{N}\times
I_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{n=2^{N}}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{I_{N}}\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|^{p}d\mu\left(x^{2}\right)$

Since

$\left\|S_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right\|_{2}\leq
c\left\|a\right\|_{2}$

if we use Holder’s inequality we can conclude that

$\displaystyle\int\limits_{I_{N}}\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|^{p}d\mu\left(x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\int\limits_{I_{N}}\left|\widehat{S}_{n}^{\left(2\right)}a\left(n,x^{2}\right)\right|d\mu\left(x^{2}\right)\right)^{p}$
$\displaystyle=$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\int\limits_{I_{N}}\left|\int\limits_{I_{N}}S_{n}^{\left(2\right)}a\left(t^{1},x^{2}\right)w_{n}\left(t^{1}\right)d\mu\left(t^{1}\right)\right|d\mu\left(x^{2}\right)\right)^{p}$
$\displaystyle=$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\int\limits_{I_{N}}\left|\int\limits_{I_{N}}\left(\int\limits_{I_{N}}a\left(t^{1},t^{2}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{2}\right)\right)w_{n}\left(t^{1}\right)d\mu\left(t^{1}\right)\right|d\mu\left(x^{2}\right)\right)^{p}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\int\limits_{I_{N}}\left(\int\limits_{I_{N}}\left|\int\limits_{I_{N}}a\left(t^{1},t^{2}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{2}\right)\right|d\mu\left(x^{2}\right)\right)d\mu\left(t^{1}\right)\right)^{p}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\frac{1}{2^{N/2}}\int\limits_{I_{N}}\left(\int\limits_{I_{N}}\left|\int\limits_{I_{N}}a\left(t^{1},t^{2}\right)D_{n}\left(x^{2}+t^{2}\right)d\mu\left(t^{2}\right)\right|^{2}d\mu\left(x^{2}\right)\right)^{1/2}d\mu\left(t^{1}\right)\right)^{p}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\frac{1}{2^{N/2}}\int\limits_{I_{N}}\left(\int\limits_{I_{N}}\left|a\left(t^{1},t^{2}\right)\right|^{2}d\mu\left(t^{2}\right)\right)^{1/2}d\mu\left(t^{1}\right)\right)^{p}$
$\displaystyle\leq$
$\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\frac{\left\|a\right\|_{\infty}}{2^{N/2}}\frac{1}{2^{N}}\frac{1}{2^{N/2}}\right)^{p}\leq\frac{c_{p}}{2^{N\left(1-p\right)}}\left(\frac{2^{2N/p}}{2^{2N}}\right)^{p}\leq
c_{p}2^{N\left(1-p\right)}.$

Hence,

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{\overline{I}_{N}\times
I_{N}}{\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|}^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$ $\displaystyle
c_{p}\sum\limits_{n=2^{N}}^{\infty}\frac{1}{n^{3-2p}}2^{N\left(1-p\right)}$
$\displaystyle\leq$ $\displaystyle\frac{c_{p}}{2^{N\left(1-p\right)}}\leq
c_{p}<\infty.$

Analogously, we can prove that

$\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{I_{N}\times\overline{I}_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{p}d\mu\left(x^{1},x^{2}\right)\leq
c_{p}<\infty.$ (4.8)

Let $\left(x^{1},x^{2}\right)\in I_{N}\times I_{N}$. Then by the definition of
$p$-atom we get that

$\displaystyle\int\limits_{I_{N}\times
I_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{p}d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{2^{N\left(2-p\right)}}\left(\int\limits_{I_{N}\times
I_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|^{2}d\mu\left(x^{1},x^{2}\right)\right)^{p/2}$
$\displaystyle\leq$
$\displaystyle\frac{1}{2^{N\left(2-p\right)}}\left(\int\limits_{I_{N}\times
I_{N}}\left|a\left(x^{1},x^{2}\right)\right|^{2}d\mu\left(x^{1},x^{2}\right)\right)^{p/2}$
$\displaystyle\leq$
$\displaystyle\frac{\left\|a\right\|_{\infty}^{p}}{2^{N\left(2-p\right)}}\frac{1}{2^{Np}}$
$\displaystyle\leq$ $\displaystyle
c_{p}\frac{1}{2^{N\left(2-p\right)}}2^{2N}\frac{1}{2^{Np}}\leq c_{p}<\infty.$

It follows that

$\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\int\limits_{I_{N}\times
I_{N}}\left|S_{n,n}a\left(x^{1},x^{2}\right)\right|d\mu\left(x^{1},x^{2}\right)$
$\displaystyle\leq$ $\displaystyle
c_{p}\sum\limits_{n=1}^{\infty}\frac{1}{n^{3-2p}}\leq c_{p}<\infty.$

By combining (4.4-4.3) we get that Theorem 4.49 is proved.

Let $0<p<1$ and $\Phi\left(n\right)$ satisfies condition 4.3. Then there
exists increasing sequence of natural numbers $\left\\{\alpha_{k}:\text{
}k\in\mathbb{N}_{+}\right\\}$ such that

$\alpha_{0}\geq 2$

and

$\sum_{k=0}^{\infty}{\Phi^{-p/4}\left(2^{2\alpha_{k}}\right)}<\infty.$ (4.10)

Let $f=\left(f_{n},\text{ }n\in\mathbb{N}_{+}\right)$ be a martingale

$f_{n}(x^{1},x^{2})={\sum_{k=1}^{n}}\lambda_{k}{a_{k}(x^{1},x^{2})}$

where

$a_{k}(x^{1},x^{2})=2^{2\alpha_{k}(1/p-1)}\left(D_{2^{2\alpha_{k}+1}}(x^{1})-D_{2\alpha_{k}}(x^{1})\right)\left(D_{2^{2\alpha_{k}+1}}(x^{2})-D_{2^{2\alpha_{k}}}(x^{2})\right)$

and

$\lambda_{k}={\Phi^{-1/4}\left(2^{2\alpha_{k}}\right)}.$

By applying (4.10) and Lemma 4.44 we obtain that $f\in H_{p}(G^{2}).$

It is evident that

$\widehat{f}(i,j)=\left\\{\begin{array}[]{l}\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)},\\\
\text{ თუ }\left(i,j\right)\in\left\\{2^{2\alpha_{k}},...,\text{
~{}}2^{2\alpha_{k}+1}-1\right\\}^{2},k\in\mathbb{N}_{+},\\\ 0,\text{
\thinspace}\\\ \text{\thinspace\thinspace თუ
\thinspace}\left(i,j\right)\notin\bigcup\limits_{k=1}^{\infty}\left\\{2^{2\alpha_{k}},...,\text{
~{}}2^{2\alpha_{k}+1}-1\right\\}^{2}.\end{array}\right.$ (4.11)

Let $2^{2\alpha_{k}}<n<2^{2\alpha_{k}+1}$. By combining (4.11) and first
equality of Lemma 2.2 we get that

$\displaystyle S_{n,n}f\left(x^{1},x^{2}\right)$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{2^{\alpha_{k-1}+1}-1}\sum_{j=0}^{2^{\alpha_{k-1}+1}-1}\widehat{f}(i,j)w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle+$
$\displaystyle\sum_{i=2^{\alpha_{k}}}^{n-1}\sum_{j=2^{\alpha_{k}}}^{n-1}\widehat{f}(i,j)w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{\eta=0}^{k-1}\sum_{i=2^{\alpha_{\eta}}}^{2^{\alpha_{\eta}+1}-1}\sum_{j=2^{\alpha_{\eta}}}^{2^{\alpha_{\eta}+1}-1}\widehat{f}(i,j)w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle+$
$\displaystyle\sum_{i=2^{\alpha_{k}}}^{n-1}\sum_{j=2^{\alpha_{k}}}^{n-1}\widehat{f}(i,j)w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{\eta=0}^{k-1}\sum_{i=2^{\alpha_{\eta}}}^{2^{\alpha_{\eta}+1}-1}\sum_{j=2^{\alpha_{\eta}}}^{2^{\alpha_{\eta}+1}-1}\frac{2^{\alpha_{\eta}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{\alpha_{\eta}}\right)}w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle+$
$\displaystyle\sum_{i=2^{\alpha_{k}}}^{n-1}\sum_{j=2^{\alpha_{k}}}^{n-1}\frac{2^{\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{\alpha_{k}}\right)}w_{i}\left(x^{1}\right)w_{j}\left(x^{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{\eta=0}^{k-1}\frac{2^{\alpha_{\eta}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{\alpha_{\eta}}\right)}\left(D_{2^{\alpha_{\eta}+1}}\left(x^{1}\right)-D_{2^{\alpha_{\eta}}}\left(x^{1}\right)\right)\left(D_{2^{\alpha_{\eta}+1}}\left(x^{2}\right)-D_{2^{\alpha_{\eta}}}\left(x^{2}\right)\right)$
$\displaystyle+$
$\displaystyle\frac{2^{\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{\alpha_{k}}\right)}\left(D_{{}_{n}}\left(x^{1}\right)-D_{2^{\alpha_{k}}}\left(x^{1}\right)\right)\left(D_{n}\left(x^{2}\right)-D_{2^{\alpha_{k}}}\left(x^{2}\right)\right)$
$\displaystyle=$ $\displaystyle I+II.$

Let $\left(x^{1},x^{2}\right)\in\left(G\backslash I_{1}\right)^{2}$ and $n\ $
is odd number. Since $n-2^{2\alpha_{k}}\ $ is also odd, according to both
equality of Lemma 2.3 we get that

$\displaystyle\left|II\right|$ $\displaystyle=$
$\displaystyle\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\left|w_{2^{2\alpha_{k}}}\left(x^{1}\right)D_{n-2^{2\alpha_{k}}}\left(x^{1}\right)w_{2^{2\alpha_{k}}}\left(x^{2}\right)D_{n-2^{2\alpha_{k}}}\left(x^{2}\right)\right|$
$\displaystyle=$
$\displaystyle\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\left|w_{2^{2\alpha_{k}}}\left(x^{1}\right)w_{n-2^{2\alpha_{k}}}\left(x^{1}\right)D_{{}_{1}}\left(x^{1}\right)w_{2^{2\alpha_{k}}}\left(x^{2}\right)w_{n-2^{2\alpha_{k}}}\left(x^{2}\right)D_{{}_{1}}\left(x^{2}\right)\right|$
$\displaystyle=$
$\displaystyle\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}.$

If we apply again second equality of Lemma 2.3 according to condition
$\alpha_{n}\geq 2$ $\left(n\in\mathbb{N}\right)$ for $I$ we can conclude that

$I=\sum_{\eta=0}^{k-1}\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{\eta}}\right)}\left(D_{2^{2\alpha_{\eta}+1}}\left(x^{1}\right)-D_{2^{2\alpha_{\eta}}}\left(x^{1}\right)\right)\left(D_{2^{\alpha_{\eta}+1}}\left(x^{2}\right)-D_{2^{\alpha_{\eta}}}\left(x^{2}\right)\right)=0.$
(4.14)

Hence,

$\displaystyle\left\|S_{n,n}f\left(x^{1},x^{2}\right)\right\|_{weak-
L_{p}(G^{2})}$ $\displaystyle\geq$
$\displaystyle\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{2\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\left(\mu\left\\{\left(x^{1},x^{2}\right)\in\left(G\backslash
I_{1}\right)^{2}:\left|S_{n,n}f\left(x^{1},x^{2}\right)\right|\geq\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{2\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\right\\}\right)^{1/p}$
$\displaystyle\geq$
$\displaystyle\frac{2^{2\alpha_{k}\left(2/p-2\right)}}{2\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\left|(G\backslash
I_{1}\right)^{2}|\geq\frac{c_{p}2^{2\alpha_{k}\left(2/p-2\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}.$

By using (4.3) we get that

$\displaystyle\underset{n=1}{\overset{2^{2\alpha_{k}+1}-1}{\sum}}\frac{\left\|S_{n,n}f\right\|_{weak-
L_{p}(G^{2})}^{p}\Phi\left(n\right)}{n^{3-2p}}$ $\displaystyle\geq$
$\displaystyle\underset{n=2^{2\alpha_{k}}+1}{\overset{2^{2\alpha_{k}+1}-1}{\sum}}\frac{\left\|S_{n,n}f\right\|_{weak-
L_{p}(G^{2})}^{p}\Phi\left(n\right)}{n^{3-2p}}$ $\displaystyle\geq$
$\displaystyle
c_{p}\Phi\left(2^{2\alpha_{k}}\right)\underset{n=2^{\alpha_{k}-1}+1}{\overset{2^{\alpha_{k}}-1}{\sum}}\frac{\left\|S_{2n+1,2n+1}f\right\|_{weak-
L_{p}(G^{2})}^{p}}{\left(2n+1\right)^{3-2p}}$ $\displaystyle\geq$
$\displaystyle
c_{p}\Phi\left(2^{2\alpha_{k}}\right)\frac{2^{2\alpha_{k}\left(1-p\right)}}{\Phi^{1/4}\left(2^{2\alpha_{k}}\right)}\underset{n=2^{2\alpha_{k}-1}+1}{\overset{2^{2\alpha_{k}}-1}{\sum}}\frac{1}{\left(2n+1\right)^{3-2p}}$
$\displaystyle\geq$ $\displaystyle
c_{p}\Phi^{3/4}\left(2^{2\alpha_{k}}\right)\rightarrow\infty,\text{ \quad as
\quad}k\rightarrow\infty.$

Theorem is proved.

###  4.4 Strong convergence of Marcinkiewicz means with respect to the two-
dimensional Walsh-Fourier series on the martingale Hardy spaces

In this section we consider strong convergence of Marcinkiewicz means with
respect to the two-dimensional Walsh-Fourier series in the martingale Hardy
spaces for $p=2/3$ (for details see Nagy and Tephnadze [35]).

###### Theorem 4.50

Let $f\in H_{2/3}\left(G^{2}\right)$. Then there exists an absolute constant
$c,$ such that

$\frac{1}{\log
n}\sum_{m=1}^{n}\frac{\left\|\mathscr{M}_{m}f\right\|_{H_{2/3}(G^{2})}^{2/3}}{m}\leq
c\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}.$

Proof: Suppose that

$\frac{1}{\log
n}{\sum_{m=1}^{n}}\frac{\left\|\mathscr{M}_{m}f\right\|_{2/3}^{2/3}}{m}\leq
c\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}.$ (4.17)

By combining Lemma 4.48 and inequalities (1.18), (1.22), (4.17) we get that

$\displaystyle\frac{1}{\log
n}{\sum_{m=1}^{n}}\frac{\left\|\mathscr{M}_{m}f\right\|_{H_{2/3}(G^{2})}^{2/3}}{m}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}{\sum_{m=1}^{n}}\frac{\left\|\mathscr{M}_{m}f\right\|_{2/3}^{2/3}}{m}$
$\displaystyle+$
$\displaystyle\left\|\widetilde{\mathscr{M}}_{\\#}^{\ast}f\right\|_{2/3}^{2/3}+\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{2/3}^{2/3}$
$\displaystyle\leq$ $\displaystyle\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}.$

Since $\mathscr{M}_{n}$ is (see Lemma 4.40) bounded from $L_{\infty}(G^{2})$
to $L_{\infty}(G^{2})$, if we use Lemma 4.45 we only have to prove that

$\frac{1}{\log
n}\sum_{m=1}^{n}\frac{\left\|\mathscr{M}_{m}a\right\|_{2/3}^{2/3}}{m}<c<\infty,$

for every $2/3$-atom $a$.

Let $a$ be $2/3$-atom with support $I^{2}$, where $\mu(I^{2})=2^{-2N}$.
Without loss the generality we may assume that $I^{2}:=I_{N}^{2}$. It is easy
to show that ${\mathscr{M}}_{n}a=0$ for $n\leq 2^{N}.$ So, we may assume that
$n>2^{N}$.

We can write that

$\displaystyle\frac{1}{\log
n}\sum_{m=1}^{n}\frac{\left\|\mathscr{M}_{m}a\right\|_{2/3}^{2/3}}{m}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\frac{\left\|{\mathscr{M}_{m}a}\right\|_{2/3}^{2/3}}{m}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\int_{I_{N}\times
I_{N}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$ $\displaystyle+$
$\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\int_{I_{N}\times\overline{I_{N}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle+$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\int_{\overline{I_{N}}\times{I_{N}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle+$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\int_{\overline{I_{N}}\times\overline{I_{N}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle=:$ $\displaystyle I_{1}+I_{2}+I_{3}+I_{4}.$

By applying Lemma 4.40 we have that

$\displaystyle I_{1}$ $\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{\infty}\int_{I_{N}\times
I_{N}}\frac{\left|\mathscr{M}_{m}a\right|^{2/3}}{m}d\mu$ $\displaystyle\leq$
$\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{\infty}\frac{1}{m}\left\|a\right\|_{\infty}^{2/3}/2^{2N}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\frac{1}{m}<c<\infty.$

Now, we estimate $I_{2}$. Set

$J_{t}:=I_{t}\backslash I_{t+1},\text{ \ \ }(t\in{\mathbb{N}}).$

We introduce $\overline{I_{N}}$ and $J_{m^{2}}$ as the following disjoint
union:

$\overline{I_{N}}=\bigcup_{m^{2}=0}^{N-1}J_{m^{2}},\quad
J_{m^{2}}=\bigcup_{q^{2}=m^{2}+1}^{N}I_{N}^{m^{2},q^{2}},$ (4.19)

where

$I_{N}^{m^{2},q^{2}}:=\begin{cases}I_{q^{2}+1}(0,...,0,x_{m^{2}}=1,0,...,0,x_{q^{2}}=1),&\text{
where }\text{ \ \ }m^{2}<q^{2}<N,\\\
I_{N}(0,...,0,x_{m^{2}}=1,0,...,0),&\text{ where }\text{ \ \
}q^{2}=N.\end{cases}$

Let $(x^{1},x^{2})\in I_{N}\times I_{N}^{m^{2},q^{2}}.$ According to Lemma
4.42 we can conclude that

$\displaystyle{|{\mathscr{M}}_{n}a(x^{1},x^{2})|}$ $\displaystyle\leq$
$\displaystyle\|a\|_{\infty}\int_{I_{N}\times
I_{N}}|K_{n}(x^{1}+t^{1},x^{2}+t^{2})|d\mu(t^{1},t^{2})$ $\displaystyle\leq$
$\displaystyle
c2^{3N}\frac{2^{m^{2}}}{n2^{N}}\sum_{s=m^{2}}^{q^{2}}D_{2^{s}}(x^{2}+e_{m^{2}})$
$\displaystyle\leq$
$\displaystyle\frac{c2^{2N+m^{2}}}{n}\sum_{s=m^{2}}^{q^{2}}2^{s}$
$\displaystyle\leq$ $\displaystyle\frac{c2^{2N+m^{2}+q^{2}}}{n}.$

Hence,

$\displaystyle I_{2}$ $\displaystyle\leq$ $\displaystyle\frac{c2^{4N/3}}{\log
n}\sum_{m=2^{N}}^{n}\sum_{m^{2}=0}^{N-1}\sum_{q^{2}=m^{2}+1}^{N}\int_{I_{N}\times
I_{N}^{m^{2},q^{2}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle\leq$ $\displaystyle\frac{c2^{4N/3}}{\log
n}\sum_{m=2^{N}}^{n}\sum_{m^{2}=0}^{N-1}\sum_{q^{2}=m^{2}+1}^{N}\int_{I_{N}\times
I_{N}^{m^{2},q^{2}}}\frac{2^{2(m^{2}+q^{2})/3}}{m^{5/3}}d\mu$
$\displaystyle\leq$ $\displaystyle\frac{c2^{4N/3}}{\log
n}\sum_{m=2^{N}}^{n}\sum_{m^{2}=0}^{N-1}\sum_{q^{2}=m^{2}+1}^{N}\frac{2^{2(m^{2}+q^{2})/3}}{m^{5/3}}2^{-N-q^{2}}$
$\displaystyle\leq$ $\displaystyle\frac{c2^{N/3}}{\log
n}\sum_{m=2^{N}}^{\infty}\frac{1}{m^{5/3}}\sum_{m^{2}=0}^{N-1}2^{2m^{2}/3}\sum_{q^{2}=m^{2}+1}^{N}2^{-q^{2}/3}$
$\displaystyle\leq$ $\displaystyle\frac{c2^{N/3}}{\log
n}\sum_{m=2^{N}}^{\infty}\frac{2^{N/3}}{m^{5/3}}$ $\displaystyle\leq$
$\displaystyle\frac{c2^{2N/3}}{\log
n}\sum_{m=2^{N}}^{\infty}\frac{1}{m^{5m/3}}$ $\displaystyle\leq$
$\displaystyle\frac{c}{N}.$

Analogously, we can prove that $I_{3}\leq c<\infty$.

Next we prove boundedness of $I_{4}$. If we apply (4.19) we get that

$\displaystyle I_{4}$ $\displaystyle\leq$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\sum_{l^{1}=0}^{N-1}\sum_{m^{2}=0}^{l^{1}-1}\int_{J_{l^{1}}\times
J_{m^{2}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle+$ $\displaystyle\frac{1}{\log
n}\sum_{m=2^{N}}^{n}\sum_{l^{1}=0}^{N-1}\sum_{m^{2}=l^{1}}^{N-1}\int_{J_{l^{1}}\times
J_{m^{2}}}\frac{\left|{\mathscr{M}_{m}a}\right|^{2/3}}{m}d\mu$
$\displaystyle=:$ $\displaystyle I_{4,1}+I_{4,2}.$

Let consider $I_{4,2}$ (Analogously we can estimate $I_{4,1}$). For
$(x^{1},x^{2})\in J_{l^{1}}\times J_{m^{2}}$ if we apply Lemma 4.41 we obtain
that

$\displaystyle{|{\mathscr{M}}_{n}a(x^{1},x^{2})|}$
$\displaystyle\leq\|a\|_{\infty}\int_{I_{N}\times
I_{N}}|K_{n}(x^{1}+t^{1},x^{2}+t^{2})|d\mu(t^{1},t^{2})$ $\displaystyle\leq$
$\displaystyle\frac{2^{N+l^{1}-m^{2}}}{n}\sum_{r^{1}=l^{1}+1}^{m^{2}+1}2^{r^{1}}D_{2^{m^{2}+1}}(x^{1}+e_{l^{1}}+e_{r^{1}})\sum_{s=m^{2}+1}^{N}D_{2^{s}}(x^{2}+e_{m^{2}}+x_{m^{2}+1,s-1}^{1})$
$\displaystyle+\frac{2^{N+l^{1}+m^{2}}}{n}\sum_{s=l^{1}}^{m^{2}}\sum_{r^{1}=l^{1}+1}^{s}D_{2^{s}}(x^{1}+e_{l^{1}}+e_{r^{1}}).$

It is evident that

$\displaystyle\int_{J_{l^{1}}\times
J_{m^{2}}}D_{2^{m^{2}+1}}^{2/3}(x^{1}+e_{l^{1}}+e_{r^{1}})D_{2^{s}}^{2/3}(x^{2}+e_{m^{2}}+x_{m^{2}+1,s-1}^{1})d\mu(x^{1},x^{2})$
$\displaystyle\leq c2^{2s/3-m^{2}/3-l^{1}}\leq c2^{-\left(m^{2}+s\right)/3}$

and

$\displaystyle\int_{J_{l^{1}}\times
J_{m^{2}}}D_{2^{s}}^{2/3}(x^{1}+e_{l^{1}}+e_{r^{1}})d\mu(x^{1},x^{2})\leq
c2^{2s/3-m^{2}-l^{1}}\leq c2^{-m^{2}-s/3}.$

Hence,

$\displaystyle\int_{J_{l^{1}}\times
J_{m^{2}}}\left|{\mathscr{M}_{m}a}\right|^{2/3}d\mu$
$\displaystyle\leq\frac{c2^{2\left(N+l^{1}-m^{2}\right)/3}}{m^{2/3}}\sum_{r^{1}=l^{1}+1}^{m^{2}+1}\sum_{s=m^{2}+1}^{N}2^{2r^{1}/3}\times$
$\displaystyle\times\int_{J_{l^{1}}\times
J_{m^{2}}}D_{2^{m^{2}+1}}^{2/3}(x^{1}+e_{l^{1}}+e_{r^{1}})D_{2^{s}}^{2/3}(x^{2}+e_{m^{2}}+x_{m^{2}+1,s-1}^{1})d\mu\left(x^{1},x^{2}\right)$
$\displaystyle+\frac{c2^{2\left(N+l^{1}+m^{2}\right)/3}}{m^{2/3}}\sum_{s=l^{1}}^{m^{2}}\sum_{r^{1}=l^{1}+1}^{s}\int_{J_{l^{1}}\times
J_{m^{2}}}D_{2^{s}}^{2/3}(x^{1}+e_{l^{1}}+e_{r^{1}})d\mu(x^{1},x^{2})$
$\displaystyle\leq$
$\displaystyle\frac{c2^{2(N+l^{1}-m^{2})/3}}{m^{2/3}}\sum_{r^{1}=l^{1}+1}^{m^{2}+1}2^{2r^{1}/3}\sum_{s=m^{2}+1}^{N}2^{-\left(m^{2}+s\right)/3}+\frac{c2^{2(N+l^{1}+m^{2})/3}}{m^{2/3}}\sum_{s=l^{1}}^{m^{2}}\sum_{r^{1}=l^{1}+1}^{s}2^{-m^{2}-s/3}$
$\displaystyle\leq$
$\displaystyle\frac{c2^{2(N+l^{1}-m^{2})/3}}{m^{2/3}}2^{-2m^{2}/3}\sum_{r^{1}=l^{1}+1}^{m^{2}+1}2^{2r^{1}/3}+\frac{c2^{2\left(N+l^{1}+m^{2}\right)/3}}{m^{2/3}}\sum_{s=l_{1}}^{m^{2}}\left(s-l_{1}-1\right)2^{-m^{2}-s/3}$
$\displaystyle\leq$
$\displaystyle\frac{c2^{2\left(N+l^{1}-m^{2}\right)/3}}{m^{2/3}}+\frac{c2^{\left(2N+l^{1}-m^{2}\right)/3}}{m^{2/3}}.$

and

$\begin{split}I_{4,2}&\leq\frac{c}{\log
n}\sum_{m=2^{N}}^{n}\frac{1}{m}\sum_{l^{1}=0}^{N-1}\sum_{m^{2}=l^{1}}^{N-1}\frac{2^{2(N+l^{1}-m^{2})/3}+2^{\left(2N-m^{2}+l^{1}\right)/3}}{m^{2/3}}\\\
&\leq\frac{c2^{2N/3}}{\log
n}\sum_{m=2^{N}}^{n}\frac{1}{m^{5/3}}\sum_{l^{1}=0}^{N-1}1\leq c.\end{split}$

###### Corollary 4.51

Let $f\in H_{2/3}\left(G^{2}\right).$ Then

$\lim_{n\rightarrow\infty}\frac{1}{\log
n}\sum_{m=1}^{n}\frac{\left\|\mathscr{M}_{m}f-f\right\|_{H_{2/3}\left(G^{2}\right)}^{2/3}}{m}=0$

and

$\lim_{n\rightarrow\infty}\frac{1}{\log
n}\sum_{m=1}^{n}\frac{\left\|\mathscr{M}_{m}f\right\|_{H_{2/3}\left(G^{2}\right)}^{2/3}}{m}=\left\|f\right\|_{H_{2/3}\left(G^{2}\right)}^{2/3}.$

###  4.5 Modulus of continuity and convergence in norm of Marcinkiewicz means
with respect to the two-dimensional Walsh-Fourier series on the martingale
Hardy spaces

In this section we investigate necessary and sufficient conditions for modulus
of continuity, which provide convergence in norm of Marcinkiewicz means with
respect to the two-dimensional Walsh-Fourier series in $H_{2/3}$-norm (For
details see [37]).

###### Theorem 4.52

a) Let $f\in H_{2/3}(G^{2})$ and

$\omega_{H_{2/3}(G^{2})}\left(\frac{1}{2^{k}},f\right)=o\left(\frac{1}{k^{3/2}}\right),\text{\quad
as \quad}k\rightarrow\infty.$ (4.20)

Then

$\left\|\mathscr{M}_{n}f-f\right\|_{H_{2/3}(G^{2})}\rightarrow 0,\text{ \quad
as \quad}n\rightarrow\infty.$

b) There exists a martingale $f\in H_{2/3}(G^{2}),$ such that

$\omega_{H_{2/3}(G^{2})}\left(\frac{1}{2^{2^{k}}},f\right)=O\left(\frac{1}{2^{3k/2}}\right),\text{
\quad as \quad}k\rightarrow\infty$

and

$\left\|\mathscr{M}_{n}f-f\right\|_{2/3}\nrightarrow 0\text{ \quad as
\quad}n\rightarrow\infty.$

Proof: In [34] it was proved that (see inequality (1.23)) the following
inequality is true:

$\left\|\mathscr{M}_{n}f\right\|_{2/3}\leq
c\log^{2/3}\left(n+1\right)\left\|f\right\|_{H_{2/3}(G^{2})}.$ (4.21)

If we apply inequality (1.22) and Lemma 4.48 according to (4.21) we get the
following estimation

$\displaystyle\left\|\mathscr{M}_{n}f\right\|_{H_{2/3}(G^{2})}^{2/3}$ (4.22)
$\displaystyle\leq\left\|\mathscr{M}_{n}f\right\|_{2/3}^{2/3}+\left\|\widetilde{\mathscr{M}}_{\\#}^{\ast}f\right\|_{2/3}^{2/3}+\left\|\widetilde{S}_{\\#}^{\ast}f\right\|_{2/3}^{2/3}$
$\displaystyle\leq
c\log\left(n+1\right)\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}+c\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle\leq
c\log\left(n+1\right)\left\|f\right\|_{H_{2/3}(G^{2})}^{2/3}.$

Let $2^{N}<n\leq 2^{N+1}.$ If we use (4.22) by simple calculations we have
that

$\displaystyle\left\|\mathscr{M}_{n}f-f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle\leq\left\|\mathscr{M}_{n}f-\mathscr{M}_{n}S_{2^{N},2^{N}}f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle+\left\|\mathscr{M}_{n}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle+\left\|S_{2^{N},2^{N}}f-f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle=\left\|\mathscr{M}_{n}\left(S_{2^{N},2^{N}}f-f\right)\right\|_{H_{2/3}}^{2/3}$
$\displaystyle+\left\|\mathscr{M}_{n}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle+\left\|S_{2^{N},2^{N}}f-f\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle\leq
c\left(\log\left(n+1\right)+1\right)\omega_{H_{2/3}(G^{2})}^{2/3}\left(\frac{1}{2^{N}},f\right)$
$\displaystyle+\left\|\mathscr{M}_{n}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f\right\|_{H_{2/3}(G^{2})}^{2/3}.$

Let $2^{N}<n\leq 2^{N+1}.$ Then it is evident that

$\displaystyle\mathscr{M}_{n}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f$
$\displaystyle=$
$\displaystyle\frac{1}{n}\sum_{k=0}^{2^{N}}S_{k,k}S_{2^{N},2^{N}}f$
$\displaystyle+$
$\displaystyle\frac{1}{n}\sum_{k=2^{N}+1}^{n}S_{k,k}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f$
$\displaystyle=$ $\displaystyle\frac{1}{n}\sum_{k=0}^{2^{N}}S_{k,k}f$
$\displaystyle+$
$\displaystyle\frac{\left(n-2^{N}\right)S_{2^{N},2^{N}}f}{n}-S_{2^{N},2^{N}}f$
$\displaystyle=$
$\displaystyle\frac{2^{N}}{n}\left(\mathscr{M}_{2^{N}}f-S_{2^{N},2^{N}}f\right)$
$\displaystyle=$
$\displaystyle\frac{2^{N}}{n}\left(S_{2^{N},2^{N}}\mathscr{M}_{2^{N}}f-S_{2^{N},2^{N}}f\right)$
$\displaystyle=$
$\displaystyle\frac{2^{N}}{n}S_{2^{N},2^{N}}\left(\mathscr{M}_{2^{N}}f-f\right).$

By combining (1.19) and (1.21) we get that

$\displaystyle\left\|\mathscr{M}_{n}S_{2^{N},2^{N}}f-S_{2^{N},2^{N}}f\right\|_{H_{2/3}(G^{2})}^{2/3}$
(4.23)
$\displaystyle\leq\left(\frac{2^{N}}{n}\right)^{2/3}\left\|S_{2^{N},2^{N}}\left(\mathscr{M}_{2^{N}}f-f\right)\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle\leq\left\|S_{2^{N},2^{N}}\left(\mathscr{M}_{2^{N}}f-f\right)\right\|_{H_{2/3}(G^{2})}^{2/3}$
$\displaystyle\leq\left\|\mathscr{M}_{2^{N}}f-f\right\|_{H_{2/3}(G^{2})}^{2/3}\rightarrow
0,\text{ where }k\rightarrow\infty,$

Hence, we immediately get that if

$\omega_{H_{2/3}(G^{2})}\left(\frac{1}{2^{n}},f\right)=o\left(\frac{1}{n^{3/2}}\right),\text{
as }n\rightarrow\infty,$

then

$\left\|\mathscr{M}_{n}f-f\right\|_{H_{2/3}(G^{2})}\rightarrow 0,\text{ as
}n\rightarrow\infty.$

Now, prove part b) of Theorem 4.52. Let

$a_{i}(x^{1},x^{2})=2^{2^{i}}\left(D_{2^{2^{i}+1}}(x^{1})-D_{2^{2^{i}}}(x^{1})\right)\left(D_{2^{2^{i}+1}}(x^{2})-D_{2^{2^{i}}}(x^{2})\right)$

and

$f_{n}(x^{1},x^{2})={\sum_{i=1}^{n}}\frac{a_{i}(x^{1},x^{2})}{2^{3i/2}}.$

Since

${\sum_{i=1}^{\infty}}\left(\frac{1}{2^{3i/2}}\right)^{2/3}<c<\infty$
$\displaystyle
S_{2^{n},2^{n}}a_{k}(x^{1},x^{2})=\begin{cases}a_{k}(x^{1},x^{2}),&\text{ \ if
\ }2^{k}\leq n,\\\ 0,&\text{ \ if \ }2^{k}>n\end{cases}$

and

$\displaystyle\text{supp }a_{k}=I_{2^{k}}^{2},$
$\displaystyle\int_{I_{2^{k}}^{2}}a_{k}d\mu=0,$
$\displaystyle\left\|a_{k}\right\|_{\infty}\leq\mu(\text{supp }a_{k})^{-3/2},$

by using Lemma 4.44, we conclude that $f\in H_{2/3}.$

On the other hand, if we apply Remark 4.39 we immediately get that

$\displaystyle f-S_{2^{n},2^{n}}f$ $\displaystyle=$
$\displaystyle\left(f^{\left(1\right)}-S_{2^{n},2^{n}}f^{\left(1\right)},...,f^{\left(n\right)}-S_{2^{n},2^{n}}f^{\left(n\right)},...,f^{\left(n+k\right)}-S_{2^{n},2^{n}}f^{\left(n+k\right)},...\right)$
$\displaystyle=$
$\displaystyle\left(0,...,0,f^{\left(n+1\right)}-f^{\left(n\right)},...,f^{\left(n+k\right)}-f^{\left(n\right)},...\right)$
$\displaystyle=$ $\displaystyle\left(0,...,0,...,\sum_{i=\log n+1}^{\log
n+k}\frac{a_{i}(x)}{2^{3i/2}},...\right),\text{ \ }k\in\mathbb{N}_{+},$

Hence,

$\displaystyle\omega_{H_{2/3}(G^{2})}\left(\frac{1}{2^{n}},f\right)\leq\sum_{i=\left[\log
n\right]}^{\infty}\frac{1}{2^{3i/2}}=O\left(\frac{1}{n^{3/2}}\right)\text{
\quad as \quad}k\rightarrow\infty.$

where $\left[\log n\right]$ denotes integer part of $\log n$.

Set

$\displaystyle n_{2^{A-2}}$ $\displaystyle=$ $\displaystyle 2^{4\cdot
2^{A-2}}+2^{4\cdot 2^{A-2}-4}+...+2^{4}+2^{0}$ $\displaystyle=$ $\displaystyle
2^{2^{A}}+2^{2^{A}-4}+...+2^{4}+2^{0}.$

If we use Lemma 4.43 we get that

$\displaystyle\mathscr{M}_{n_{2^{k-2}}}f-f$ $\displaystyle=$
$\displaystyle\frac{2^{2^{k}}\mathscr{M}_{2^{2^{k}}}f}{n_{2^{k-2}}}+\frac{1}{n_{2^{k-2}}}\sum_{j=2^{2^{k}}+1}^{n_{2^{k-2}}}S_{j,j}f$
$\displaystyle-$
$\displaystyle\frac{2^{2^{k}}f}{n_{2^{k-2}}}-\frac{n_{2^{k-2}-1}f}{n_{2^{k-2}}}.$

It is evident that

$\widehat{f}(i,j)=\begin{cases}\frac{2^{2^{k}}}{2^{3k/2}},&\text{ თუ
}\left(i,j\right)\in\left\\{2^{2^{k}},...,2^{2^{k}+1}-1\right\\}^{2},\
k\in\mathbb{N}\\\ 0,&\text{ თუ
}\left(i,j\right)\notin\bigcup\limits_{k=0}^{\infty}\left\\{2^{2^{k}},...,2^{2^{k}+1}-1\right\\}^{2}.\end{cases}$
(4.25)

Let $2^{2^{k}}<j\leq n_{2^{k-1}}.$ Since $w_{v+2^{2^{k}}}=w_{2^{2^{k}}}w_{v},$
when $v<2^{2^{k}}$, if we apply (4.25) and first equality of Lemma 2.2 we
obtain that

$\displaystyle S_{j,j}f\left(x^{1},x^{2}\right)$ $\displaystyle=$
$\displaystyle S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)$
$\displaystyle+$
$\displaystyle\sum_{v=2^{2^{k}}}^{j-1}\sum_{s=2^{2^{k}}}^{j-1}\widehat{f}(v,s)w_{v,s}\left(x^{1},x^{2}\right)$
$\displaystyle=$ $\displaystyle
S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)$ $\displaystyle+$
$\displaystyle\frac{2^{2^{k}}}{2^{3k/2}}\sum_{v=0}^{j-2^{2^{k}}-1}\sum_{s=0}^{j-2^{2^{k}}-1}w_{v+2^{2^{k}}}\left(x^{1}\right)w_{s+2^{2^{k}}}\left(x^{2}\right)$
$\displaystyle=$ $\displaystyle
S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)$ $\displaystyle+$
$\displaystyle\frac{2^{2^{k}}w_{2^{2^{k}}}\left(x^{1}\right)w_{2^{2^{k}}}\left(x^{2}\right)}{2^{3k/2}}\sum_{v=0}^{j-2^{2^{k}}-1}\sum_{s=0}^{j-2^{2^{k}}-1}w_{v}\left(x^{1}\right)w_{s}\left(x^{2}\right)$
$\displaystyle=$ $\displaystyle
S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)$ $\displaystyle+$
$\displaystyle\frac{2^{2^{k}}w_{2^{2^{k}}}\left(x^{1}\right)w_{2^{2^{k}}}\left(x^{2}\right)D_{{}_{j-2^{2^{k}},j-2^{2^{k}}}}\left(x^{1},x^{2}\right)}{2^{3k/2}}.$

Hence,

$\displaystyle\frac{1}{n_{2^{k-2}}}\sum_{j=2^{2^{k}}+1}^{n_{2^{k-2}}}S_{j,j}f\left(x^{1},x^{2}\right)$
$\displaystyle=$
$\displaystyle\frac{n_{2^{k-2}-1}S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)}{n_{2^{k-2}}}$
$\displaystyle+\frac{2^{2^{k}}w_{2^{2^{k}}}\left(x^{1}\right)w_{2^{2^{k}}}\left(x^{2}\right)}{n_{2^{k-2}}2^{3k/2}}\sum_{j=1}^{n_{2^{k-2}-1}}D_{{}_{j,j}}\left(x^{1},x^{2}\right)$
$\displaystyle=$
$\displaystyle\frac{n_{2^{k-2}-1}S_{2^{2^{k}},2^{2^{k}}}f\left(x^{1},x^{2}\right)}{n_{2^{k-2}}}$
$\displaystyle+\frac{2^{2^{k}}w_{2^{2^{k}}}\left(x^{1}\right)w_{2^{2^{k}}}\left(x^{2}\right)n_{2^{k-2}-1}K_{n_{2^{k-2}-1}}\left(x^{1},x^{2}\right)}{n_{2^{k-1}}2^{3k/2}}.$

By applying (4.5) we have that

$\displaystyle\|\mathscr{M}_{n_{2^{k-2}}}f-f\|_{2/3}^{2/3}$ (4.26)
$\displaystyle\geq\frac{c}{2^{k}}\|{n_{2^{k-2}-1}}K_{n_{2^{k-2}-1}}\|_{2/3}^{2/3}$
$\displaystyle-\left(\frac{2^{2^{k}}}{n_{2^{k-2}}}\right)^{2/3}\|\mathscr{M}_{2^{2^{k}}}f-f\|_{2/3}^{2/3}$
$\displaystyle-\left(\frac{n_{2^{k-2}-1}}{n_{2^{k-2}}}\right)^{2/3}\|S_{2^{2^{k}},2^{2^{k}}}f-f\|_{2/3}^{2/3}$
$\displaystyle\geq\frac{c}{2^{k}}\|{n_{2^{k-2}-1}}K_{n_{2^{k-2}-1}}\|_{2/3}^{2/3}$
$\displaystyle-\|\mathscr{M}_{2^{2^{k}}}f-f\|_{2/3}^{2/3}$
$\displaystyle-\|S_{2^{2^{k}},2^{2^{k}}}f-f\|_{2/3}^{2/3}.$

Set

$\displaystyle x^{1}\in I_{2^{k-2}}^{m,l}$ $\displaystyle=:$ $\displaystyle
I_{2^{k-2}}\left(0,...,0,x_{4m}^{1}=1,0,...,0,x_{4l}^{1}=1,x_{4l+1}^{1},...,x_{2^{k-2}-1}^{1}\right)$

and

$\displaystyle x^{2}\in J_{2^{k-2}}^{l,q}$ $\displaystyle=:$ $\displaystyle
I_{2^{k-2}}\left(0,...,0,x_{4l}^{2}=1,x_{4l+1}^{1}...,x_{4q-1}^{1},1-x_{4q}^{1},x_{4q+1}^{2},...,x_{2^{k-2}-1}^{2}\right).$

According to Lemma 4.43 we get that

$n_{{}_{2^{k-2}-1}}\left|K_{n_{2^{k-2}-1}}\left(x^{1},x^{2}\right)\right|\geq
2^{4q+4l+4m-3}.$

Hence,

$\displaystyle\int_{G}(n_{{}_{2^{k-2}-1}}|K_{n_{2^{k-2}-1}}(x^{1},x^{2})|)^{2/3}d\mu(x^{1},x^{2})$
$\displaystyle\geq
c\sum_{m=1}^{2^{k-2}-3}\sum_{l=m+1}^{2^{k-2}-2}\sum_{q=l+1}^{2^{k-2}-1}\sum_{x_{4l+1}^{1}=0}^{1}...\sum_{x_{{}_{2^{k-2}-1}}^{1}=0}^{1}\sum_{x_{4q+1}^{2}=0}^{1}...\sum_{x_{{}_{2^{k-2}-1}}^{2}=0}^{1}$
$\displaystyle\int_{I_{2^{k-2}}^{m,l}\times
J_{2^{k-2}}^{l,q}}({n_{2^{k-2}-1}}|K_{n_{2^{k-2}-1}}(x^{1},x^{2})|)^{2/3}d\mu(x^{1},x^{2})$
$\displaystyle\geq
c\sum_{m=1}^{2^{k-2}-3}\sum_{l=m+1}^{2^{k-2}-2}\sum_{q=l+1}^{2^{k-2}-1}\sum_{x_{4l+1}^{1}=0}^{1}...\sum_{x_{{}_{2^{k-2}-1}}^{1}=0}^{1}\sum_{x_{4q+1}^{2}=0}^{1}...\sum_{x_{{}_{2^{k-2}-1}}^{2}=0}^{1}$
$\displaystyle\mu\left(I_{2^{k-2}}^{m,l}\times
J_{2^{k-2}}^{l,q}\right)2^{\left(8q+8l+8m\right)/3}$ $\displaystyle\geq
c\sum_{m=1}^{2^{k-2}-3}\sum_{l=m+1}^{2^{k-2}-2}\sum_{q=l+1}^{2^{k-2}-1}2^{\left(8q+8l+8m\right)/3}2^{2^{k-2}-4l}2^{2^{k-2}-4q}\left(\frac{1}{2^{2^{k-2}}}\right)^{2}$
$\displaystyle\geq
c\sum_{m=1}^{2^{k-2}-3}2^{8m/3}\sum_{l=m+1}^{2^{k-2}-2}2^{-4l/3}\sum_{q=l+1}^{2^{k-2}-1}2^{-4q/3}$
$\displaystyle\geq c\sum_{m=1}^{2^{k-2}-3}1\geq c2^{k}.$

By combining (1.17), (1.21) and (4.26) we can conclude that

${\limsup_{k\rightarrow\infty}}\|\mathscr{M}_{n_{2^{k-2}}}f-f\|_{2/3}\geq
c>0.$

Theorem 4.52 is proved.

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