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

Let $f=\left(f_{n},\text{ }n\in\mathbb{N}_{+}\right)\in H_{p}(G)$ be a
martingale from the Example 1, 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 (19) and (122) we get that $f\in H_{p}(G).$ According to (20) we
have that

(125)
$\displaystyle\widehat{f}(j)=\left\\{\begin{array}[]{l}\Phi^{1/2p}\left(2^{\left|\alpha_{k}\right|+1}\right)\text{
\ if \
}j\in\left\\{2^{\left|\alpha_{k}\right|},...,2^{\left|\alpha_{k}\right|+1}-1\right\\},\text{
}k\in\mathbb{N}_{+},\\\ 0\text{ },\text{ \ if \
}j\notin\bigcup\limits_{k=0}^{\infty}\left\\{2^{\left|\alpha_{k}\right|},...,2^{\left|\alpha_{k}\right|+1}-1\right\\}.\text{
}\end{array}\right.$

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

(126)
$\sigma_{{}_{n}}f=\frac{1}{n}\sum_{j=1}^{2^{\left|\alpha_{k}\right|}}S_{j}f+\frac{1}{n}\sum_{j=2^{\left|\alpha_{k}\right|}+1}^{n}S_{j}f=III+IV.$

It is evident that

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

Let $2^{\left|\alpha_{s}\right|}<j\leq 2^{\left|\alpha_{s}\right|+1},$ where
$s=1,2,...,k.$ If we apply (1) 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 (21) we can conclude that

(129)
$S_{j}f=\sum_{\eta=0}^{s}\Phi^{1/2p}\left(2^{\left|\alpha_{\eta}\right|+1}\right)\left(D_{2^{\left|\alpha_{\eta}\right|+1}}-D_{2^{\left|\alpha_{\eta}\right|}}\right).$

Let $x\in I_{2}\left(e_{0}+e_{1}\right).$ Since (see Lemmas 2 and 9)

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

by combining (121) and (127-130) we get that

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

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

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

By combining (121) and (130) we can conclude that

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

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 1 and 8 and
(130) follows that

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

Let $0<p<1/2$ and $n\in\mathbb{A}_{0,2}.$ By combining (126-3.5) we get that

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

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 10 is complete.

######  Theorem 11.

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 1 and Lemma 6 we obtain that

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

It is easy to easy to show that

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

and

(136)
$S_{i}f_{m}=\left\\{\begin{array}[]{l}2^{m}\left(D_{i}-D_{2^{m}}\right),\text{
\ if }i=2^{m}+1,...,2^{m+1}-1,\\\ \text{ }f_{m},\text{ \ if }i\geq 2^{m+1},\\\
0,\text{ \ otherwise.}\end{array}\right.$

Let $0<n<2^{m}.$ By using first equality of Lemma 1 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 12 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 3 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.

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