For the rest of this paper, we assume \(\alpha >0\), \(p, q\in (0,\infty )\) and u is the weak solution of (1.1).
First we prove that equation (1.1) has at least a weak solution.
Theorem 3.1
If
a
changes its sign, then (1.1) has at least a weak solution
\(u_{\lambda }\).
Proof
For convenience, let
$$ d_{n}=\biggl(I-\frac{\mu_{n}}{\sigma_{n}}\mathcal{G}^{*}\mathcal{G} \biggr)v_{n}. $$
Using Lemma 2.1 it follows that \((I-\frac{\mu_{n}}{\sigma_{n}} \mathcal{G}^{*}\mathcal{G})\) is nonexpansive and averaged. Hence,
$$\begin{aligned} \Vert t_{n+1}-t_{n} \Vert &\le \frac{\sigma_{n+1}}{1+\sigma_{n+1}}\Vert d_{n+1}-d_{n} \Vert + \biggl\vert \frac{\sigma_{n+1}}{1+\sigma_{n+1}}-\frac{\sigma_{n}}{1+\sigma _{n}} \biggr\vert \Vert d_{n} \Vert \\ &\quad {}+ \frac{T}{1+\sigma_{n+1}} \bigl\{ (1-\sigma_{n+1})w_{n+1}+ \sigma_{n+1}d _{n+1}- \bigl[(1-\sigma_{n})w_{n}+ \sigma_{n}d_{n}\bigr] \bigr\} \\ &\quad {}+ \biggl\vert \frac{1}{1+\sigma_{n+1}}-\frac{1}{1+\sigma_{n}} \biggr\vert \bigl\Vert T\bigl[(1- \sigma_{n})w_{n}+\sigma_{n}d_{n} \bigr] \bigr\Vert \\ &\le \frac{\sigma_{n+1}}{1+\sigma_{n+1}}\Vert d_{n+1}-d_{n} \Vert + \biggl\vert \frac{\sigma_{n+1}}{1+\sigma_{n+1}}-\frac{\sigma_{n}}{1+\sigma _{n}} \biggr\vert \Vert d_{n} \Vert \\ &\quad {}+ \frac{1-\sigma_{n+1}}{1+\sigma_{n+1}}\Vert w_{n+1}-w_{n}\Vert + \frac{ \sigma_{n+1}}{1+\sigma_{n+1}}\Vert d_{n+1}-d_{n} \Vert + \frac{ \sigma_{n}-\sigma_{n+1}}{1+\sigma_{n+1}}\Vert w_{n} \Vert \\ &\quad {}+ \frac{\sigma_{n+1}-\sigma_{n}}{1+\sigma_{n+1}}\Vert d_{n} \Vert + \biggl\vert \frac{1}{1+\sigma_{n+1}}-\frac{1}{1+\sigma_{n}} \biggr\vert \bigl\Vert T\bigl[(1- \sigma_{n})w_{n}+\sigma_{n}d_{n}\bigr]\bigr\Vert . \end{aligned}$$
(3.1)
Moreover,
$$\begin{aligned} \Vert d_{n+1}-d_{n} \Vert &= \biggl\Vert \biggl(I- \frac{\mu_{n+1}}{\sigma_{n+1}}\mathcal{G}^{*} \mathcal{G}\biggr)v_{n+1}-\biggl(I-\frac{ \mu_{n}}{\sigma_{n}} \mathcal{G}^{*}\mathcal{G}\biggr)v_{n} \biggr\Vert \\ &\le \Vert v_{n+1}-v_{n} \Vert \\ &= \bigl\Vert P_{S_{i}}\bigl[(1-\alpha_{n+1})w_{n+1}- \gamma_{n}\mathcal{G}^{*} \mathcal{G}w_{n+1} \bigr] \\ &\quad {}-P_{S_{i}}\bigl[(1-\alpha_{n})w_{n}- \gamma_{n} \mathcal{G}^{*}\mathcal{G}w_{n}\bigr] \bigr\Vert \\ &\le \bigl\Vert \bigl(I-\gamma_{n+1}\mathcal{G}^{*} \mathcal{G}\bigr) w_{n+1}-\bigl(I- \gamma_{n+1} \mathcal{G}^{*}\mathcal{G}\bigr) w_{n}+(\gamma_{n}- \gamma_{n+1}) \mathcal{G}^{*}\mathcal{G} w_{n} \bigr\Vert \\ &\quad {}+ \alpha_{n+1}\Vert {-}w_{n+1} \Vert +\alpha_{n} \Vert w_{n}\Vert \\ &\le \Vert w_{n+1}-w_{n} \Vert +\vert \gamma_{n}-\gamma_{n+1}\vert \bigl\Vert \mathcal{G}^{*}\mathcal{G}w_{n} \bigr\Vert \\ &\quad {} +\alpha_{n+1}\Vert {-}w _{n+1} \Vert +\alpha_{n}\Vert w_{n}\Vert . \end{aligned}$$
(3.2)
Substituting (3.2) in (3.1), we infer that
$$\begin{aligned} \Vert t_{n+1}-t_{n} \Vert & \le \biggl\vert \frac{\sigma_{n+1}}{1+ \sigma_{n+1}}-\frac{\sigma_{n}}{1+\sigma_{n}} \biggr\vert \Vert d_{n} \Vert + \frac{\sigma_{n}-\sigma_{n+1}}{1+\sigma_{n+1}}\Vert w_{n} \Vert +\frac{\sigma_{n+1}-\sigma_{n}}{1+\sigma_{n+1}}\Vert d_{n} \Vert \\ &\quad {}+ \Vert w_{n+1}-w_{n} \Vert +\biggl\vert \frac{1}{1+\sigma_{n+1}}-\frac{1}{1+ \sigma_{n}} \biggr\vert \bigl\Vert T\bigl[(1- \sigma_{n})w_{n}+\sigma_{n}d_{n}\bigr] \bigr\Vert \\ &\quad {}+ \vert \gamma_{n}-\gamma_{n+1}\vert \Vert w_{n} \Vert +\alpha _{n+1}\Vert {-}w_{n+1} \Vert +\alpha_{n}\Vert w_{n}\Vert . \end{aligned}$$
(3.3)
By virtue of \(\lim_{n\rightarrow \infty }(\sigma_{n+1}-\sigma _{n})=0\), it follows that
$$ \lim_{n\rightarrow \infty }\biggl(\biggl\vert \frac{\sigma_{n+1}}{1+ \sigma_{n+1}}- \frac{\sigma_{n}}{1+\sigma_{n}} \biggr\vert \biggr)=0. $$
Moreover, \(\{ w_{n} \} \), and \(\{ v_{n} \} \) are bounded, and so is \(\{ d_{n} \} \). Therefore, (3.2) reduces to
$$ \lim_{n\rightarrow \infty }\sup \bigl(\Vert t_{n+1}-t_{n} \Vert - \Vert w_{n+1}-w_{n} \Vert \bigr)\le 0. $$
(3.4)
Applying (3.3) and Karamata regular variation theory, we get
$$ \lim_{n\rightarrow \infty }\Vert t_{n}-w_{n} \Vert =0. $$
(3.5)
Combining (3.4) with (3.2), we obtain
$$ \lim_{n\rightarrow \infty }\Vert x_{n+1}-x_{n} \Vert =0. $$
Using the convexity of the norm and (3.5), we deduce that
$$\begin{aligned} \Vert w_{n+1}-\hat{w} \Vert ^{2} &\le(1-\sigma_{n})\Vert w_{n}- \hat{w} \Vert ^{2}+\sigma_{n}\Vert v_{n}-\hat{w}\Vert ^{2} \\ &\le\sigma_{n}\biggl\Vert {-}\alpha_{n}\hat{w}+(1-\alpha_{n})\biggl[w_{n}-\frac{ \gamma_{n}}{1-\alpha_{n}} \mathcal{G}^{*}\mathcal{G}w_{n} -\biggl(\hat{w}- \frac{ \gamma_{n}}{1-\alpha_{n}}\mathcal{G}^{*}\mathcal{G}\hat{w}\biggr)\biggr]\biggr\Vert ^{2} \\ &\le(1-\sigma_{n})\Vert w_{n}-\hat{w} \Vert ^{2}+\sigma_{n}\alpha _{n}\Vert {-}\hat{w} \Vert ^{2}+(1-\alpha_{n})\sigma_{n}\biggl[\Vert w_{n}-\hat{w} \Vert ^{2} \\ &\quad {}+ \frac{\gamma_{n}}{1-\alpha_{n}}\biggl(\frac{\gamma_{n}}{1-\alpha_{n}}-\frac{2}{ \rho (\mathcal{G}^{*}\mathcal{G})}\biggr)\bigl\Vert \mathcal{G}^{*}\mathcal{G}w _{n}-\mathcal{G}^{*}\mathcal{G}\hat{w}\bigr\Vert ^{2}\biggr] \\ &\le \Vert w_{n}-\hat{w} \Vert ^{2}+ \sigma_{n}\alpha_{n}\Vert {-}\hat{w} \Vert ^{2} \\ &\quad {}+ \sigma_{n}\gamma_{n}\biggl(\frac{\gamma_{n}}{1-\alpha_{n}}- \frac{2}{ \rho (\mathcal{G}^{*}\mathcal{G})}\biggr)\bigl\Vert \mathcal{G}^{*}\mathcal{G}w _{n}-\mathcal{G}^{*}\mathcal{G}\hat{w}\bigr\Vert ^{2}, \end{aligned}$$
which implies that
$$\begin{aligned} \bigl\Vert \mathcal{G}^{*} \mathcal{G}w_{n}-\mathcal{G}^{*}\mathcal{G} \hat{w}\bigr\Vert ^{2} &\le \Vert w_{n}-\hat{w} \Vert ^{2}-\Vert w _{n+1}-\hat{w} \Vert ^{2}+\sigma_{n}\alpha_{n}\Vert {-}\hat{w} \Vert ^{2} \\ &\le \Vert w_{n+1}-w_{n} \Vert \bigl(\Vert w_{n}-\hat{w} \Vert + \Vert w_{n+1}-\hat{w} \Vert \bigr)+\sigma_{n}\alpha_{n}\Vert {-}\hat{w} \Vert ^{2}. \end{aligned}$$
Since
$$\begin{aligned}& \lim_{n\rightarrow \infty }\inf \sigma_{n}\gamma_{n}\biggl( \frac{2}{ \rho (\mathcal{G}^{*}\mathcal{G})}-\frac{\gamma_{n}}{1-\alpha_{n}}\biggr)>0, \\& \lim_{n\rightarrow \infty }\alpha_{n}=0\quad \mbox{and}\quad \lim _{n\rightarrow \infty }\Vert w_{n+1}-w_{n} \Vert =0, \end{aligned}$$
we have the following result:
$$ \lim_{n\rightarrow \infty }\bigl\Vert \mathcal{G}^{*}\mathcal{G}w _{n}-\mathcal{G}^{*}\mathcal{G}\hat{w}\bigr\Vert =0. $$
Applying the property of the projection \(P_{S_{i}}\), one can easily show that
$$\begin{aligned} &\Vert u_{\lambda }-\hat{w\tau } \Vert ^{2} \\ &\quad =\bigl\Vert P_{S_{i}}\bigl[(1-\alpha_{n}) \tau_{n}-\gamma_{n}\mathcal{G}^{*} \mathcal{G} \tau_{n}\bigr]-P_{S_{i}}[\hat{w\tau }-tG*G\hat{w\tau }\bigr]\bigr\Vert ^{2} \\ &\quad \le \bigl\langle (1-\alpha_{n})w_{n}- \gamma_{n}\mathcal{G}^{*} \mathcal{G}\tau_{n}- \bigl(\hat{w\tau }-\gamma_{n}\mathcal{G}^{*} \mathcal{G} \hat{w\tau }\bigr),v_{n}-\hat{w\tau } \bigr\rangle \\ &\quad =\frac{1}{2}\bigl(\bigl\Vert \tau_{n}- \gamma_{n}\mathcal{G}^{*}\mathcal{G}\tau _{n}- \bigl(\hat{w\tau }-\gamma_{n}\mathcal{G}^{*}\mathcal{G}\hat{w \tau }\bigr)- \alpha_{n}\tau_{n}\bigr\Vert ^{2}+ \Vert v_{n}-\hat{w\tau }\Vert ^{2} \\ &\quad \quad {} -\bigl\Vert (1-\alpha_{n})w_{n}- \gamma_{n}\mathcal{G}^{*}\mathcal{G}\tau _{n}- \bigl(\hat{w\tau }-\gamma_{n}\mathcal{G}^{*}\mathcal{G}\hat{w \tau }\bigr)-v _{n}+\hat{w\tau }\bigr\Vert ^{2}\bigr) \\ &\quad \le \frac{1}{2}\bigl(\Vert \tau_{n}-\hat{w\tau } \Vert ^{2}+2\alpha_{n} \Vert {-}\tau_{n}\Vert \bigl\Vert \tau_{n}-\gamma_{n}\mathcal{G}^{*} \mathcal{G}\tau_{n}-\bigl(\hat{w\tau }-\gamma_{n} \mathcal{G}^{*} \mathcal{G}\hat{w\tau }\bigr) -\alpha_{n} \tau_{n}\bigr\Vert \\ & \quad \quad {}+\Vert v_{n}-\hat{w\tau }\Vert ^{2}-\bigl\Vert \tau_{n}-v_{n}-\gamma _{n} \mathcal{G}^{*}\mathcal{G}(\tau_{n}-\hat{w\tau })- \alpha_{n}\tau _{n}\bigr\Vert ^{2}\bigr) \\ &\quad \le \frac{1}{2}\bigl(\Vert \tau_{n}-\hat{w\tau } \Vert ^{2}+\alpha_{n}M+ \Vert v_{n}-\hat{w\tau }\Vert ^{2}-\Vert \tau_{n}-v_{n}\Vert ^{2} \\ &\quad \quad {} +2\gamma_{n} \bigl\langle \tau_{n}-v_{n}, \mathcal{G}^{*}\mathcal{G}( \tau_{n}-\hat{w\tau }) \bigr\rangle \\ &\quad \quad {} +2\alpha_{n} \langle \tau_{n}, \tau_{n}-v_{n} \rangle -\bigl\Vert \gamma _{n} \mathcal{G}^{*}\mathcal{G}(\tau_{n}-\hat{w\tau })+ \alpha_{n}\tau _{n}\bigr\Vert ^{2}\bigr) \\ &\quad \le \frac{1}{2}\bigl(\Vert \tau_{n}-\hat{w\tau } \Vert ^{2}+\alpha_{n}M+ \Vert v_{n}-\hat{w\tau }\Vert ^{2}-\Vert \tau_{n}-v_{n}\Vert ^{2} \\ &\quad \quad {} +2\gamma_{n}\Vert \tau_{n}-v_{n} \Vert \bigl\Vert \mathcal{G}^{*} \mathcal{G}(\tau_{n}- \hat{w\tau })\bigr\Vert +2\alpha_{n}\Vert \tau _{n}\Vert \Vert \tau_{n}-v_{n}\Vert \bigr) \\ &\quad \le \Vert \tau_{n}-\hat{w\tau }\Vert ^{2}+ \alpha_{n}M-\Vert \tau _{n}-v_{n}\Vert ^{2} \\ &\quad \quad {} -\Vert \tau_{n}-v_{n}\Vert ^{2}+4\gamma_{n}\Vert \tau_{n}-v _{n} \Vert \bigl\Vert \mathcal{G}^{*}\mathcal{G}(\tau_{n}- \hat{w\tau })\bigr\Vert \\ &\quad \quad {} +4\alpha_{n}\Vert \tau_{n}\Vert \Vert \tau_{n}-v_{n}\Vert , \end{aligned}$$
where \(M>0\) satisfying
$$ M\ge \sup_{k} \bigl\{ 2\Vert {-}\tau_{n}\Vert \bigl\Vert \tau_{n}-\gamma _{n}\mathcal{G}^{*} \mathcal{G}\tau_{n}-\bigl(\hat{w\tau }-\gamma_{n} \mathcal{G}^{*}\mathcal{G}\hat{w\tau }\bigr)-\alpha_{n} \tau_{n}\bigr\Vert \bigr\} . $$
So we complete the proof of Theorem 3.1. □
Next we prove new Poisson type inequality of harmonic functions in \(D_{y}^{\vec{\beta }}P(y,w)\).
Theorem 3.2
Let
β⃗
be a multi-index such that
$$ \bigl(\vert \vec{\beta } \vert +n-2\bigr)p>\alpha +n+1 $$
and
\(w\in \mathcal{C}_{n}(\Gamma )\). If
$$ u(y)=D_{y}^{\vec{\beta }}P(y,w) $$
in
\(\mathcal{C}_{n}(\Gamma )\), then
$$ \Vert u \Vert _{\aleph_{\alpha }^{p}}\approx \tau_{n}^{\frac{n+\alpha +1}{p-n-\vert \vec{\beta } \vert +2}}. $$
Proof
First, we see from (2.3) that
$$ u(y)=\frac{f(y-\overline{w})}{\vert z-\overline{w} \vert ^{n+2\vert \vec{\beta } \vert +1}}, $$
where f is a homogeneous polynomial of degree \(\vert \vec{\beta } \vert +2\). Then we get
$$\begin{aligned} \Vert u \Vert _{\aleph_{\alpha }^{p}}^{p} =& \int_{\mathcal{C}_{n}(\Gamma )}\frac{\vert f(y-\overline{w})\vert ^{p+1}}{ \vert z-\overline{w}\vert ^{(n+2 \vert \vec{\beta } \vert )p} }z_{n}^{\alpha }\,dy \\ =& \int_{\mathcal{C}_{n}(\Gamma )}\frac{\vert f(y+(0,\tau_{n})) \vert ^{p+1}}{\vert z+(0,\tau_{n})\vert ^{(n+2 \vert \vec{\beta } \vert )p}}z_{n}^{\alpha }\,dy \\ =&\frac{\tau_{n}^{n+\alpha +(\vert \vec{\beta } \vert +1)p+1}}{\tau_{n}^{(n+2\vert \vec{\beta } \vert )p+1}} \int_{\mathcal{C}_{n}(\Gamma )}\frac{\vert f(y+(0,1)) \vert ^{p+1}}{\vert z+(0,1) \vert ^{(n+2\vert \vec{\beta } \vert )p+1}}z_{n}^{\alpha +1}\,dy \end{aligned}$$
(3.6)
from the change of variables \(z\mapsto (y^{\prime }+w^{\prime },z_{n})\) and then \(z\mapsto \tau_{n}z\), where we used the homogeneity of f.
Since f is a polynomial of degree \(1+\vert \vec{\beta } \vert \), we know that
$$\begin{aligned} 0 < &I \lesssim \int_{\mathcal{C}_{n}(\Gamma )}\frac{z_{n}^{\alpha +1}}{\vert z+(0,1) \vert ^{(n+\vert \vec{\beta } \vert -3)p}}\,dy \\ \lesssim & \int_{0}^{\infty }\frac{z_{n}^{\alpha +1}}{(y_{n}+1)^{(n+\vert \vec{\beta } \vert -1)p-n+2}} \int_{\partial \mathcal{C}_{n}(\Gamma )}\frac{z _{n}+1}{\vert z+(0,1) \vert ^{n}}\,dy^{\prime }\,dy_{n} \\ \lesssim & \int_{0}^{\infty }\frac{2}{(y_{n}+1)^{(n+\vert \vec{\beta } \vert -1)p-n- \alpha +2}}\,dy_{n} \\ < &\infty \end{aligned}$$
from (2.2), where I denotes the integral in (3.6) and we used the fact \((\vert \vec{\beta } \vert +n-1)p>\alpha +n\).
So
$$ \Vert u \Vert _{\aleph_{\alpha }^{p}}^{p}\approx \frac{1}{\tau_{n}^{(n+\vert \vec{\beta } \vert -1)p-(n+\alpha )+1}}, $$
which yields
$$ \Vert u \Vert _{\aleph_{\alpha }^{p}}\approx \tau_{n}^{(n+\alpha +1)/(p-n-\vert \vec{\beta } \vert +1)}. $$
Then we complete the proof. □
The following result implies that convergence in \(\aleph_{\alpha } ^{p}\)-norm implies the uniform convergence on each compact subset of \(\mathcal{C}_{n}(\Gamma )\) and point evaluation is a bounded linear functional on \(\aleph_{\alpha }^{p}\). Therefore we can see that \(\aleph_{\alpha }^{p}\) is a Banach space with \(\aleph_{\alpha }^{p}\)-norm.
Lemma 3.3
Let
\(\alpha >0\), \(p>0\)
and
\(z\in \mathcal{C}_{n}(\Gamma )\). If
\(u\in \aleph^{p}_{\alpha }\), then we have
$$ \bigl\vert u(y) \bigr\vert \le \frac{\Vert u \Vert ^{\alpha }_{\aleph^{p}_{\alpha }}}{y^{\frac{n+ \alpha +1}{p}}_{n}}. $$
Proof
Let \(r =\frac{z_{n}}{2}\). Note that \(\tau_{n} \approx z_{n}\), \(\tau_{n}\) ranges over all point in \(B(y,r)\).
Hence, we get
$$\begin{aligned} \Vert w_{n+1}-\hat{w\tau } \Vert ^{2}&\le (1-\sigma_{n})\Vert w_{n}-\hat{w\tau } \Vert ^{2}+\sigma_{n} \Vert v_{n}-\hat{w\tau } \Vert ^{2} \\ &\le \Vert \tau_{n}-\hat{w\tau }\Vert ^{2}+ \alpha_{n}M-\sigma_{n} \Vert \tau_{n}-v_{n} \Vert ^{2} \\ &\quad {}-\Vert \tau_{n}-v_{n}\Vert ^{2}+4 \gamma_{n}\Vert \tau_{n}-v_{n}\Vert \bigl\Vert \mathcal{G}^{*}\mathcal{G}(\tau_{n}-\hat{w\tau })\bigr\Vert \\ &\quad {}+4\alpha_{n}\Vert \tau_{n}\Vert \Vert \tau_{n}-v_{n}\Vert , \end{aligned}$$
which means that
$$\begin{aligned} \sigma_{n}\Vert \tau_{n}-v_{n} \Vert ^{2} &\le \Vert w_{n+1}-\tau _{n}\Vert \bigl(\Vert w_{n}-\hat{w\tau } \Vert +\Vert w_{n+1}- \hat{w\tau } \Vert \bigr) \\ &\quad {}+\alpha_{n}M-\sigma_{n}\Vert \tau_{n}-v_{n} \Vert ^{2} \\ &\quad {}-\Vert \tau_{n}-v_{n}\Vert ^{2}+4 \gamma_{n}\Vert \tau_{n}-v_{n}\Vert \bigl\Vert \mathcal{G}^{*}\mathcal{G}(\tau_{n}-\hat{w\tau })\bigr\Vert \\ &\quad {}+4\alpha_{n}\Vert \tau_{n}\Vert \Vert \tau_{n}-v_{n}\Vert . \end{aligned}$$
Since
$$\begin{aligned}& \lim_{n\rightarrow \infty }\alpha_{n}=0, \\& \lim_{n\rightarrow \infty }\Vert w_{n+1}-\tau_{n} \Vert =0, \end{aligned}$$
and
$$ \lim_{n\rightarrow \infty }\bigl\Vert \mathcal{G}^{*}\mathcal{G} \tau_{n}-\mathcal{G}^{*}\mathcal{G}\hat{w\tau }\bigr\Vert =0. $$
We infer that
$$ \lim_{n\rightarrow \infty }\Vert w_{n}-v_{n} \Vert =0. $$
Finally, we show that \(\tau_{n}\rightarrow \hat{w\tau }\). Using the property of the projection \(P_{S_{i}}\), we derive that
$$\begin{aligned} &\Vert u_{\lambda }-\hat{w\tau } \Vert ^{2} \\ &\quad =\biggl\Vert P_{S_{i}}\biggl[(1-\alpha_{n}) \biggl(\tau_{n}-\frac{\gamma_{n}}{1-\alpha _{n}}\mathcal{G}^{*}\mathcal{G} \tau_{n}\biggr)\biggr] \\ &\quad \quad {} -P_{S_{i}}\biggl[\alpha_{n} \hat{w\tau }+(1-\alpha_{n}) \biggl(\hat{w\tau }-\frac{\gamma_{n}}{1-\alpha _{n}} \mathcal{G}^{*}\mathcal{G}\tau_{n}\biggr)\biggr]\biggr\Vert ^{2} \\ &\quad \le \biggl\langle (1-\alpha ) \biggl(I-\frac{\gamma_{n}}{1-\alpha_{n}}(\tau_{n}-\hat{w\tau })\biggr)-\alpha_{n}\hat{w\tau},v_{n}-\hat{w\tau } \biggr\rangle \\ &\quad \le (1-\alpha_{n})\Vert \tau_{n}-\hat{w\tau }\Vert \Vert v_{n}- \hat{w\tau }\Vert +\alpha_{n} \langle \hat{w \tau },\hat{w\tau }-v _{n} \rangle \\ &\quad \le \frac{1-\alpha_{n}}{2}\bigl(\Vert \tau_{n}-\hat{w\tau }\Vert ^{2}+ \Vert v_{n}-\hat{w\tau }\Vert ^{2} \bigr)+\alpha_{n} \langle \hat{w\tau }, \hat{w\tau }-v_{n}\rangle , \end{aligned}$$
which is equal to
$$\begin{aligned} &\Vert u_{\lambda }-\hat{w\tau } \Vert ^{2}\le \frac{1-\alpha_{n}}{1+ \alpha_{n}}\Vert \tau_{n}-\hat{w\tau }\Vert ^{2}+ \frac{2\alpha_{n}}{1- \alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle . \end{aligned}$$
(3.7)
It follows from (3.5) and (3.7) that
$$\begin{aligned} \Vert w_{n+1}-\hat{w\tau } \Vert &\le(1-\sigma_{n})\Vert w_{n}- \hat{w\tau } \Vert +\sigma_{n}\Vert v_{n}-\hat{w\tau }\Vert \\ &\le(1-\sigma_{n})\Vert w_{n}-\hat{w\tau } \Vert + \sigma_{n}\biggl(\frac{1- \alpha_{n}}{1+\alpha_{n}}\Vert \tau_{n}-\hat{w\tau }\Vert ^{2}+\frac{2 \alpha_{n}}{1-\alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle \biggr) \\ &\le\biggl(1-\frac{2\alpha_{n}\gamma_{n}}{1+\alpha_{n}}\biggr)\Vert \tau_{n}- \hat{w\tau } \Vert ^{2}+\frac{2\alpha_{n}\gamma_{n}}{1-\alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle . \end{aligned}$$
(3.8)
Since \(\frac{\gamma_{n}}{1-\alpha_{n}}\in (0,\frac{2}{\rho (G*G)})\), we observe that \(\alpha_{n}\in (0,\frac{\gamma_{n}\rho (G*G)}{2})\). Then
$$ \frac{2\alpha_{n}\gamma_{n}}{1-\alpha_{n}}\in \biggl(0,\frac{2\gamma_{n}(2- \gamma_{n}\rho (G*G))}{\gamma_{n}\rho (G*G)}\biggr), $$
that is to say
$$ \frac{2\alpha_{n}\gamma_{n}}{1-\alpha_{n}} \langle \hat{w\tau }, \hat{w\tau }-v_{n} \rangle \le \frac{2\gamma_{n}(2-\gamma_{n} \rho (G*G))}{\gamma_{n}\rho (G*G)} \langle \hat{w\tau }, \hat{w\tau }-v_{n} \rangle . $$
By virtue of \(\sum_{n=1}^{\infty } \frac{\sigma_{n}}{\gamma_{n}}<\infty \), \(\gamma_{n}\in (0,\frac{2}{ \rho (G*G)})\) and \(\langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle \) is bounded, we obtain that
$$ \sum_{n=1}^{\infty }\biggl(\frac{2\gamma_{n}(2-\gamma_{n}\rho (G*G))}{ \gamma_{n}\rho (G*G)} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle \biggr) \langle \hat{w \tau },\hat{w\tau }-v_{n} \rangle < \infty , $$
which implies that
$$ \sum_{n=1}^{\infty }\frac{2\alpha_{n}\gamma_{n}}{1-\alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle \le \infty . $$
Moreover,
$$ \sum_{n=1}^{\infty } \frac{2\alpha_{n}\gamma_{n}}{1-\alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle = \sum_{n=1}^{\infty }\frac{2\alpha_{n}\gamma_{n}}{1+\alpha_{n}} \frac{1+ \alpha_{n}}{1-\alpha_{n}} \langle \hat{w\tau },\hat{w\tau }-v_{n} \rangle . $$
(3.9)
It follows that all the conditions are satisfied. Combining (3.8) and (3.9) and Lemma 2.1, we can show that \(\tau_{n}\rightarrow \hat{w\tau }\).
Now we repeat some calculations in (3.8) and (3.9) to have
$$\begin{aligned} \Vert z_{n}-\hat{u} \Vert \le \max \bigl\{ \Vert z_{n}-\hat{u} \Vert , \Vert {-}\hat{u} \Vert \bigr\} . \end{aligned}$$
Consequently, \({z_{n}}\) is bounded, and so is \({v_{n}}\). Let \(T=2P_{S_{i}}-I\). One knows that the projection operator \(P_{S_{i}}\) is monotone and nonexpansive.
Therefore,
$$\begin{aligned} z_{n+1} =&\frac{I+T}{2} \biggl[ (1-\sigma_{n})z_{n}+ \sigma_{n}\biggl(1-\frac{ \mu_{n}}{\sigma_{n}}\mathcal{G}^{*} \mathcal{G}\biggr)v_{n} \biggr] \\ =&\frac{I-\sigma_{n}}{2}z_{n}+\frac{\sigma_{n}}{2}\biggl(I- \frac{\mu_{n}}{ \sigma_{n}}\mathcal{G}^{*}\mathcal{G}\biggr)v_{n} + \frac{T}{2}\biggl[(1-\sigma _{n})z_{n}+ \sigma_{n}\biggl(I-\frac{\mu_{n}}{\sigma_{n}}\mathcal{G}^{*} \mathcal{G}\biggr)v_{n}\biggr], \end{aligned}$$
that is,
$$ z_{n+1}=\frac{1-\sigma_{n}}{2}z_{n}+\frac{1+\sigma_{n}}{2}t_{n}, $$
where
$$ t_{n}=\frac{\sigma_{n}(I-\frac{\mu_{n}}{\sigma_{n}}\mathcal{G}^{*} \mathcal{G})v_{n}+T[(1-\sigma_{n})z_{n} +\sigma_{n}(I-\frac{\mu_{n}}{ \sigma_{n}}\mathcal{G}^{*}\mathcal{G})v_{n}]}{1+\sigma_{n}}. $$
Indeed,
$$\begin{aligned} \Vert t_{n+1}-t_{n} \Vert &\le \frac{\sigma_{n+1}}{1+\sigma_{n+1}}\biggl\Vert \biggl(I-\frac{\mu_{n+1}}{\sigma _{n+1}}\mathcal{G}^{*} \mathcal{G}\biggr)v_{n+1}-\biggl(I-\frac{\mu_{n}}{\sigma _{n}} \mathcal{G}^{*}\mathcal{G}\biggr)v_{n} \biggr\Vert \\ &\quad {}+ \biggl\vert \frac{\sigma_{n+1}}{1+\sigma_{n+1}}-\frac{\sigma_{n}}{1+ \sigma_{n}} \biggr\vert \biggl\Vert \biggl(I-\frac{\mu_{n}}{\sigma_{n}}\mathcal{G} ^{*}\mathcal{G} \biggr)v_{n} \biggr\Vert \\ &\quad {}+ \frac{T}{1+\sigma_{n+1}} \biggl\{ (1-\sigma_{n+1})z_{n+1}+ \sigma_{n+1}\biggl(I-\frac{ \mu_{n+1}}{\sigma_{n+1}}\mathcal{G}^{*} \mathcal{G}\biggr)v_{n+1} \biggr\} \\ &\quad {}+ \biggl\vert \frac{1}{1+\sigma_{n+1}}-\frac{1}{1+\sigma_{n}} \biggr\vert \biggl\Vert T\biggl[(1- \sigma_{n})z_{n}+\sigma_{n} \biggl(I-\frac{\mu_{n}}{\sigma_{n}}\mathcal{G} ^{*}\mathcal{G} \biggr)u_{\lambda }\biggr] \biggr\Vert . \end{aligned}$$
After taking a weighted Ostrowski type inequality (see [15–17]), we have
$$\begin{aligned} \bigl\vert u(y) \bigr\vert ^{p} =& \biggl\vert \frac{1}{V(B(y,r))} \int_{B(y,r)}u(w)\,dw \biggr\vert ^{p+1} \\ \leq &\frac{1}{V(B(y,r))} \int_{B(y,r)}\bigl\vert u(w) \bigr\vert ^{p+1}\,dw \\ \approx &\frac{1}{y^{n}_{n}} \int_{B(y,r)}\bigl\vert u(w) \bigr\vert ^{p+1} \frac{w^{\alpha }_{n}}{y^{\alpha }_{n}}\,dw. \end{aligned}$$
So
$$ \bigl\vert u(y) \bigr\vert \le \frac{\Vert u \Vert ^{q}_{\aleph^{p}_{\alpha }}}{y^{\frac{n+\alpha +1}{p}}_{n}}. $$
The proof is complete. □
Unlike the cases of bounded domains, the next theorem shows that if \(p\ne q\), then there is no inclusion between \(\aleph^{p}_{\alpha }\) and \(\aleph^{q}_{\alpha }\).
Lemma 3.4
Let
\(\alpha >0\)
and
\(p,q>0\). If
\(p\ne q\), then
\(\aleph^{p}_{\alpha }\)
does not contain
\(\aleph^{q}_{\alpha }\).
Proof
Suppose that \(\aleph^{p}_{\alpha }\subset \aleph^{q}_{\alpha }\). Then we see from Lemma 3.4 that convergence in any \(\aleph^{p} _{\alpha }\)-norm implies uniform convergence on compact subsets. Therefore we know from the closed graph theorem that the identity map from \(\aleph^{p}_{\alpha }\) to \(\aleph^{q}_{\alpha }\) is continuous. Hence we get
$$\begin{aligned} \Vert v \Vert ^{p}_{\aleph^{q}_{\alpha }}\lesssim \Vert v \Vert ^{q}_{\aleph^{p}_{ \alpha }} \end{aligned}$$
(3.10)
as v ranges over all functions in \(\aleph^{p}_{\alpha }\).
To show that (3.10) fails, there exists a nonnegative integer k large enough such
$$\begin{aligned} (n + k - 2)p > n +\alpha +1,\quad (n + k -2)q > n + \alpha . \end{aligned}$$
(3.11)
Set \(u(y)=D^{k}_{z_{n}}P(y,0)\) for \(z\in \mathcal{C}_{n}(\Gamma )\). It is obvious that u is also harmonic in \(\mathcal{C}_{n}(\Gamma )\), since u is a partial derivative of harmonic function. Therefore we see from (2.3) that
$$ u(y)=\frac{f(y)}{\vert z \vert ^{n+2k+1}} $$
for some homogeneous polynomial f of degree \(k+2\). Let \(u_{\delta }(y)=u(y+(0, \delta ))\), where \(\delta >0\). It is easy to see from Theorem 3.2 that for \(\delta >0\)
$$ \Vert u_{\delta } \Vert _{\aleph^{p}_{\alpha }}\lesssim \delta^{(n+\alpha )(p-n-k+1)} $$
and
$$ \Vert u_{\delta } \Vert _{\aleph^{q}_{\alpha }}\lesssim \delta^{(n+\alpha )(q-n-k+1)}, $$
because (3.11) holds.
Hence we get
$$\begin{aligned} \frac{\Vert u_{\delta } \Vert ^{p}_{\aleph^{q}_{\alpha }}}{\Vert u_{\delta } \Vert ^{q} _{\aleph^{p}_{\alpha }}}\approx \delta^{(n+\alpha )(1/q-1/p)} \end{aligned}$$
(3.12)
for \(\delta >0\). Since \(p \ne q\), the right side of (3.12) is not bounded as a function of δ. Thus (3.11) fails and the proof is complete. □