Proof of Theorem 3.1
By virtue of the fixed point theorem with respect to the stationary Schrödinger operator in [8], we have
$$2\pi ig(z)=\bigl\langle u^{+}(t),(t-z)^{-1}\bigr\rangle $$
for any \(u\in\mathcal{F}^{\prime}_{\mathcal{L}^{p}}\), which shows that
$$\begin{aligned} 2\pi ig(z) =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t) \biggl(-\frac{\partial }{\partial t} \biggr)^{(l-1)}(t-z)^{-1}\,dt \\ =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t) (-1)^{l-1}(l-1)!(t-z)^{-l} \,dt, \end{aligned}$$
where r is a nonnegative integer and \(u_{l}\in L_{p}\).
So
$$2\pi \bigl\vert g(t+ix) \bigr\vert \le\sum_{l=1}^{r} \int_{\mathbb{R}} \bigl\vert u_{l}(t) \bigr\vert (l-1)!(t-z)^{-l}\,dt, $$
which shows that
$$\bigl\vert g(t+ix) \bigr\vert \le\frac{1}{2\pi}\sum _{l=1}^{r}(l-1)! \biggl( \int_{\mathbb{R}} \bigl\vert u_{l}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{1}{p}} \biggl( \int_{R} (t-z)^{-lq}\,dt \biggr)^{\frac{1}{q}} $$
from the Hölder inequality.
Put
$$I= \int_{\mathbb{R}} (t-z)^{-lq}\,dt. $$
So
$$\begin{aligned} I =& \int_{\mathbb{R}} \frac{1}{[(t-x)^{2}+t^{2}])^{\frac{lq}{2}}}\,dt \\ =& \int_{\mathbb{R}} \frac{1}{t^{lq}[(\frac{t}{y})^{2}+1]^{\frac {lq}{2}}}\,dt \\ =&\frac{1}{t^{lq-1}} \int_{\mathbb{R}}\frac{1}{(1+t^{2})^{\frac{lq}{y}}}\,dt, \end{aligned}$$
which yields
$$I\le\frac{C}{\delta^{lq-1}}< \infty, $$
where C is a positive constant.
Since \(u_{l} \in\mathcal{L}^{p}\), then
$$\sup_{t\in\mathbb{R},t\ge\delta>0} \bigl\Vert g(t+ix) \bigr\Vert =C_{\delta}< \infty, $$
where M is a positive constant,
$$\begin{aligned} &2\pi A_{\delta}=\sum_{l=1}^{r}(l-1)! \frac{MC^{\frac{1}{q}}}{\frac{k\prime -1}{\delta q}}, \\ &\bigl\vert t^{\frac{1}{p}}g(t+ix) \bigr\vert \le\sum _{l=1}^{r}(l-1)! \Vert u_{l} \Vert _{L}^{p}\frac {1}{t^{lq-1-\frac{1}{p}}} \int_{\mathbb{R}}\frac{1}{(1+t^{2})^{\frac{lq}{x}}}\,dt \end{aligned}$$
and
$$lq-1-\frac{1}{p}=p^{2}(1-l)-1< 0. $$
So
$$\lim_{x\to\infty}\sup_{t\in\mathbb{R}} \bigl\vert g(t+ix) \bigr\vert =O\bigl(t^{-\frac{1}{p}}\bigr). $$
By virtue of the structure formula, we have
$$\begin{aligned} 2\pi ig(z) =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t) \biggl(-\frac{\partial }{\partial t} \biggr)^{(l-1)}(t-z)^{-1}\,dt \\ =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t) \biggl(\frac{\partial}{\partial z} \biggr)^{(l-1)}(t-z)^{-1}\,dt \\ =&\sum_{l=1}^{r}\biggl(\frac{\partial}{\partial z} \biggr)^{(l-1)} \int_{\mathbb{R}} (t-z)^{-1}u_{l}(t)\,dt \\ =&\sum_{l=1}^{r}\biggl(\frac{\partial}{\partial z} \biggr)^{(l-1)}G_{k}(z), \end{aligned}$$
where
$$2\pi iG_{k}(z)= \int_{\mathbb{R}} u_{l}(t) (t-z)^{-1}\,dt. $$
So we obtain \(G_{k}(z)\in\mathfrak{H}^{p}(C_{+})\) from Lemma 4.1, which shows that
$$G^{(j)}=\sum_{l=1}^{r}\biggl( \frac{\partial}{\partial z}\biggr)^{(k+j-1)}G_{k}. $$
Proof of Theorem 3.2
Since \(G_{k}(z)\in\mathfrak{H}^{p}(C_{+})\), where \(G_{k}(t+ix)\in\mathcal {L}^{p}\) for fixed x, there exists the solution \(u_{l}(t)\in\mathcal {L}^{p}\), where \(u_{l}\) is the nontangential limit of \(g(z)\).
Since \(D_{\mathcal{L}^{q}}\in\mathcal{L}^{q}\), we see that \(u_{l}(t)\in D^{\prime}_{\mathcal{L}^{p}}\) and
$$\begin{aligned} \bigl\vert G_{k}(t+ix) \bigr\vert ^{p} \leq& \frac{1}{\pi t^{2}} \int_{D(t+ix,x)} \bigl\vert G_{k}(\tau+i\eta ) \bigr\vert ^{p}\,d\lambda \\ \leq&\frac{1}{\pi t^{2}} \int_{t-x}^{t+x} \int_{0}^{2x} \bigl\vert G_{k}(\tau+i \eta ) \bigr\vert ^{p}\,d\eta \,d\zeta \\ \leq&\frac{2}{\pi x} \Vert G_{k} \Vert _{\mathfrak{H}^{p}}^{p}. \end{aligned}$$
So
$$\bigl\vert G_{k}(t+ix) \bigr\vert \le\biggl(\frac{1}{x} \Vert G_{k} \Vert _{\mathfrak{H}^{p}}^{p} \biggr)^{\frac {1}{p}}=t^{\frac{1}{p}} \bigl( \Vert G_{k} \Vert _{\mathfrak{H}^{p}}^{p}\bigr)^{\frac{1}{p}}, $$
which shows that
$$G_{k}(t+ix)=O\bigl(t^{-\frac{1}{p}}\bigr), $$
where \(x>0\) and
$$\sup_{t\in\mathbb{R},t\ge\delta>0} \bigl\Vert g(t+ix) \bigr\Vert \le \frac{1}{\delta} \Vert G_{k} \Vert _{\mathfrak{H}^{p}}^{p}=C_{\delta}< \infty. $$
We know that \(G_{k}(z)\) can be represented as follows:
$$2\pi iG_{k}(t)=\bigl\langle u_{l}(t),(t-z)^{-1} \bigr\rangle , $$
from Lemma 4.2, which yields
$$\begin{aligned} 2\pi ig(z) =&\sum_{l=1}^{r}\biggl( \frac{\partial}{\partial z}\biggr)^{l-1}G_{k}(z) \\ =&\sum_{l=1}^{r}\biggl(\frac{\partial}{\partial z} \biggr)^{l-1}\bigl\langle u_{l}(t),(t-z)^{-1}\bigr\rangle \\ =&\sum_{l=1}^{r}\biggl\langle u_{l}(t),\biggl(\frac{\partial}{\partial z}\biggr)^{l-1}(t-z)^{-1} \biggr\rangle \\ =&\sum_{l=1}^{r}\bigl\langle D^{(l-1)}u_{l}(t),(t-z)^{-1}\bigr\rangle \\ =&\Biggl\langle \sum_{l=1}^{r} D^{(l-1)}u_{l}(t),(t-z)^{-1}\Biggr\rangle . \end{aligned}$$
Put
$$u=\sum_{l=1}^{r} D^{(l-1)}u_{l}, $$
where \(u\in D^{\prime}_{\mathcal{L}^{p}}\), which shows that \(g(z)\) is one of the analytic representations of u.
Proof of Corollary 1
By virtue of the fixed point theorem with respect to the stationary Schrödinger operator in [8], we know that
$$\begin{aligned} 2\pi ig(z) =&\bigl\langle u(t),(t-z)^{-j}\bigr\rangle \\ =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t) \biggl(-\frac{\partial}{\partial t} \biggr)^{(l-1)}(t-z)^{-j}\,dt \\ =&\sum_{l=1}^{r} \int_{\mathbb{R}} u_{l}(t)\frac{(l+j-2)!}{(j-1)!(t-z)^{l+j-1}}\,dt. \end{aligned}$$
The rest of the proof of the corollary is similar to the proof of Theorem 3.1. So we omit the details here for the sake of brevity.
The proof of Corollary 1 is complete.