It is well known that,:
\noindent Theorem 1.Theorem 1. For $f\in L^{2}(\mathbb H_{n}=\text{The Heisenberg group of dimesion } 2n+1)$ we have the expansion $$f(z, s)= (2\pi)^{-n} \sum_{k=0}^{\infty} \int_{0}^{\infty} f \ast e_{k}^{\lambda} (z, s) |\lambda |^{n} d\lambda . $$;$$f(z, s)= (2\pi)^{-n} \sum_{k=0}^{\infty} \int_{0}^{\infty} f \ast e_{k}^{\lambda} (z, s) |\lambda |^{n} d\lambda $$ where $\lambda \in \mathbb R^{\ast}$,$e_{k}^{\lambda}(z, s):= e^{i\lambda s}\phi_{k}(\sqrt{\mid \lambda \mid}z );$ $e_{k}^{\lambda}(z, s):= e^{i\lambda s}\phi_{k}(\sqrt{\mid \lambda \mid}z )$; where $\phi_{k}(z)= (2\pi)^{\frac {n}{2}} \sum_{|\alpha | = k} \Phi_{\alpha, \alpha}(z).$$\phi_{k}(z)= (2\pi)^{\frac {n}{2}} \sum_{|\alpha | = k} \Phi_{\alpha, \alpha}(z)$, and for each $\alpha, \beta \in \mathbb N^{n}$ and $z\in \mathbb C^{n}$, we define the special Hermite functions $\Phi_{\alpha, \beta }$ by $$\Phi_{\alpha, \beta}(z):= (2\pi)^{-\frac{n}{2}} \int_{\mathbb R^{n}} e^{i x\cdot \xi} \Phi_{\alpha}(\xi + \frac{y}{2}) \Phi_{\beta}(\xi- \frac{y}{2}) d\xi .$$
\noindent Theorem 2.Theorem 2. For $f\in L^{2}(\mathbb H_{n})$ we have $$\parallel f \parallel_{2}^{2} = (2\pi)^{-n-1} \sum_{k=0}^{\infty} \int_{-\infty}^{\infty} \int_{\mathbb C^{n}} \mid f \ast e_{k}^{\lambda} (z, 0)\mid^{2} \lambda ^{2n} dz d\lambda . $$
\textbf{My question is}:My question is: Put $\eta= r+it, r>0$, and $ T_{\eta}f(z,s):= (2\pi)^{-n} \sum_{k=0}^{\infty} \int_{0}^{\infty} e^{-\eta(2k+n)|\lambda |} f \ast e_{k}^{\lambda} (z, s) |\lambda |^{n} d\lambda $.$$ T_{\eta}f(z,s):= (2\pi)^{-n} \sum_{k=0}^{\infty} \int_{0}^{\infty} e^{-\eta(2k+n)|\lambda |} f \ast e_{k}^{\lambda} (z, s) |\lambda |^{n} d\lambda. $$ Can we say this operator is bounded on $L^{2}(\mathbb H_{n})$, furthermore, on $L^{p}(\mathbb H_{n}), 1\leq p \leq \infty $ ?
