Path-connected components of the space of generalized integration operators on Fock spaces

Proof. The proof follows arguments similar to those used in the proof of Theorem 1.3 in [16]; for the sake of completeness, we include the details here.

(i) If the operator \(\mathcal{T}_{g}^{n, m}:\mathcal{F}_p^ {\alpha}\to \mathcal{F}_q^{\beta}\) is bounded, then using the inclusion \(\mathcal{F}_q^{\beta}\subseteq \mathcal{F}_{\infty}^{\beta}\) for \(q\leq \infty\), and the estimate (1), \[\begin{aligned} \infty>\left\|\mathcal{T}_{g}^{n, m}\right\| &\geq \left\|\mathcal{T}_{g}^{n, m}k_{(w,\alpha)}\right\|_{(q, \beta) }\succeq \sup_{z\in \mathbb {C}}\left|\mathcal{T}_{g}^{n, m}k_{(w,\alpha)}(z)\right|e^{-\frac{\beta }{2}|z|^2}\\ &\succeq \frac{|\alpha\overline{w}|^{m}\left|g^{(n-m)}(z)\right|}{(1+|z|)^n}\left|e^{\alpha z\overline{w}-\frac{\alpha |w|^2}{2}}\right|e^{-\frac{\beta }{2}|z|^2} \\ &\succeq \frac{(1+|w|)^{m}\left|g^{(n-m)}(z)\right|}{(1+|z|)^n}\left|e^{\alpha z\overline{w}-\frac{\alpha |w|^2}{2}}\right|e^{-\frac{\beta }{2}|z|^2}, \end{aligned}\] for all \(w\in \mathbb {C}\). Putting \(w=z\), we obtain \(M_{(\alpha\beta, g, nm)}\in L^{\infty}(\mathbb{C})\). On the other hand, if \(M_{(\alpha\beta, g, nm)}\) is bounded over \(\mathbb {C}\), for \(q <\infty\), \[\begin{aligned} \left\|\mathcal{T}_{g}^{n, m}f\right\|_{(q, \beta) }^q &\asymp \int_{\mathbb{C}}\frac{\left|f^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &\leq \left(\sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}^q(z)\right)\int_{\mathbb{C}}\frac{\left|f^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}dA(z)\\ &\preceq \left(\sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}^q(z)\right)\|f\|_{(q, \alpha) }^q\preceq \left(\sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}^q(z)\right)\|f\|_{(p, \alpha)}^q. \end{aligned}\]

Thus, \(\left\|\mathcal{T}_{g}^{n, m}\right\|\preceq \sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}(z)\) and hence \(\mathcal{T}_{g}^{n, m}\) is bounded. The case \(q=\infty\) is similar to the above, which only needs to replace the integral in the above estimate by supremum, we omit it.

Next, if \(\mathcal{T}_{g}^{n, m}\) is compact, then using the fact that \(k_{(w, \alpha)}\) is uniformly bounded on \(\mathcal{F}_p^{\alpha}\) and converges to zero on a compact subsets of \(\mathbb {C}\) as \(|w|\) goes to \(\infty\), \[\lim_{|w|\to \infty}M_{(\alpha\beta, g, nm)}(w)\preceq \lim_{|w|\to \infty}\left\|\mathcal{T}_{g}^{n, m}k_{(w,\alpha)}\right\|_{(q, \beta) }= 0.\]

Therefore \(\lim_{|w|\to \infty}M_{(\alpha\beta, g, nm)}(w)=0\). For the other direction, we let \(f_l\) to be bounded sequence in \(\mathcal{F}_p^{\alpha}\) that converges to 0 uniformly on a compact subsets of \(\mathbb{C}\) as \(l \to \infty\). Then, for \(R>0\) and \(q<\infty\), using (1), \[\begin{aligned} \left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}^q &\asymp \int_{\mathbb{C}}\frac{\left|f_l^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &=\left(\int_{|z|\leq R}+\int_{|z|>R}\right)\frac{\left|f_l^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q \int_{|z|\leq R}\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ & \ \ \ \ \ \ \ +\left(\sup_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)\right)\int_{|z|>R}\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}dA(z)\\ &\preceq \|\mathcal{T}_{g}^{n, m}z^m\|_{(q, \beta)}^q\max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\|f_l\|_{(q, \alpha)}^q\left(\sup_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)\right)\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\|f_l\|_{(p, \alpha)}^q\left(\sup_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)\right)\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\sup_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z). \end{aligned}\]

Letting \(l\to \infty\) and then \(R\to \infty\), we get \[\left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}\to 0,\] and hence \(\mathcal{T}_{g}^{n, m}\) is compact. The case \(q=\infty\) is very similar to the above. Thus, we omit it.

(ii) We first show that the integral conditions imply compactness of the operator. Let \((f_l)\) be a uniformly bounded sequence in \(\mathcal{F}_p^{\alpha}\) and \(f_l\to 0\) uniformly on compact subsets of \(\mathbb {C}\) as \(l\to \infty\). Then, for \(R>0\) \[\label{comp5} \begin{aligned} \left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}^q &\asymp \int_{\mathbb{C}}\frac{\left|f_l^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &=\left(\int_{|z|\leq R}+\int_{|z|>R}\right)\frac{\left|f_l^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q \int_{|z|\leq R}\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ & \ \ \ \ \ \ \ +\int_{|z|>R}\left(M_{(\alpha\beta, g, nm)}^q(z)\right)\left(\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}\right)dA(z) \\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q +\int_{|z|>R}\left(M_{(\alpha\beta, g, nm)}^q(z)\right)\left(\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}\right)dA(z). \end{aligned} \tag{2}\]

Now, for \(p<\infty\), applying Hölder’s inequality, and then using (1), the integral in (2) is further estimated as follows; \[\begin{aligned} &\int_{|z|>R}\left(M_{(\alpha\beta, g, nm)}^q(z)\right)\left(\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}\right)dA(z)\\ &\leq \left(\int_{|z|>R}\frac{\left|f_l^{(m)}(z)\right|^p}{(1+|z|)^{mp}}e^{-\frac{p\alpha}{2}|z|^2}dA(z)\right)^{\frac{q}{p}}\left(\int_{|z|>R} M_{(\alpha\beta, g, nm)}^{\frac{pq}{p-q}}(z)dA(z)\right)^{\frac{p-q}{p}}\\ &\preceq \|f_l\|_{(p, \alpha)}^q\left(\int_{|z|>R} M_{(\alpha\beta, g, nm)}^{\frac{pq}{p-q}}(z)dA(z)\right)^{\frac{p-q}{p}} \preceq \left(\int_{|z|>R} M_{(\alpha\beta, g, nm)}^{\frac{pq}{p-q}}(z)dA(z)\right)^{\frac{p-q}{p}}, \end{aligned}\] and hence \[\label{comp4} \left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}^q\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\left(\int_{|z|>R} M_{(\alpha\beta, g, nm)}^{\frac{pq}{p-q}}(z)dA(z)\right)^{\frac{p-q}{p}}. \tag{3}\]

Similarly, for \(p=\infty\), from (2) we have \[\begin{aligned} \label{comp3} \left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}^q&\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\int_{|z|>R}\left(M_{(\alpha\beta, g, nm)}^q(z)\right)\left(\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}\right)dA(z)\notag\\ &\leq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\left(\sup_{|z|>R}\frac{\left|f_l^{(m)}(z)\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}\right)\int_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)dA(z)\notag\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\|f_l\|_{(\infty, \alpha)}^q\int_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)dA(z)\notag\\ &\preceq \max_{|z|\leq R}\left|f_l^{(m)}(z)\right|^q+\int_{|z|>R}M_{(\alpha\beta, g, nm)}^q(z)dA(z). \end{aligned} \tag{4}\]

Letting \(l\to \infty\), and then \(R\to \infty\), in (3) and (4), \(\left\|\mathcal{T}_{g}^{n, m}f_l\right\|_{(q, \beta)}\to 0\) as \(l\to \infty\), which implies that \(\mathcal{T}_{g}^{n, m}\) is compact.

Next, we show that boundedness of the operator implies the integral conditions. For this, using (1), \[\begin{aligned} \left\|\mathcal{T}_{g}^{n, m}\right\|^q\|f\|_{(p, \alpha)}^p&\geq \left\|\mathcal{T}_{g}^{n, m}f\right\|_{(q, \beta)}^q\\ &\asymp \int_{\mathbb{C}}\frac{\left|f^{(m)}(z)\right|^q\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{-\frac{q\beta}{2}|z|^2}dA(z)\\ &=\int_{\mathbb {C}}\left(\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2}\right)\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{\frac{(\alpha-\beta)q}{2}|z|^2}dA(z)\\ &=\int_{\mathbb {C}}\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2}d\mu_{(\alpha \beta, g, nm, q)}(z), \end{aligned}\] where \[d\mu_{(\alpha \beta, g, nm, q)}(z):=\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{\frac{(\alpha-\beta)q}{2}|z|^2}dA(z).\]

Thus, if the operator \(\mathcal{T}_{g}^{n, m}\) is bounded, then the differential operator \(D^mf=f^{(m)}\), \[D^m:\mathcal{F}_p^ {\alpha}\to L_w^q(\mathbb {C}, d\mu_{(\alpha \beta, g, nm, q)}),\] is bounded, where \(L_w^q(\mathbb {C}, d\mu_{(\alpha \beta, g, nm, q)})\) denote the space of all measure function \(f\) such that \(\left|f(z)\right|e^{-\frac{q\alpha}{2}|z|^2}\in L^p(\mathbb {C}, d\mu_{(\alpha \beta, g, nm, q)})\). By Theorem 3.3 and 3.4 of [14], \(D^m\) is bounded if and only if \[\tilde{\mu}(w):=\int_{\mathbb {C}}(1+|z|)^{mq}e^{-\frac{\alpha q}{2}|z-w|^2}d\mu_{(\alpha \beta, g, nm, q)}(z),\] belongs to \(L^{\frac{p}{p-q}}(\mathbb {C}, dA)\) for \(p<\infty\), and belongs to \(L^1(\mathbb {C}, dA)\) for \(p=\infty\). But, \[\begin{aligned} \tilde{\mu}(w)&=\int_{\mathbb {C}}(1+|z|)^{mq}e^{-\frac{\alpha q}{2}|z-w|^2}d\mu_{(\alpha \beta, g, nm, q)}(z)\\ &=\int_{\mathbb {C}}(1+|z|)^{mq}e^{-\frac{\alpha q}{2}|z-w|^2}\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}} e^{\frac{(\alpha-\beta)q}{2}|z|^2}dA(z)\\ &\geq \int_{D(w,1)}e^{-\frac{\alpha q}{2}|z-w|^2}\frac{\left|g^{(n-m)}(z)\right|^q}{(1+|z|)^{(n-m)q}} e^{\frac{(\alpha-\beta)q}{2}|z|^2}dA(z)\\ &\succeq \frac{\left|g^{(n-m)}(w)\right|^q}{(1+|w|)^{(n-m)q}} e^{\frac{(\alpha-\beta)q}{2}|w|^2}=M_{(\alpha\beta, g, nm)}^q(w). \end{aligned}\]

Therefore, \(M_{(\alpha\beta, g, nm)}\) belongs to \(L^{\frac{pq}{p-q}}(\mathbb {C}, dA)\) for \(p<\infty\), and belongs to \(L^q(\mathbb {C}, dA)\) for \(p=\infty\). ◻

Proof. If \(\mathcal{T}_{g}^{n, m}:\mathcal{F}_p^ {\alpha}\to \mathcal{F}_q^{\beta}\) is bounded, then, by Theorem 1, \(M_{(\alpha\beta, g, nm)}\) is uniformly bounded over \(\mathbb{C}\), and hence there exists a positive real number \(N\) such that \[M_{(\alpha\beta, g, nm)}(z)=\frac{\left|g^{(n-m)}(z)\right|e^{\frac{\alpha-\beta}{2}|z|^2}}{(1+|z|)^{n-m}}\leq N,\] which implies that \[\left|g^{(n-m)}(z)\right|\leq \frac{N(1+|z|)^{n-m}}{e^{\frac{\alpha-\beta}{2}|z|^2}}.\]

The right hand side of the above inequality goes to zero as \(|z|\to \infty\). Thus, \(g^{(n-m)}\) is identically zero function, and hence \(g\) is a polynomial of degree at most \((n-m)-1\). On the other hand, if \(g\) is a polynomial of degree at most \((n-m)-1\), then clearly \(M_{(\alpha\beta, g, nm)}(z)\) goes to zero, and by Theorem 1, \(\mathcal{T}_{g}^{n, m}:\mathcal{F}_p^ {\alpha}\to \mathcal{F}_q^{\beta}\) is compact. Since compactness of the operator implies boundedness, we have the desired equivalence. ◻

Proof. (i) Suppose \(\mathcal{T}_{g}^{n, m}:\mathcal{F}_p^ {\alpha}\to \mathcal{F}_q^{\beta}\) is bounded. Then, by Theorem 1, there exists a positive real number \(N\) such that \[\left|g^{(n-m)}(z)\right|\leq N(1+|z|)^{n-m},\] which by Liouville’s Theorem gives that \(g^{(n-m)}\) is a polynomial of degree at most \(n-m\), and hence \(g\) is a polynomial of degree at most \(2(n-m)\). On the other hand, if \(g\) is a polynomial of degree at most \(2(n-m)\), then \(M_{(\alpha\beta, g, nm)}\) is clearly uniformly bounded over \(\mathbb{C}\), and by Theorem 1, \(\mathcal{T}_{g}^{n, m}:\mathcal{F}_p^ {\alpha}\to \mathcal{F}_q^{\beta}\) is bounded. The compactness case also follows from similar arguments.

(ii) By Theorem 1, boundedness or compactness of the operator is equivalent with \(M_{(\alpha\beta, g, nm)}\in L^{r}(\mathbb{C}, dA)\), where \(r=\frac{pq}{p-q}\) for \(p<\infty\) and \(r=q\) for \(p=\infty\). That is, the operator is bounded or compact if and only if there exists a real number \(N\) such that \[\int_{\mathbb{C}}\left(\frac{\left|g^{(n-m)}(z)\right|}{(1+|z|)^{n-m}}\right)^{r}dA(z)\leq N,\] from which the restriction on exponents \(p\) and \(q\), and \(g\) follows. ◻

Proof. The proof follows arguments similar to those used in the proof of Proposition 1.5 of [16], with the only modification being the replacement of the function \(g'\) there by \(g^{(n-m)}\). ◻

Proof. (i) (a) First we assume the operator is bounded. Since by Proposition 1, \(\rho(g)\leq 2\) and \(g\) is zero-free on \(\mathbb {C}\), Hadamard’s product formula gives \(g(z)=e^{bz+az^2}\) for some \(a, b\in \mathbb {C}\). Thus, we only show that the restrictions on \(a\) and \(b\) hold. If \(a=0\), then clearly \(|a|<\frac{\beta-\alpha}{2}\). Assume \(a\neq 0\), then \(\rho(g)=2\) and by Proposition 1, \(\sigma(g)=|a|\leq \frac{\beta-\alpha}{2}\). In particular, if \(|a|=\frac{\beta-\alpha}{2}\), then using Faá di Bruno’s formula, \[\label{Faa di} g^{(n-m)}(z)=e^{bz+az^2}\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}, \tag{6}\] where \(m_1\) and \(m_2\) are nonnegative integers such that \(1m_1+2m_2=n-m\) and the sum is over all such a partition \((m_1,m_2)\) of \(n-m\), we obtain \[\begin{aligned} M_{(\alpha\beta, g, nm)}(z)&=\frac{\left|g^{(n-m)}(z)\right|e^{\frac{\alpha-\beta}{2}|z|^2}}{(1+|z|)^{n-m}}\\ &=\frac{\left|e^{bz+az^2}\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{-\frac{(\beta-\alpha)}{2}|z|^2}\\ &=\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}, \end{aligned}\] for all \(z\neq 0\). setting \(a=|a|e^{-2i\theta}=\frac{(\beta-\alpha)}{2}e^{-2i\theta}\), \(0\leq\theta<\pi\), and replacing \(z\) by \(e^{i\theta}w\) in the above equation, we obtain \[M_{(\alpha\beta, g, nm)}(we^{i\theta})=\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2awe^{i\theta}+b)^{m_1}a^{m_2}\right|}{(1+|w|)^{n-m}}e^{\Re(bwe^{i\theta})+\frac{(\beta-\alpha)}{2}(\Re(w^2)-|w|^2)},\] for all \(w\in \mathbb {C}\). Particularly, for a real number \(w\), we get \[\label{new1} M_{(\alpha\beta, g, nm)}(we^{i\theta})=\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2awe^{i\theta}+b)^{m_1}a^{m_2}\right|}{(1+|w|)^{n-m}}e^{w\Re(be^{i\theta})}. \tag{7}\]

Since \(\mathcal{T}_g^{n, m}:\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) is bounded, by Theorem 1, the function \(M_{(\alpha\beta, g, nm)}(we^{i\theta})\) above is uniformly bounded, which implies that \(\Re(be^{i\theta})=0\). Thus, either \(b=0\) or \(e^{-i\theta}=\pm \frac{ib}{|b|}\). Therefore, either \(b=0\) or \(a=\frac{(\beta-\alpha)}{2}e^{-2i\theta}=-\frac{(\beta-\alpha)b^2}{2|b|^2}\).

Conversely, if \(g\) has the form in (5) with the given restrictions \(|a|<\frac{\beta-\alpha}{2}\), or \(|a|=\frac{\beta-\alpha}{2}\) and either \(b=0\) or \(a=-\frac{(\beta-\alpha)b^2}{2|b|^2}\), then \[\begin{aligned} \label{new} \sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}&(z) =\sup_{z\in \mathbb {C}}\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\notag\\ &\leq \left(\sup_{z\in \mathbb {C}}\frac{\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2|a||z|+|b|)^{m_1}|a|^{m_2}}{(1+|z|)^{n-m}}\right)\left(\sup_{z\in \mathbb {C}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\right)\notag\\ &\preceq \sup_{z\in \mathbb {C}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}. \end{aligned} \tag{8}\]

Now, if \(|a|<\frac{\beta-\alpha}{2}\), then \[\sup_{z\in \mathbb {C}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\leq \sup_{z\in \mathbb {C}}e^{|b||z|)-(\frac{(\beta-\alpha)}{2}-|a|)|z|^2}<\infty.\]

If \(|a|=\frac{\beta-\alpha}{2}\) and \(b=0\), then \(\sup_{z\in \mathbb {C}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\leq 1\). If \(|a|=\frac{\beta-\alpha}{2}\) and \(a=-\frac{(\beta-\alpha)b^2}{2|b|^2}\), then \[\begin{aligned} \sup_{z\in \mathbb {C}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}&=\sup_{z\in \mathbb {C}}e^{\Re(bz)-\frac{(\beta-\alpha)}{2|b|^2}\Re((bz)^2)-\frac{(\beta-\alpha)}{2}|z|^2}\\ &=\sup_{z\in \mathbb {C}}e^{\Re(bz)-\frac{(\beta-\alpha)}{2|b|^2}((\Re(bz))^2-(\Im(bz))^2)-\frac{(\beta-\alpha)}{2}|z|^2}\\ &=\sup_{z\in \mathbb {C}}e^{\Re(bz)-\frac{(\beta-\alpha)}{2|b|^2}\left((\Re(bz))^2-(\Im(bz))^2\right)-\frac{(\beta-\alpha)}{2}|z|^2}\\ &\preceq \sup_{z\in \mathbb {C}}e^{\frac{(\beta-\alpha)}{2|b|^2}(\Im(bz))^2-\frac{(\beta-\alpha)}{2}|z|^2}\leq 1. \end{aligned}\]

From the above estimates and (8), we have \(M_{(\alpha\beta, g, nm)}\) is uniformly bounded over \(\mathbb {C}\). Thus, by Theorem 1, \(\mathcal{T}_g^{n, m}:\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) is bounded.

(b) Suppose \(\mathcal{T}_g^{n, m}:\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) is compact. Then, it is bounded and by \((a)\) above \(|a|\leq \frac{\beta-\alpha}{2}\). If \(|a|=\frac{\beta-\alpha}{2}\), then since \(\mathcal{T}_g^{n, m}\) is compact, by Theorem 1, (7) must goes to zero as \(w\to \infty\). But, \(e^{\Re(be^{i\theta})}=0\) and \[M_{(\alpha\beta, g, nm)}(we^{i\theta})=\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2awe^{i\theta}+b)^{m_1}a^{m_2}\right|}{(1+|w|)^{n-m}}\asymp 1,\] which is a contradiction. Thus, \(|a|< \frac{\beta-\alpha}{2}\). On the other hand, if \(g\) has the form in (5) with the given conditions, then \[M_{(\alpha\beta, g, nm)}(z) =\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\to 0,\] as \(|z|\to \infty\), and by Theorem 1, the operator is compact.

(ii) Suppose \(g\) has the form in (5) with \(|a|<\frac{\beta-\alpha}{2}\), then clearly the function \(M_{(\alpha\beta, g, nm)}\) belongs to \(L^{r}(\mathbb{C}, dA)\), where \(r=\frac{pq}{p-q}\) for \(p<\infty\) and \(r=q\) for \(p=\infty\). Hence, by Theorem 1, the operator is bounded (compact). Now, suppose \(\mathcal{T}_g^{n, m}:\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) is bounded (compact). Then by Proposition 1 and what we have shown in \((i)\) above, \(g\) has the form in 5 and \(\sigma(g)=|a|<\frac{\beta-\alpha}{2}\). ◻

Proof. To prove the lower estimate, let \(\mathcal{C} :\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) be a compact operator. Then, since \(k_{(w, \alpha)}\) weakly converges to zero on \(\mathcal{F}_{p}^\alpha\), we have \[\begin{aligned} \|\mathcal{T}_g^{n, m}-\mathcal{C}\| &\geq \limsup_{|w| \to \infty } \left\|\mathcal{T}_g^{n, m}k_{(w, \alpha)}-\mathcal{C}k_{(w, \alpha)}\right\|_{(q, \beta)}\\ &\geq \limsup_{|w| \to \infty } \left\|\mathcal{T}_g^{n, m}k_{(w, \alpha)}\right\|_{(q, \beta)}-\left\|\mathcal{C}k_{(w, \alpha)}\right\|_{(q, \beta)}\\ &= \limsup_{|w|\to \infty} \left\|\mathcal{T}_g^{n, m}k_{(w, \alpha)}\right\|_{(q, \beta)} \succeq \limsup_{|w|\to \infty}M_{(\alpha\beta, g, nm)}(w). \end{aligned}\]

For the upper estimate, we consider a sequence \(\Phi_i(z)=\frac{i}{i+1}z\) for each \(i\in \mathbb{N}\). Since \(\frac{i}{i+1}<1\), by Corollary \(3.5\) of [17], the composition operator \(C_{\Phi_i}\) is compact. Using the estimate in (1), for some fixed positive number \(R>0\) and \(q<\infty\), \[\begin{aligned} \left\|\mathcal{T}_g^{n, m}\right\|_e&\leq \left\|\mathcal{T}_g^{n, m}-\mathcal{T}_g^{n, m}\circ C_{\Phi_i}\right\|\leq \sup_{\|f\|_{(p, \alpha)}\leq 1}\left\|(\mathcal{T}_g^{n, m}-\mathcal{T}_g^{n, m}\circ C_{\Phi_i})f\right\|_{(q, \beta)}\\ &\asymp \sup_{\|f\|_{(p, \alpha)}\leq 1}\left(\int_{\mathbb {C}} \frac{\left|g^{(n-m)}(z)\right|^q\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|^q}{(1+|z|)^{nq}}e^{-\frac{q\beta}{2}|z|^2}dA(z) \right)^{\frac{1}{q}}\\ &\leq \sup_{\|f\|_{(p, \alpha)}\leq 1}\left(\int_{|z|\leq R} \frac{\left|g^{(n-m)}(z)\right|^q\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|^q}{(1+|z|)^{nq}}e^{-\frac{q\beta}{2}|z|^2}dA(z) \right)^{\frac{1}{q}} \\ & \ \ \ \ \ \ +\sup_{\|f\|_{(p, \alpha)}\leq 1}\left(\int_{|z|>R} \frac{\left|g^{(n-m)}(z)\right|^q\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|^q}{(1+|z|)^{nq}}e^{-\frac{q\beta}{2}|z|^2}dA(z) \right)^{\frac{1}{q}} \\ &:=Int_1+Int_2. \end{aligned}\]

We now estimate each integrals above separately. Using the estimate in (1), \(Int_2\) is bounded from above by

\[\label{est7} \begin{aligned} \sup_{|z|>R}M_{(\alpha\beta, g, nm)}(z)&\sup_{\|f\|_{(p, \alpha)}\leq 1}\left(\int_{|z|>R} \frac{\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}dA(z) \right)^{\frac{1}{q}}\\ &\asymp \sup_{|z|>R}M_{(\alpha\beta, g, nm)}(z)\sup_{\|f\|_{(p, \alpha)}\leq 1}\|f-f(\Phi_i)\|_{(q, \alpha)}\\ &\preceq \sup_{|z|>R}M_{(\alpha\beta, g, nm)}(z)\left(\sup_{\|f\|_{(p, \alpha)}\leq 1}\|f\|_{(q, \alpha)}+\sup_{\|f\|_{(p, \alpha)}\leq 1}\|C_{\Phi_i}f\|_{(q, \alpha)}\right)\\ &\preceq \sup_{|z|>R}M_{(\alpha\beta, g, nm)}(z), \end{aligned} \tag{9}\] where the last in equality is by boundedness of composition operator \(C_{\Phi_i}: \mathcal{F}_{p}^\alpha \to \mathcal{F}_{q}^\alpha\) and the inclusion \(\mathcal{F}_{p}^\alpha\subseteq \mathcal{F}_{q}^\alpha\) for \(p\leq q\). On the other hand, \[\begin{aligned} \label{est9} Int_1&\preceq \sup_{|z|\leq R}M_{(\alpha\beta, g, nm)}(z)\sup_{\|f\|_{(p, \alpha)}\leq 1}\left(\int_{|z|\leq R} \frac{\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|^q}{(1+|z|)^{mq}}e^{-\frac{q\alpha}{2}|z|^2}dA(z) \right)^{\frac{1}{q}}\notag\\ &\leq \sup_{|z|\leq R}M_{(\alpha\beta, g, nm)}(z) \left(\int_{|z|\leq R} \frac{e^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{mq}}dA(z) \right)^{\frac{1}{q}} \times \left(\sup_{\|f\|_{(p, \alpha)}\leq 1}\sup_{|z|\leq R}\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|\right)\notag\\ &\preceq \sup_{z\in \mathbb {C}}M_{(\alpha\beta, g, nm)}(z)\sup_{\|f\|_{(p, \alpha)}\leq 1}\sup_{|z|\leq R}\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|\notag\\ &\preceq \sup_{\|f\|_{(p, \alpha)}\leq 1}\sup_{|z|\leq R}\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right| \end{aligned} \tag{10}\]

Integrating the function \(f^{(m+1)}\) along the radial segment \([\frac{iz}{i+1}z, z]\), we get \[\label{est5} \Big|f^{(m)}(z)-f^{(m)}(\Phi_i(z))\Big|\leq \frac{|z|\left|f^{(m+1)}(z^*)\right|}{i+1} , \tag{11}\] for some \(z^*\) in the segment \([\frac{iz}{i+1}z, z]\). Using Cauchy’s estimate for \(f^{(m+1)}\), we also have \[\label{est8} \left|f^{(m+1)}(z^*)\right|\leq \frac{1}{R}\max_{|z|=2R}\left|f^{(m)}(z)\right|. \tag{12}\]

From inequalities (11) and (12), and using the estimate \[\label{local} \left|f^{(m)}(z)\right|\preceq (1+|z|)^me^{\frac{\alpha}{2}|z|^2}\|f\|_{(p, \alpha)}, \tag{13}\] which follows from (1) (see also, Lemma 2.2 of [14]), we get \[\begin{aligned} \left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|&=\left(\frac{i}{i+1}\right)^m\left|\left(\frac{(i+1)^m}{i^n}\right)f^{(m)}(z)-f^{(m)}(\Phi_i(z))\right|\\ &\preceq \frac{\left|f^{(m)}(z)\right|}{(i+1)^m}+\left|f^{(m)}(z)-f^{(m)}(\Phi_i(z))\right|\\ &\leq \frac{\left|f^{(m)}(z)\right|}{(i+1)^m}+\frac{|z|}{R(i+1)}\max_{|z|=2R}\left|f^{(m)}(z)\right|\\ &\preceq \left(\frac{(1+|z|)^me^{\frac{\alpha}{2}|z|^2}}{(i+1)^m}+\frac{2(1+2R)^me^{2\alpha R^2}}{i+1}\right)\|f\|_{(p, \alpha)}. \end{aligned}\]

Using the above estimate, inequality (10) is bounded from above by \[\label{est6} \sup_{\|f\|_{(p, \alpha)}\leq 1}\sup_{|z|\leq R}\left|f^{(m)}(z)-\left(\frac{i}{i+1}\right)^mf^{(m)}(\Phi_i(z))\right|\preceq \frac{(1+|z|)^me^{\frac{\alpha}{2}|z|^2}}{(i+1)^m}+\frac{2(1+2R)^me^{2\alpha R^2}}{i+1}. \tag{14}\]

Therefore, from (10) and (14), we have \[Int_1\preceq \frac{(1+|z|)^me^{\frac{\alpha}{2}|z|^2}}{(i+1)^m}+\frac{2(1+2R)^me^{2\alpha R^2}}{i+1}.\]

Letting \(i\) goes to \(\infty\) and then \(R\to \infty\) in the above estimate and estimate (9), we get \[\left\|\mathcal{T}_g^{n, m}\right\|_e\preceq \limsup_{|z|\to \infty}M_{(\alpha\beta, g, nm)}(z).\]

The case \(q=\infty\) is very similar to the above proof, which only needs to replace the integral by supremum. ◻

Proof. (i) From Theorem (2), \[\begin{aligned} \left\|\mathcal{T}_g^{n, m}\right\|_e&\asymp \limsup_{|z|\to \infty}M_{(\alpha\beta, g, nm)}(z)=\limsup_{|z|\to \infty}\frac{\left|g^{(n-m)}(z)\right|}{(1+|z|)^{n-m}} \\ &=\limsup_{|z|\to \infty}\frac{\left|a_{2(n-m)}\left(\frac{(2(n-m))!}{(n-m)!}\right)z^{n-m}+…+a_{n-m}(n-m)!\right|}{(1+|z|)^{n-m}}\\ &\asymp |a_{2(n-m)}|. \end{aligned}\]

(ii) Using Theorem 2 and the formula (6), \[\begin{aligned} \left\|\mathcal{T}_g^{n, m}\right\|_e&\asymp \limsup_{|z|\to \infty}M_{(\alpha\beta, g, nm)}(z)=\limsup_{|z|\to \infty}\frac{\left|g^{(n-m)}(z)\right|}{(1+|z|)^{n-m}}e^{-\frac{(\beta-\alpha)}{2}|z|^2}\\ &= \limsup_{|z|\to \infty}\frac{\left|e^{bz+az^2}\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{-\frac{(\beta-\alpha)}{2}|z|^2}\\ &\preceq \limsup_{|z|\to \infty}\frac{\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}|2az+b|^{m_1}|a|^{m_2}}{(1+|z|)^{n-m}}e^{\Re(bz)+\Re(az^2)-\frac{(\beta-\alpha)}{2}|z|^2}\\ &\preceq\limsup_{|z|\to \infty}\frac{\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}|2az+b|^{m_1}|a|^{m_2}}{(1+|z|)^{n-m}}\\ &\leq \sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}\left(\limsup_{|z|\to \infty}\frac{|2az+b|^{m_1}|a|^{m_2}}{(1+|z|)^{n-m}}\right) \asymp |a|^{n-m}. \end{aligned}\]

Similarly, from lower we have \[\begin{aligned} \left\|\mathcal{T}_g^{n, m}\right\|_e&\asymp \limsup_{|z|\to \infty}\frac{\left|e^{bz+az^2}\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}e^{-\frac{(\beta-\alpha)}{2}|z|^2}\\ &\succeq \limsup_{|z|\to \infty}\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(2az+b)^{m_1}a^{m_2}\right|}{(1+|z|)^{n-m}}\asymp |a|^{n-m}. \end{aligned}\] ◻

Proof. For the proof of this proposition, we follow the same procedures as in [17]. If \(g_1^{n, m}(0)\) or \(g_2^{n, m}(0)\) is zero, or \(g_1^{n, m}(0)=g_2^{n, m}(0)\), then trivially \(\mathcal{T}_{g_1(0)}^{n, m}\) and \(\mathcal{T}_{g_2(0)}^{n, m}\) are in the same path-connected components. Thus, we assume \[g_1^{n, m}(0)=\lambda g_2^{n, m}(0),\] where \(\lambda\in\mathbb {C}\backslash \{0, 1\}\). Consider a continuous function \(\gamma:[0,1]\to \mathbb {C}\) such that \[\gamma(0)=0, \gamma(1)=1, \gamma(t)\neq \frac{1}{1-\lambda},\] for all \(t\in [0,1]\). Define a sequence of functions \(h_t\) by \[h_t:=(1-\gamma(t))g_1^{n, m}(0)+\gamma(t)g_2^{n, m}(0)\neq 0,\] for all \(t\in [0,1]\). Define a map \(\mathcal{T}_{h_t}^{n, m}:[0,1] \to \mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\). Then, \(\mathcal{T}_{h_0}^{n, m}=\mathcal{T}_{g_1(0)}^{n, m}\), \(\mathcal{T}_{h_1}^{n, m}=\mathcal{T}_{g_2(0)}^{n, m}\) and \(\mathcal{T}_{h_t}^{n, m}:\mathcal{F}_{p}^\alpha\to \mathcal{F}_{q}^\beta\) are compact for all \(t\in [0, 1]\). Since \[\lim_{t \rightarrow s}\left\|\mathcal{T}_{h_t}^{n, m}-\mathcal{T}_{h_s}^{n, m}\right\|=\lim_{t \rightarrow s}\left|\gamma(t)-\gamma(s)\right|\left\|\mathcal{T}_{(g_2-g_1)(0)}^{n, m}\right\|=0,\] the map \(\mathcal{T}_{h_t}^{n, m}\) is continuous. This completes the proof. ◻

Proof. Since \(\mathcal{T}_{g}^{n, m}\) is compact, by Corollary 2, for \(0<p\leq q\leq \infty\), \(g\) is a polynomial of degree at most \(2(n-m)-1\), and for \(0<q<p\leq \infty\), \(g\) is a polynomial of degree at most \(2(n-m)-1\) and \(q>\left\{ \begin{array}{lr} \frac{2p}{p+2},& p<\infty \\ 2, & p=\infty \end{array} \right.\). In both cases, \(g\) has the form \[g(z)=a_lz^l+a_{l-1}z^{l-1}+…+a_0,\] where \(l=2(n-m)-1\). If \(a_k=0\) for all \(k\in[\frac{l+1}{2},l]\), then \(\mathcal{T}_{g}^{n, m}=\mathcal{T}_{g(0)}^{n, m}\) are zero operators and the result holds trivially. We assume \(a_k\neq 0\) for some \(k\in[\frac{l+1}{2},l]\). Define functions \(g_{t}:[0,1]\to \mathbb {C}\) by \[g_{t}(z)=g(tz).\]

Consider a map \(\mathcal{T}_{g_{t}}^{n, m}:[0,1] \to \mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\). Then, \(\mathcal{T}_{g_{1}}^{n, m}=\mathcal{T}_{g}^{n, m}\), \(\mathcal{T}_{g_{0}}^{n, m}=\mathcal{T}_{g(0)}^{n, m}\) and \(\mathcal{T}_{g_{t}}^{n, m}:\mathcal{F}_{p}^\alpha \to \mathcal{F}_{q}^\beta\) are compact for all \(t\). We claim that this map is continuous, that is, \[\label{enough} \lim_{t \rightarrow s}\|\mathcal{T}_{g_{t}}^{n, m}-\mathcal{T}_{g_{s}}^{n, m}\|=0, \tag{15}\] for every \(s\in [0,1]\).

Case \(0<p\leq q<\infty\) or \(0<q<p\leq \infty\).

For each \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\), an application of formula (1) to the \(\mathcal{F}_{q}^\beta\) norm of a function \(\mathcal{T}_{g_{t}-g_{s}}^{n, m}f\) gives \[\label{new2} \begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f-\mathcal{T}_{g_{s}}^{n, m}f\|_{(q, \beta)}^q &\asymp \int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z) \\ &= |t-s|^q\int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|g_{(t,s)}^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z), \end{aligned} \tag{16}\] where \[g_{(t,s)}^{(n-m)}(z)=\sum\limits_{i=(n-m)+1}^{2(n-m)-1}a_i \frac{i!}{(i-(n-m))!} z^{i-(n-m)} \sum\limits_{j=(n-m)}^{i-1}s^{j-(n-m)}t^{i-1-j}.\]

For \(s, t\in [0, 1]\) and a positive integer \(i\) with \((n-m)+1\leq i\leq 2(n-m)-1\), \[\sum\limits_{j=(n-m)}^{i-1}s^{j-(n-m)}t^{i-1-j}\leq \sum\limits_{j=(n-m)}^{i-1}1=i-(n-m).\]

Thus, \(|g_{(t,s)}^{(n-m)}(z)|\leq |h^{(n-m)}(z)|\), where \[h^{(n-m)}(z):=\sum\limits_{i=(n-m)+1}^{2(n-m)-1}a_i \frac{i!}{(i-(n-m)-1)!} z^{i-(n-m)}.\]

Clearly, \(\deg(h^{(n-m)})\leq (n-m)-1\) and hence \(\deg(h)\leq 2(n-m)-1\). Moreover, we have a restriction \(q>\left\{ \begin{array}{lr} \frac{2p}{p+2},& p<\infty \\ 2, & p=\infty \end{array} \right.\) for \(0<q<p\leq \infty\). By Corollary 2, the operator \(\mathcal{T}_{h}^{n, m}:\mathcal{F}_{p}^\alpha \to \mathcal{F}_{q}^\beta\) is bounded. Thus, \[\begin{aligned} \int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|g_{(t,s)}^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)&\leq \int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|h^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &\asymp \|\mathcal{T}_{h}^{n, m}f\|_{(q, \beta)}^q\leq \|\mathcal{T}_{h}^{n, m}\|^q\|f\|_{(p, \alpha)}^q<\infty. \end{aligned}\]

Using this and the estimate in (16), we obtain \[\|\mathcal{T}_{g_{t}}^{n, m}-\mathcal{T}_{g_{s}}^{n, m}\|\preceq |t-s|.\] Letting \(t\to s\), (15) holds.

Case \(0<p\leq q=\infty\).

Similarly, for \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\), using formula (1) and above estimates, \[\begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f-\mathcal{T}_{g_{s}}^{n, m}f\|_{(\infty, \beta)}&\asymp \sup_{z\in \mathbb {C}}\frac{\left|f^{(m)}(z)\right|\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}}\\ &=|t-s|\left(\sup_{z\in \mathbb {C}}\frac{\left|f^{(m)}(z)\right|\left|g_{(t,s)}^{(n-m)}(z)\right|e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}} \right) \\ &\leq |t-s|\left(\sup_{z\in \mathbb {C}}\frac{\left|f^{(m)}(z)\right|\left|h^{(n-m)}(z)\right|e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}} \right) \\ &\asymp |t-s|\|\mathcal{T}_{h}^{n, m}f\|_{(\infty, \beta)} \leq |t-s|\|\mathcal{T}_{h}^{n, m}\|\|f\|_{(p, \alpha)}. \end{aligned}\]

Therefore, \(\|\mathcal{T}_{g_{t}}^{n, m}-\mathcal{T}_{g_{s}}^{n, m}\|\preceq |t-s|\). From which, (15) holds. ◻

Proof. Since \(\mathcal{T}_{g}^{n, m}\) is compact, by Corollary 3, \(g\) has the form \(g(z)=e^{bz+az^2}\) for some \(a, b\in \mathbb {C}\) with \(|a|<\frac{\beta-\alpha}{2}\). If \(a=b=0\), then \(\mathcal{T}_{g}^{n, m}=\mathcal{T}_{g(0)}^{n, m}\) are zero operators and the result holds trivially. For the remaining cases, define functions \(g_{t}:[0,1]\to \mathbb {C}\) by \(g_{t}(z)=g(tz)\). Consider a map \(\mathcal{T}_{g_{t}}^{n, m}:[0,1] \to \mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\). Then, \(\mathcal{T}_{g_{1}}^{n, m}=\mathcal{T}_{g}^{n, m}\), \(\mathcal{T}_{g_{0}}^{n, m}=\mathcal{T}_{g(0)}^{n, m}\) and \(\mathcal{T}_{g_{t}}^{n, m}:\mathcal{F}_{p}^\alpha \to \mathcal{F}_{q}^\beta\) are compact for all \(t\). We need to show (15) holds, to conclude that \(\mathcal{T}_{g_{t}}^{n, m}\) is continuous.

Case \(a=0\) and \(b\neq 0\).

We first estimate \(|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)|\), \[\label{adnan} \begin{aligned} |g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)|&\asymp |e^{btz}-e^{bsz}|\leq \sum\limits_{i=0}^{\infty}\frac{(|bz|)^i|t^i-s^i|}{i!} \\ &\leq |t-s|\sum\limits_{i=1}^{\infty}\frac{|bz|^i}{(i-1)!}=|t-s||bz|e^{|bz|},\ \forall z\in\mathbb {C}. \end{aligned} \tag{17}\]

Depending on the exponents \(p\) and \(q\), we consider the following cases.

(i) \(0<p\leq q<\infty\):

For \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\), using estimates (1) and (17), we obtain \[\label{umer} \begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f-&\mathcal{T}_{g_{s}}^{n, m}f\|_{(q, \beta)}^q \asymp \int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &\preceq |t-s|^q\int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q|bz|^qe^{q|bz|}e^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &=|t-s|^q\int_{\mathbb {C}}\left( \frac{\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2}}{(1+|z|)^{mq}} \right)\left(\frac{|z|(1+|z|)^me^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n}}\right)^qdA(z). \end{aligned} \tag{18}\]

Using \[\label{iftu} \begin{aligned} \sup_{z\in \mathbb {C}}\frac{|z|(1+|z|)^me^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n}}&\leq \sup_{z\in \mathbb {C}} \frac{e^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n-m-1}}\\ &\leq\sup_{z\in \mathbb {C}} e^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}<\infty, \end{aligned} \tag{19}\] the estimate in (1) and the inclusion \(\mathcal{F}_{p}^\alpha\subseteq \mathcal{F}_{q}^\alpha\), for \(p\leq q\), we further estimate (18) from above as follows. \[\begin{aligned} |t-s|^q\int_{\mathbb {C}}&\left( \frac{\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2}}{(1+|z|)^{mq}} \right)\left(\frac{|z|(1+|z|)^me^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n}}\right)^qdA(z)\\ &\preceq |t-s|^q\int_{\mathbb {C}} \frac{\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2}}{(1+|z|)^{mq}} dA(z) \\ &\asymp |t-s|^q\|f\|_{(q, \alpha)}^q\preceq |t-s|^q\|f\|_{(p, \alpha)}^q. \end{aligned}\]

Thus, \[\|\mathcal{T}_{g_{t}}^{n, m}-\mathcal{T}_{g_{s}}^{n, m}\|\preceq |t-s|,\] from which (15) holds.

(ii) \(0<p\leq q=\infty\):

Similarly, for \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\), using estimates (1), (17) and (19),

\[\begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f&-\mathcal{T}_{g_{s}}^{n, m}f\|_{(\infty, \beta)} \asymp \sup_{z\in\mathbb {C}}\frac{\left|f^{(m)}(z)\right|\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}} \\ &\preceq |t-s|\sup_{z\in\mathbb {C}}\frac{\left|f^{(m)}(z)\right||bz|e^{|bz|}e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}} \\ &=|t-s|\sup_{z\in\mathbb {C}}\left( \frac{\left|f^{(m)}(z)\right|e^{-\frac{\alpha}{2}|z|^2}}{(1+|z|)^{m}} \right)\left(\frac{|z|(1+|z|)^me^{|bz|-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n}}\right)\\ &\preceq |t-s|\sup_{z\in\mathbb {C}}\left( \frac{\left|f^{(m)}(z)\right|e^{-\frac{\alpha}{2}|z|^2}}{(1+|z|)^{m}} \right) \\ &\asymp |t-s|\|f\|_{(\infty, \alpha)}\preceq |t-s|\|f\|_{(p, \alpha)}. \end{aligned}\]

The last estimate is by the inclusion \(\mathcal{F}_{p}^\alpha\subseteq \mathcal{F}_{\infty}^\alpha\), for \(p\leq \infty\). From the above estimate, (15) holds for this case also.

(iii) \(0<q<p\leq \infty\):

Following similar procedures as those leading to (18) and using the estimate in (13) we obtain \[\begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f-\mathcal{T}_{g_{s}}^{n, m}f\|_{(q, \beta)}^q&\preceq |t-s|^q\int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q|z|^qe^{q|bz|}e^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &\preceq |t-s|^q\|f\|_{(p, \alpha)}^q\int_{\mathbb {C}}\frac{(1+|z|)^{mq}|z|^qe^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &\preceq |t-s|^q\|f\|_{(p, \alpha)}^q\int_{\mathbb {C}}e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z). \end{aligned}\]

Thus, the conclusion in (15) holds, if we show that \[\int_{\mathbb {C}}e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z)<\infty.\]

To show this, let \(R\) be a positive real number such that \(|b|<\frac{R(\beta-\alpha)}{2}\) and define a number \(\gamma:=\frac{\beta R-\alpha R-2|b|}{2R}<0\), then clearly \[\int_{|z|\leq R}e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z)<\infty,\] and \[\begin{aligned} \int_{|z|>R}e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z)&\leq \int_{|z|>R}e^{q|z|^2\left(\frac{|b|}{|z|}-\frac{(\beta-\alpha)}{2}\right)}dA(z)\\ &=\int_{|z|>R}e^{-q\gamma|z|^2}dA(z)<\infty. \end{aligned}\]

Hence, \[\label{int1} \int_{\mathbb {C}}e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z) =\left(\int_{|z|\leq R}+\int_{|z|>R}\right)e^{q|b||z|-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z)<\infty. \tag{20}\]

Case \(a\neq 0\).

Using (6) and functions \(g_{t}:[0,1]\to \mathbb {C}\) defined above, \[\begin{aligned} |g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)|&\asymp \Bigg|\sum\limits_{m_1,m_2}\frac{(n-m)!a^{m_2}}{m_1!m_2!}\left((2atz+b)^{m_1}e^{btz+a(tz)^2}-(2asz+b)^{m_1}e^{bsz+a(sz)^2}\right)\Bigg| \\ &\leq \sum\limits_{m_1,m_2}\frac{(n-m)!|a|^{m_2}}{m_1!m_2!}\bigg|\sum\limits_{i=0}^{\infty}\frac{(2atz+b)^{m_1}(btz+at^2z^2)^i-(2asz+b)^{m_1}(bsz+as^2z^2)^i}{i!}\bigg|, \end{aligned}\] where in the last estimate we used the power series representations of \(e^{btz+a(tz)^2}\) and \(e^{bsz+a(sz)^2}\). Inserting the binomial expansions of \((2atz+b)^{m_1}\), \((btz+at^2z^2)^i\), \((2asz+b)^{m_1}\) and \((bsz+as^2z^2)^i\), the right hand side of the above estimate is equal to;

\[\begin{aligned} &\sum\limits_{m_1,m_2}\left(\frac{(n-m)!|a|^{m_2}}{m_1!m_2!} \left| \sum\limits_{i=0}^{\infty}\frac{\left(\sum\limits_{l=0}^{m_1}\sum\limits_{j=0}^{i}\left(\begin{array}{c}m_1\\ l\end{array}\right)\left(\begin{array}{c}i\\ j\end{array}\right)2^{m_1-l}a^{m_1+i-l-j}b^{l+j}z^{m_1+2i-l-j}(t^{m_1+2i-l-j}-s^{m_1+2i-l-j}) \right)}{i!}\right|\right)\\ &\leq |t-s|\sum\limits_{m_1,m_2}\left(\frac{(n-m)!|a|^{m_2}}{m_1!m_2!}\sum\limits_{i=0}^{\infty}\frac{\big(\sum\limits_{l=0}^{m_1}\sum\limits_{j=0}^{i}\left(\begin{array}{c}m_1\\ l\end{array}\right)\left(\begin{array}{c}i\\ j\end{array}\right)2^{m_1-l}|a|^{m_1+i-l-j}|b|^{l+j}|z|^{m_1+2i-l-j}(m_1+2i-l-j) \big)}{i!}\right)\\ &\preceq |t-s|\sum\limits_{m_1,m_2}\bigg(\frac{(n-m)!|a|^{m_2}}{m_1!m_2!}\sum\limits_{i=0}^{\infty}\frac{(2|a||z|+|b|)^{m_1}(|b||z|+|a||z|^2)^i}{(i-1)!}\bigg) \\ &=|t-s|\sum\limits_{m_1,m_2}\bigg(\frac{(n-m)!|a|^{m_2}(2|a||z|+|b|)^{m_1}(|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2}}{m_1!m_2!}\bigg) \\ &\asymp |t-s|(|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2}\sum\limits_{m_1,m_2}\bigg(\frac{|a|^{m_2}(2|a||z|+|b|)^{m_1}}{m_1!m_2!}\bigg). \end{aligned}\]

Therefore, we have \[\label{iftu2} |g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)|\preceq |t-s|(|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2}\sum\limits_{m_1,m_2}\left(\frac{|a|^{m_2}(2|a||z|+|b|)^{m_1}}{m_1!m_2!}\right). \tag{21}\]

(i) \(0<p\leq q<\infty \ \ \text{or} \ \ 0<q<p\leq \infty\).

Using (21), the estimate in (1) and (13), we obtain \[\begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f&-\mathcal{T}_{g_{s}}^{n, m}f\|_{(q, \beta)}^q\asymp \int_{\mathbb {C}}\frac{\left|f^{(m)}(z)\right|^q\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|^qe^{-\frac{q\beta}{2}|z|^2}}{(1+|z|)^{nq}}dA(z)\\ &\preceq |t-s|^q\int_{\mathbb {C}}\Bigg(\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2} \left(\sum\limits_{m_1,m_2}\frac{|a|^{m_2}(2|a||z|+|b|)^{m_1}}{m_1!m_2!(1+|z|)^n} \right)^q \\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (|b||z|+|a||z|^2)^qe^{q|b||z|+q|a||z|^2-\frac{q(\beta-\alpha)}{2}|z|^2}\Bigg)dA(z)\\ &\preceq |t-s|^q\int_{\mathbb {C}}\left|f^{(m)}(z)\right|^qe^{-\frac{q\alpha}{2}|z|^2} (|b||z|+|a||z|^2)^qe^{q|b||z|+q|a||z|^2-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z) \\ &\preceq |t-s|^q\|f\|_{(p, \alpha)}^q\int_{\mathbb {C}}(1+|z|)^{mq} (|b||z|+|a||z|^2)^qe^{q|b||z|+q|a||z|^2-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z), \end{aligned}\] for \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\). The limit in (15) holds true, if we show that \[\int_{\mathbb {C}}(1+|z|)^{mq} (|b||z|+|a||z|^2)^qe^{q|b||z|+q|a||z|^2-\frac{q(\beta-\alpha)}{2}|z|^2}dA(z)<\infty.\]

But, since \(|a|<\frac{\beta-\alpha}{2}\), similar procedures as those leading to (20) shows that the integral is finite. Therefore, \(\mathcal{T}_{g_{t}}^{n, m}\) is continuous.

(ii) \(0<p\leq q=\infty\).

Similarly, using (21), for \(f\in \mathcal{F}_{p}^\alpha\) with \(\|f\|_{(p, \alpha)}\leq 1\),

\[\begin{aligned} \|\mathcal{T}_{g_{t}}^{n, m}f&-\mathcal{T}_{g_{s}}^{n, m}f\|_{(\infty, \beta)}\\&\asymp \sup_{z\in\mathbb {C}}\frac{\left|f^{(m)}(z)\right|\left|g_{t}^{(n-m)}(z)-g_{s}^{(n-m)}(z)\right|e^{-\frac{\beta}{2}|z|^2}}{(1+|z|)^{n}}\\ &\preceq |t-s|\sup_{z\in\mathbb {C}}\Bigg(\left|f^{(m)}(z)\right|e^{-\frac{\alpha}{2}|z|^2} \left(\sum\limits_{m_1,m_2}\frac{|a|^{m_2}(2|a||z|+|b|)^{m_1}}{m_1!m_2!(1+|z|)^n} \right)(|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2-\frac{(\beta-\alpha)}{2}|z|^2}\Bigg)\\ &\preceq |t-s|\sup_{z\in\mathbb {C}}\left|f^{(m)}(z)\right|e^{-\frac{\alpha}{2}|z|^2} (|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2-\frac{(\beta-\alpha)}{2}|z|^2}\\ &\preceq |t-s|\|f\|_{(p, \alpha)}\sup_{z\in\mathbb {C}}\left((1+|z|)^{m} (|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2-\frac{(\beta-\alpha)}{2}|z|^2}\right). \end{aligned}\]

Since \[\sup_{z\in\mathbb {C}}\left((1+|z|)^{m} (|b||z|+|a||z|^2)e^{|b||z|+|a||z|^2-\frac{(\beta-\alpha)}{2}|z|^2}\right)<\infty,\] we get \[\|\mathcal{T}_{g_{t}}^{n, m}-\mathcal{T}_{g_{s}}^{n, m}\|\preceq |t-s|.\]

Letting \(t\) goes to \(s\), the conclusion follows. ◻

Proof. (i) Let \(\mathcal{T}_{g_1}^{n, m}\) and \(\mathcal{T}_{g_2}^{n, m}\) be two compact operators in \(\mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\). We need to show that \(\mathcal{T}_{g_1}^{n, m}\) and \(\mathcal{T}_{g_2}^{n, m}\) belong to the same path-connected component. First, by Theorem 3, \(\mathcal{T}_{g_1}^{n, m}\) and \(\mathcal{T}_{g_1(0)}^{n, m}\) belong to the same path-connected component, and \(\mathcal{T}_{g_2}^{n, m}\) and \(\mathcal{T}_{g_2(0)}^{n, m}\) also belong to the same path-connected component. But, by Proposition 2, \(\mathcal{T}_{g_1(0)}^{n, m}\) and \(\mathcal{T}_{g_2(0)}^{n, m}\) belong to the same path-connected component. Hence, \(\mathcal{T}_{g_1}^{n, m}\) and \(\mathcal{T}_{g_2}^{n, m}\) belong to the same path-connected component of \(\mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\).

(ii) Follows from \((i)\) and Corollary 2. ◻

Proof. The forward implication follows from Proposition 3. We will prove the backward implication. Suppose \(g\) is a polynomial of degree \(2(n-m)\), that is, \[g(z)=a_{2(n-m)}z^{2(n-m)}+a_{2(n-m)-1}z^{2(n-m)-1}+…+a_1z+a_0,\] and \(a_{2(n-m)}\neq 0\). It is enough to show there exists a positive number \(\mu\) such that \[\|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\|\succeq \mu,\] for all \(\mathcal{T}_{g_1}^{n, m}\in \mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\) such that \(\mathcal{T}_{g_1}^{n, m}\neq \mathcal{T}_{g}^{n, m}\), that is, \(g_1\) has the form \[g_1(z)=b_{2(n-m)}z^{2(n-m)}+b_{2(n-m)-1}z^{2(n-m)-1}+…+b_1z+b_0,\] with \(b_k\neq a_k\) for some positive integer \(k\), \(n-m\leq k\leq 2(n-m)\). Thus, applying \(\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\) to the function \(k_{(w, \alpha)}(z)\) in \(\mathcal{F}_{p}^\alpha\), using (1) and the inclusion \(\mathcal{F}_{q}^\beta\subseteq \mathcal{F}_{\infty}^\beta\), \(0<q\leq \infty\), we obtain \[\label{iftu3} \begin{aligned} \|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\|&\geq \|\mathcal{T}_g^{n, m}k_{(w, \alpha)}-\mathcal{T}_{g_1}^{n, m}k_{(w, \alpha)}\|_{(q, \beta)} \succeq \|\mathcal{T}_g^{n, m}k_{(w, \alpha)}-\mathcal{T}_{g_1}^{n, m}k_{(w, \alpha)}\|_{(\infty, \beta)} \\ &\asymp \sup_{z\in\mathbb {C}}\frac{\left|k_{(w, \alpha)}^{(m)}(z)\right|\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|}{(1+|z|)^n}e^{-\frac{\beta}{2}|z|^2} \\ &\succeq \frac{|w|^m\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|}{(1+|z|)^n}e^{-\frac{\beta}{2}|w-z|^2} \\ &\geq \frac{|(a_{2(n-m)}-b_{2(n-m)})\left(\frac{(2(n-m))!}{(n-m)!}\right)z^{n-m}+…+(a_{n-m}-b_{n-m})(n-m)!|}{(1+|z|)^{n-m}}, \end{aligned} \tag{22}\] where the last inequality is obtained by putting \(z=w\). If \(a_{n-m}\neq b_{n-m}\), then we set \(z=0\) in the above estimate to obtain \[\|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\|\succeq \mu=|a_{n-m}- b_{n-m}|\neq 0.\]

If \(a_{n-m}= b_{n-m}\), then there exists some \(k\), \(n-m<k\leq 2(n-m)\), such that \(a_k\neq b_k\). Let \(i\) be the smallest of such \(k\). Then, from (22), \[\begin{aligned} &\frac{\left|(a_{2(n-m)}-b_{2(n-m)})\left(\frac{(2(n-m))!}{(n-m)!}\right)z^{n-m}+…+(a_{n-m}-b_{n-m})(n-m)!\right|}{(1+|z|)^{n-m}}\\ \qquad&=|z|^i\frac{\left|(a_{2(n-m)}-b_{2(n-m)})\left(\frac{(2(n-m))!}{(n-m)!}\right)z^{n-m-i}+…+(a_{i}-b_{i})\left(\frac{i!}{(2(n-m)-i)!}\right)\right|}{(1+|z|)^{n-m}}\\ \qquad &\geq \frac{\left|(a_{2(n-m)}-b_{2(n-m)})\left(\frac{(2(n-m))!}{(n-m)!}\right)z^{n-m-i}+…+(a_{i}-b_{i})\left(\frac{i!}{(2(n-m)-i)!}\right)\right|}{(1+|z|)^{n-m-i}}. \end{aligned}\]

Setting \(z=0\), we similarly obtain \[\|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\|\succeq \mu=|a_{i}- b_{i}|\neq 0.\]

This completes the proof. ◻

Proof. Similarly the forward implication follows from Proposition 4. Let \(g(z)=e^{bz+az^2}\) for some \(a, b\in \mathbb {C}\) with \(|a|=\frac{\beta-\alpha}{2}\) and either \(b=0\) or \(a=-\frac{(\beta-\alpha)b^2}{2|b|^2}\). Consider \(\mathcal{T}_{g_1}^{n, m}\in \mathcal{T}^{n, m}(\mathcal{F}_{p}^\alpha, \mathcal{F}_{q}^\beta)\) such that \(\mathcal{T}_{g_1}^{n, m}\neq \mathcal{T}_{g}^{n, m}\), that is, \(g_1\) has the form \(g_1(z)=e^{b_1z+a_1z^2}\) for some \(a_1, b_1\in \mathbb {C}\) with \(|a_1|< \frac{\beta-\alpha}{2}\) or \(|a_1|= \frac{\beta-\alpha}{2}\) and either \(b_1=0\) or \(a_1=-\frac{(\beta-\alpha)b_1^2}{2|b_1|^2}\), with \(a_1\neq a\) or \(b_1\neq b\).

(i) \(0<p\leq q<\infty\).

Applying \(\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\) to the normalized kernel function, using (1), the estimate \(1+|z|\asymp 1+|w|\) for \(z\in D(w,1)\), subharmonicity of \(|g^{(n-m)}-g_1^{(n-m)}|^q\) and the formula in (6), \[\begin{aligned} \|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\|^q &\succeq \int_{\mathbb {C}}\frac{\left|k_{(w, \alpha)}^{(m)}(z)\right|^q\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}}e^{-\frac{\beta q}{2}|z|^2}dA(z)\\ &=\int_{\mathbb {C}}\frac{|w|^{mq}\left|e^{\alpha z\overline{w}}\right|^q\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|^q}{(1+|z|)^{nq}}e^{-\frac{\beta q}{2}|z|^2-\frac{\alpha q}{2}|w|^2}dA(z)\\ &\geq \int_{D(w,1)}\frac{|w|^{mq}\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|^qe^{-\frac{(\beta-\alpha)q}{2}|z|^2}}{(1+|z|)^{nq}}e^{-\frac{\alpha q}{2}|w-z|^2}dA(z) \\ &\succeq \frac{1}{(1+|w|)^{(n-m)q}}\int_{D(w,1)}\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|^qe^{-\frac{(\beta-\alpha)q}{2}|z|^2}dA(z) \\ &\succeq \frac{\left|g^{(n-m)}(w)-g_1^{(n-m)}(w)\right|^q}{(1+|w|)^{(n-m)q}}e^{-\frac{(\beta-\alpha)q}{2}|w|^2}\\ &=\left(\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}\left(e^{bw+aw^2}(2aw+b)^{m_1}a^{m_2}-e^{b_1w+a_1w^2}(2a_1w+b_1)^{m_1}a_1^{m_2}\right)\right|^q}{(1+|w|)^{(n-m)q}}\right)e^{-\frac{(\beta-\alpha)q}{2}|w|^2}, \end{aligned}\] for all \(w\in\mathbb {C}\). After putting \(w=0\) and since \(a\neq a_1\) or \(b\neq b_1\), we obtain \[\|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\| \succeq \mu=\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(b^{m_1}a^{m_2}-b_1^{m_1}a_1^{m_2})\right|\neq 0.\]

(ii) \(0<p\leq q=\infty\):

Similarly, for this case we have \[\begin{aligned} \|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\| &\succeq \sup_{z\in\mathbb {C}}\frac{\left|k_{(w, \alpha)}^{(m)}(z)\right|\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|}{(1+|z|)^{n}}e^{-\frac{\beta }{2}|z|^2} \\ &=\sup_{z\in\mathbb {C}}\frac{|w|^{m}\left|e^{\alpha z\overline{w}}\right|\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|}{(1+|z|)^{n}}e^{-\frac{\beta }{2}|z|^2-\frac{\alpha }{2}|w|^2}\\ &\geq \frac{|w|^{m}\left|g^{(n-m)}(z)-g_1^{(n-m)}(z)\right|e^{-\frac{(\beta-\alpha)}{2}|z|^2}}{(1+|z|)^{n}}e^{-\frac{\alpha }{2}|w-z|^2} \\ &\succeq \frac{\left|g^{(n-m)}(w)-g_1^{(n-m)}(w)\right|}{(1+|w|)^{(n-m)}}e^{-\frac{(\beta-\alpha)}{2}|w|^2}, \end{aligned}\] where the last estimate is obtained by putting \(w=z\). Using the formula in (6), the right hand side of the above estimate is further equal to; \[\left(\frac{\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}\left(e^{bw+aw^2}(2aw+b)^{m_1}a^{m_2}-e^{b_1w+a_1w^2}(2a_1w+b_1)^{m_1}a_1^{m_2}\right)\right|}{(1+|w|)^{(n-m)}}\right) e^{-\frac{(\beta-\alpha)}{2}|w|^2},\] for all \(w\in\mathbb {C}\). Putting \(w=0\) and since \(a\neq a_1\) or \(b\neq b_1\), we obtain \[\|\mathcal{T}_g^{n, m}-\mathcal{T}_{g_1}^{n, m}\| \succeq \mu=\left|\sum\limits_{m_1,m_2} \frac{(n-m)!}{m_1!m_2!}(b^{m_1}a^{m_2}-b_1^{m_1}a_1^{m_2})\right|\neq 0.\] ◻