What I read in a book I could not understand, some one please help.
Let $\mathcal{F}=\{f:\mathbb{C^n}\rightarrow\mathbb{C}: \text{$f$ is holomorphic and}\int_{\mathbb{C}^n}\lvert f(z)\rvert^2e^{-\lVert z \rVert^2}d\nu(z)<\infty, \nu \text{is area measure}\}$ and $\langle f,g\rangle=\int_{\mathbb{C}^n}f\bar{g}e^{-\lVert z \rVert^2}d\nu(z)$. They want to show that a bounded linear map $T:\mathcal{F}\rightarrow \mathcal{F}$,defined by $Tf(z)=f(\Sigma z+b), \text{ $\Sigma$ is non singular diagonal squre matrix of order $n$ with $\lVert\Sigma\rVert<1$ and $b\in\mathbb{C}^n$ }$,is compact.
And they are saying that "to show that $T$ is compact,it suffices to show that a sequence {$f_n$} that is bounded in $\mathcal{F}$ and converges to zero uniformly on compact subsets of $\mathbb{C}^n$ has its image under $T$ converges to 0 in norm " . I know that $T$ is compact if {$T(x_n)$} has convergent subsequence when {$x_n$} is bounded, with respective norms.
I could not able to relate this thing with my known one.