Let $f\in L^2(\Bbb R^n)$ and $d\geq0$. Then $$\int_{\Bbb R^n}\int_{\Bbb R^n} \frac{|f(x)||\hat f(y)|}{(1+|x|+|y|)^d} e^{|x||y|} dxdy<\infty$$ implies that $f=0$ when $d\leq n$.
The analogue of this theorem in the Heisenberg group $ \Bbb H^n$ for $f\in L^2(\Bbb H^n)$ is as follows $$\int_{\Bbb H^n}\int_{\Bbb R} \frac{|f(z,t)|\quad||\hat f(\lambda)||_{HS}}{(1+|t|+| \lambda|)^d} e^{|t||\lambda|} d\mu(\lambda)dzdt<\infty$$ implies that $f=0$ when $d\leq n/2+1$. for more detail we can see this https://www.researchgate.net/publication/231999338_An_Analogue_of_Beurling's_Theorem_for_the_Heisenberg_Group
my question why the author doesen't put the following condition $$\int_{\Bbb H^n}\int_{\Bbb R} \frac{|f(z,t)|\quad||\hat f(\lambda)||_{HS}}{(1+|(z,t)|+| \lambda|)^d} e^{|(z,t)||\lambda|} d\mu(\lambda)dzdt<\infty.$$