Let $T$ be a bounded linear operator $l^2(\mathbb{N})$. By picking an orthonormal basis $B=\{b_i\}_{i\in\mathbb{N}}$ for $l^2(\mathbb{N})$, we can write $T$ as an infinite matrix $(T_{ij})_{i,j\in\mathbb{N}}$, defined by $$T_{ij}=\langle Tb_i,b_j\rangle.$$ Suppose that $(T_{ij})$ satisfies the following property: for any $k,l\in\mathbb N$, $$\sup_{i,j}i^k j^l |T_{ij}| <\infty.$$ Question 1: Is this property invariant under a change of basis? In other words, if $B'=\{b'_i\}_{i\in\mathbb{N}}$ is another orthonormal basis of $l^2(\mathbb{N})$, with respect to which $T$ takes the form $(T_{ij}')_{i,j\in\mathbb{N}}$, then is it still true that $$\sup_{i,j}i^k j^l |T'_{ij}| <\infty$$ for any $k,l\in\mathbb N$?
Perhaps an easier question is the following. Let $B$ be the standard orthonormal basis of $l^2(\mathbb{N})$, and let $x\in l^2(\mathbb{N})$ be an element whose sequence of coefficients $(a_i)$ with respect to $B$ are rapidly decreasing, in the sense that for any $k\in\mathbb N$, $$\sup_{i}i^k |a_i| <\infty.$$
Question 2: Let $B'$ is another orthonormal basis for $l^2(\mathbb{N})$, and let $(a_i')$ be the sequence of coefficients for $x$ with respect to $B'$. Then is it true that for all $k\in\mathbb N$, $$\sup_{i}i^k |a_i'| <\infty?$$
It's easy to answer the vector part. Let $B'=\{f_n\}$. Define $$ x=\frac{\sqrt6}\pi\sum_n\frac1n\,f_n. $$ Now form $B=\{e_n\}$ by taking $e_1=x$ and completing to an orthonormal basis. So $a_1=1$, $a_2=a_3=\cdots=0$. This gives $$ \sup_i i^k|a_i|=1,\qquad k\in\mathbb N. $$ Meanwhile, $$ \sup_i i^k\,|a_i'|=\sup_i\frac{i^{k}}i=\infty,\qquad k\in\mathbb N\setminus\{1\}. $$
We can then push this to the operator case. Let $T=xx^*$; that is, $T$ is the rank-one projection onto the span of $x$. Then, in the basis $B$ we have $T_{11}=1$ with $T_{ij}=0$ whenever $(i,j)\ne(1,1)$. But on $B'$, $$ T_{ij}=a_ia_j=\frac1{ij} $$ and the supremum is infinite.