Riesz basis in Hilbert space We know that a collection of vectors $\{x_{k}\}$ in a Hilbert space called Riesz basis if it is an image of orthonormal for H under invertible linear transformation. How to prove that there is constants $A,B$ such that for all $x\in H$
$$
A||x||^2\leq\sum_{k}\langle x,x_k \rangle^2\leq B||x||^2?
$$
 A: I think what you exact mean is that there exist a bounded linear operator $L$ on $H$ with bounded inverse and an orthonormal basis $\{e_k\}$ of $H$, such that $Le_k=x_k$ for every $k$. If so, then $ \langle x,x_k \rangle = \langle x,Le_k \rangle = \langle L^*x,e_k \rangle $, where $L^*$ is the adjoint operator of $L$. Therefore, $\sum_k| \langle x,x_k \rangle |^2=\sum_k| \langle L^*x,e_k \rangle |^2=\|L^*x\|^2$. Since $L$ is bounded with a bounded inverse, $L^*$ is also bounded with a bounded inverse. The conclusion follows. 
A: We assume the Hilbert space separable (otherwise, take $x$ orthogonal to all the $x_n$ and different from $0$). Let $T\colon H\to H$ linear invertible and $\{e_n\}$ an orthonormal basis such that $Te_n=x_n$ for all $n$. We have by Bessel-Parseval that 
$$\sum_k\langle x,x_k\rangle^2=\sum_k\langle x,Te_k\rangle^2=\sum_k\langle T^*x,e_k\rangle^2= \lVert T^*x\rVert^2\leq \lVert T^*\rVert^2\lVert x\rVert^2.$$
As $T$ is invertible, so is $T^*$, so $\lVert T^*x\rVert$ is below bounded by an universal constant (which doesn't depend on $x$) when $x$ is in the unit ball.
