Q:

Prove that the image of an orthonormal basis through a linear, invertible and bounded transformation is a bounded and unconditional Schauder basis.

Prove that the image of an orthonormal basis through a linear, invertible and bounded transformation is a bounded and unconditional Schauder basis.
I am having trouble finding a starting point for this proof.
I know that a Schauder basis for $X$ if to each vector $x$ in the space there corresponds a unique sequence of scalars $\{c_1,c_2,\dots\}$ such that $x=\sum_{n=1}^\infty c_nx_n$.

A:

Let $H$ be a separable Hilbert space with orthonormal basis $\{ e_n : n\in\mathbb{N}\}$, and $X$ a separable Banach space. Further let $T\colon H\to X$ be a bounded invertible linear map (aka, an isomorphism in the category of topological vector spaces).
Then you need to prove that $\{ Te_n : n\in\mathbb{N}\}$ is a bounded unconditional Schauder basis of $X$.
Let $x_n := Te_n$. It is clear that $\{x_n : n\in\mathbb{N}\}$ is a bounded family, since
$$\lVert x_n\rVert_X = \lVert Te_n\rVert_X \leqslant \lVert T\rVert\cdot\lVert e_n\rVert_H = \lVert T\rVert.$$
Now, to show that $\{ x_n : n\in\mathbb{N}\}$ is an unconditional Schauder basis, you use the fact that $\{ e_n : n\in\mathbb{N}\}$ is an unconditional Schauder basis of $H$. The orthonormality of $\{ e_n : n\in\mathbb{N}\}$ is not important, it just makes the estimates for the unconditionality a little more convenient.
We start by showing the existence of the sequence of coefficients.
Let $x\in X$. Then, since $\{e_n: n\in\mathbb{N}\}$ is a Schauder basis in $H$, there exists a (unique) sequence $(c_n)$ of scalars such that
$$T^{-1}x = \lim_{N\to\infty}\sum_{n=0}^N c_n\cdot e_n.$$
Now, since $T$ is continuous,
$$x = T(T^{-1}x) = T\left(\lim_{N\to\infty} \sum_{n=0}^N c_n\cdot e_n\right) = \lim_{N\to\infty} \sum_{n=0}^N c_n\cdot Te_n = \lim_{N\to\infty} \sum_{n=0}^N c_n\cdot x_n.$$
Next, for the uniqueness of the coefficient sequence, suppose
$$\lim_{N\to\infty} \sum_{n=0}^N c_n \cdot x_n = 0.$$
Use the continuity of $T^{-1}$ and the fact that $\{e_n : n\in\mathbb{N}\}$ is a Schauder basis of $H$ to conclude $c_n = 0$ for all $n$.
For the unconditionality of $\{ x_n : n\in\mathbb{N}\}$, you use the unconditionality of $\{ e_n : n \in \mathbb{N}\}$ and the fact that $T$ as well as $T^{-1}$ are continuous, which gives you constants $c_1, c_2 > 0$ with
$$c_1 \lVert u\rVert_H \leqslant \lVert Tu\rVert_X \leqslant c_2\lVert u\rVert_H$$
for all $u\in H$.

