# 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 $x=\sum_{n=1}^\infty c_nx_n$.

• Let $H$ be your Hilbert space with ONB $\{ e_n : n \in \mathbb{N}\}$. Let $T\colon H \to X$ be the isomorphism. [You need not assume $X = H$.] Let $x_n = T(e_n)$ for $n\in\mathbb{N}$. $\{x_n : n \in \mathbb{N}\}$ is bounded because …. Start with the existence of the coefficient sequence. Next, attack uniqueness. Finally, show that it's unconditional, so you can reorder as you please. Commented Nov 24, 2014 at 18:52
• Well $T$ is bounded if and only if $T$ is continuous. Commented Nov 24, 2014 at 20:37

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 $u\in H$.