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

  • $\begingroup$ 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. $\endgroup$ Commented Nov 24, 2014 at 18:52
  • $\begingroup$ Well $T$ is bounded if and only if $T$ is continuous. $\endgroup$ Commented Nov 24, 2014 at 20:37

1 Answer 1


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 $u\in H$.


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