Limit operations and Schauder combinations

Question

Let $V$ be a Banach (Hilbert if needed) space with a countable (orthonormal if needed) Schauder basis $\{v^i:i\in\mathbb{R}\}$. Does this basis always work like the Fourier basis $\{1,\cos(x),\sin(x),\cos(2x),\sin(2x),\cdots\}$ of $L^2(S^1)$ when it comes to interchange limit operations?

For example, if $v_n$ is sequence on $V$ converging to $v$ and $v_n=\sum_i \alpha^n_i v^i$ and $v=\sum_i \alpha_iv^i$, then $\lim_n \alpha^n_i=\alpha_i$? Or even, if $v(t)=\sum_i \alpha_i(t)v^i$ is a smooth path on $V$, $v'(t)=\sum_i \alpha'_i(t)v^i$? Or even for integrations and etc...

Answer 1

Yes. Let's first consider the case where $V$ is a Hilbert space and the basis is orthonormal. With your notation we have

$a_i^n = \langle v^i,v_n\rangle$ and $a_i=\langle v^i,v\rangle$ so that

$$\begin{align}\vert a_i^n -a_i\vert &=\vert \langle v^i, v_n-v \rangle\vert\\ & \leq \|v^i\|\|v_n-v\|\to 0\end{align}$$

More generally, if $V$ is a Banach space, the component functionals $v^i_*:V\to\mathbb{F}$ are continuous and defined

$$v_*^i(v) = a_i$$ with your notation. So we can make a similar argument

$$\begin{align}\vert a_i^n -a_i\vert &=\vert v_*^i(v_n-v)\vert\\ & \leq \|v_*^i\|\|v_n-v\|\to 0\end{align}$$ where the first norm is the operator norm of the component functional.

Answer 2

Hint ... For each $j$ you need to know that $L_j$ is a continuous linear functional, where: $$L_j\left(\sum_i a_i v^i\right) = a_j .$$For this look up the definition of Schauder basis. Or (easier) in Hilbert space with orthonormal basis use the inner product.