If $\{v_n\}_{n \in \mathbb{N}}$ is a basic sequence then is $\{v_n+v_0\}_{n \in \mathbb{N}}$ a basic sequence too?

Question

Let $V$ be a Banach space over $\mathbb{C}$

Let $\{v_n\}_{n \in \mathbb{N}} \subset V$ be a basic sequence in $V$

Let $v_0 \in V$ such that $v_0 \notin \{v_n\}_{n \in \mathbb{N}}$

My question is:

Is the sequence $\{v_n+v_0\}_{n \in \mathbb{N}}$ a basic sequence ?

Thanks.

Answer 1

Not necessarily.

Let $(e_i)$ be the standard unit vector basis of $\ell_2$. Take $v_n=e_{n+1}$ for $n\ge1$ and $v_0=e_1$. Then $(v_n)_{n=1}^\infty$ is a basic sequence whose closed linear span does not contain $e_1$.

Let $y_n=v_n+v_0$ for $n\ge1$. Note that for each $n$, the norm of $$e_1-\sum_{i=1}^n{1\over n} y_n=(0,\underbrace{ 1/n,1/n,\ldots,1/n,}_{(n-1)\rm - terms}\ 0,\ldots)$$ is $\sqrt{(n-1)/n^2}$. It follows from this that $e_1$ is in the closed linear span of the $y_i$.

But, a moment's thought reveals that $e_1$ cannot be written in the form $\sum\alpha_i y_i$; thus, $(y_i)$ is not basic.

(Alternatively, one can show the projections associated to $(y_i)$ are not uniformly bounded; see my comment above.)