Are coefficients of a sequence of vector in Banach space limited? Let $V$ be a Banach separable space over $\mathbb{C}$ and $B$ be a Hamel basis of $V$
Let $U=\operatorname{span}(B)$
Let $\{v_m\} \in U$ be a sequence of $U$ with
$$
v_m = \sum_j a_{m,j}b_{m,j}
$$
and $a_{m,j} \in \mathbb{C}$ and $b_{m,j} \in B$ and the sum is finite and such that
$$
\lim_{m \to \infty} v_m = b
$$
with $b \in B$
My question is if $|a_{m,j}|$ is limited, that is $\exists M>0 : \forall m,j |a_{m,j}| < M$
Thanks.
 A: In general, the set of coefficients has no reason to be bounded. I'll look only at sequences $(v_m)$ in $\operatorname{span} B$ with $\lim\limits_{m\to \infty} v_m = 0$, if you want a different limit in $U := \operatorname{span} B$, that changes only a finite number of coefficients overall, by a constant amount, so doesn't change whether the coefficients remain bounded or not.
If there is a sequence $(\beta_k)$ in $B$ with $\lVert \beta_k\rVert \to 0$, then $v_m = \lVert \beta_m\rVert^{-1/2}\cdot \beta_m$ is a sequence in $U$ converging to $0$ with unbounded coefficient set. So a necessary condition to have bounded coefficient sets $\{ a_{m,j}\}$ is that there is a $c > 0$ with $\lVert b\rVert \geqslant c$ for all $b \in B$. But that is not a sufficient condition. If there is a $b\in B$ with $b \in \overline{\operatorname{span} (B \setminus \{b\})}$, we can find $w_m \in \operatorname{span}(B\setminus \{b\})$ with $\lVert w_m - b\rVert < m^{-2}$, and $v_m = mw_m - mb$ is a sequence in $U$ with $v_m \to 0$ whose coefficient set is unbounded. We can combine these two examples: if there is a sequence $(\beta_k)$ in $B$ such that
$$\operatorname{dist}\bigl(\beta_k, \operatorname{span}(B\setminus \{\beta_k\})\bigr) \to 0,$$
then we can find a sequence $(v_m)$ in $U$ with $v_m \to 0$ such that the coefficient set $\{a_{m,j}\}$ is unbounded. So a necessary condition is that
$$\delta_B := \inf \Bigl\{\operatorname{dist}\bigl(b, \operatorname{span}(B\setminus \{b\})\bigr) : b \in B\Bigr\} > 0.\tag{$\ast$}$$
This condition is also sufficient. For suppose $(\ast)$ holds, and $v_m$ is a sequence in $U$ such that $\{a_{m,j}\}$ is unbounded. Extracting a subsequence, we can assume that
$$s_m := \sup_j \{\lvert a_{m,j}\rvert\} > m$$
for every $m$. Picking $j_m$ so that $\lvert a_{m,j_m}\rvert = s_m$, we see that
$$\lVert v_m\rVert = \lvert a_{m,j_m}\rvert\cdot \Biggl\lVert b_{j_m} - \sum_{j\neq j_m} \frac{-a_{m,j}}{a_{m,j_m}} b_j\Biggr\rVert \geqslant \lvert a_{m,j_m}\rvert\cdot \delta_B > m\cdot \delta_B,$$
so the sequence $(v_m)$ is unbounded, and hence not convergent.
