Are coefficients of a sequence of vector in a space generated by a sequence, bounded? Let $V$ be a Banach space over $\mathbb{C}$
Let $\{b_n\}_{n \in \mathbb{N}} \subset V$ such that $b_n \to b_0$ and let $B=\{b_n\}_{n \in \mathbb{N}} \cup \{b_0\}$
Let $B$ linearly independent and $\forall b \in B: \left\| b \right\|> m$
Let $U=\operatorname{span}(B)$
Let $\{v_m\}_{m \in \mathbb{N}}$ be a sequence of U such that $v_m \to b_1$
We have that 
$$
v_m=\sum_j a_{m,j}b_{m,j}
$$
with $a_{m,j} \in \mathbb{C}$ and $b_{m,j} \in B$
My question is if $|a_{m,j}|$ is limited, that is $\exists M>0 : \forall m,j |a_{m,j}| < M$
Thanks for any suggestion
 A: No, this is false. To see this, choose $j_m$ such that $\|b_{j_m}-b_0\| < 2^{-m}$ and put
$$
a_{m,j} :=
\begin{cases}
0 &\text{for }j\notin\{0,j_m\}\\
2^{m/2} &\text{for }j = 0\\
-2^{m/2} &\text{for }j = j_m.
\end{cases}
$$
Then the $a_{m,j}$'s are definitely not bounded, but
$$
\|v_m\| = \left\|\sum_{j=0}^\infty a_{m,j}b_j\right\| = \|a_{m,0}b_0 + a_{m,j_m}b_{j_m}\| = 2^{m/2}\|b_{j_m} - b_0\| < 2^{-m/2},
$$
which tends to zero. Hence $v_m\to 0$ as $m\to\infty$.
A: The answer is no.
Take $V = (L^1[0,1],\|\cdot\|_1)$.
Let $B = \{1, 1+x, 1+x^2, \ldots\}\subseteq V$. Notice that $B$ is linearly independent, $\|b\|_1 \ge 1, \forall b \in B$, and we have $1+x^n \xrightarrow{n\to\infty} 1$ since:
$$\big\|(1+x^n) - 1\big\|_1 = \int_0^1 x^n \,dx =\frac{1}{n+1}\xrightarrow{n\to\infty} 0$$
Also, $\mathrm{span}\,B$ is in fact $\mathcal{P}[0,1]$, the space of all polynomials on $[0,1]$.
Define a sequence $(f_n)_{n=1}^\infty$ in $\mathrm{span}\,B$ as $$f_n(x) = 1 + x + (1-x)^n, \quad x \in [0,1]$$
We have $f_n \xrightarrow{n\to\infty} 1+x$:
$$\big\|1 + x + (1-x)^n - (1+x)\big\|_1 = \int_0^1 (1-x)^n \,dx =\frac{1}{n+1}\xrightarrow{n\to\infty} 0$$
However, we have:
$$1 + x + (1-x)^n = 1 + x + \sum_{k=0}^n {n\choose k}(-1)^kx^k$$
so for $n > 2$ the coefficient of the element $1+x^{n-1}$ in $f_n$ must be be $(-1)^{n-1}n$, since $1+x^{n-1}$ is the only element of $B$ with the power $x^{n-1}$.
Thus, the coefficients are unbounded.
