Question about linear bounded operator in Banach spaces. Let $T:X\to Y$ a linear bounded operator between Banach spaces. Let $U$ a neighbourhood of $0\in Y$, $t\in(0,1)$ and suppose that $\forall u\in U$ $\exists \bar x\in X$ with $\|\bar x\|\le1$ and $\bar u\in U$ such that $$u=T\bar x +t\bar u.$$
Then if I take $u\in U$, I can write
$$u=T\left(\sum_{i=0}^{\infty} t^ix_i\right), \ \ \ \|x_i\|\le1$$ and that series has sense since $$\|\sum_{i=0}^{\infty} t^ix_i\|\le \sum_{i=0}^{\infty} t^i=C<+\infty.$$
In this way I obtain that $$U\subset T\left(B_X(0,C)\right).$$
Is it correct?
 A: For each $u\in U$, there is some sequence $(x_n)_{n\geq 0}\subset X$ with $\lVert x_i\rVert\leq 1$, and some sequence $(u_n)_{n\geq 0}\subset U$ with
$$u_n=Tx_{n}+t\cdot u_{n+1},$$
where $u=u_0$.
Then:
\begin{align}
u=u_0
&=Tx_0+t\cdot u_1&&
=T\left(\sum_{i=0}^0t^{i}x_i\right)+t^1\cdot u_1\\
&=Tx_0+t\cdot \left(Tx_1+t\cdot u_2\right)&&
=T\left(\sum_{i=0}^1t^{i}x_i\right)+t^2\cdot u_2\\
&=T\left(\sum_{i=0}^1t^{i}x_i\right)+t^2\left(Tx_2+t\cdot u_3\right)&&
=T\left(\sum_{i=0}^2t^{i}x_i\right)+t^3\cdot u_3\\
&= \dots\,,
\end{align}
and more generally, for each $k\in\mathbb N$ we'll have
$$u=T\left(\sum_{i=0}^kt^{i}x_i\right)+t^{k+1}\cdot u_{k+1}.$$
This does not quite imply that $u=T\left(\sum_{i=0}^{\infty}t^{i}x_i\right)$.
Indeed, this happens if and only if
$$\lim_{k\to\infty} \left\lVert u-T\left(\sum_{i=0}^kt^{i}x_i\right) \right\rVert
=\lim_{k\to\infty} \left\lVert t^{k+1}\cdot u_{k+1}\right\rVert=
\lim_{k\to\infty} t^k\,\lVert  u_k\rVert $$
equals $0$.
It is not clear to me that this happens as per the hypotheses.
