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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?

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  • $\begingroup$ NO. Assuming your last line means "$ U\subset$...", not " $U\in$ ...". Let $T=0$ and $U=B_Y(0,1)$.... For $u\in U$ take $s>1$ such that $s\|u\|<1$... Let $t=s^{-1}$ and $\bar x=0$ and $\bar u=su$.... Assuming $\dim Y>0.$ $\endgroup$ – DanielWainfleet Feb 10 '18 at 22:18
  • $\begingroup$ Thank you, I have edited the last line. However, in your comment $t$ depends on $u$. I fix a $t$ which is good for every $u\in U$. $\endgroup$ – aleio1 Feb 10 '18 at 23:04
  • $\begingroup$ Sorry. I mis-read it. $\endgroup$ – DanielWainfleet Feb 10 '18 at 23:17
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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.

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  • $\begingroup$ If we assume that $U$ is bounded, then $\|u_k\|<M \ \ \forall k$ and so the last limit is 0. Isn't it? $\endgroup$ – aleio1 Feb 10 '18 at 22:17
  • $\begingroup$ Yes, if we assume that $U$ is bounded then the last limit is $0$. $\endgroup$ – Fimpellizieri Feb 10 '18 at 22:19
  • $\begingroup$ Notice however that this assumption is important. If we take $U=X=Y$ and $T=\text{Id}$, then it's not hard to cook up some examples. With $t=\frac12$ and $\alpha>0$ we can say that $u=-\alpha\,u + \frac12 (2+2\alpha)\, u$ and making a recurrence out of this we can see that the limit above diverges for $u \neq 0$. $\endgroup$ – Fimpellizieri Feb 10 '18 at 22:26

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