$X,Y$ be Banach , $T \in \mathcal B(X,Y)$ be onto ; then , for every sequence $y_n \to y \in Y$ , $\exists x_n \to x\in X$ s.t. $T(x_n)=y_n , T(x)=y$?

Question

Let $X,Y$ be Banach spaces , $T:X \to Y$ be a surjective continuous linear transformation , then is it true that for every convergent sequence $\{y_n\}$ in $Y$ , converging to $y \in Y$ , there exist a sequence $\{x_n\}$ in $X$ , converging to $x \in X$ , such that $T(x_n)=y_n , \forall n \in \mathbb N$ and $T(x)=y$ ?

Answer 1

If $T$ is onto, this is the open mapping theorem:

You first choose $x$ with $Tx=y$. Then you look at the images of small open balls around $x$. These images are open and contain $y$ and thus contain all but finitely many $y_n$. Hence, you can choose $x_n$ in the balls as needed.

Answer 2

We have the following result which answers the question.

Let $X$ and $Y$ be Banach spaces and $T\colon X\to Y$ a surjective bounded linear operator. Let $(y_{n})_{n\in\mathbb{N}}$ be a sequence in $Y$ converging to some $y\in Y$. Then for every $x\in X$ where $Tx = y$ there is a sequence $(x_{n})_{n\in\mathbb{N}}$ in $X$ such that $(x_{n})_{n\in\mathbb{N}}$ converges to $x$ with $Tx_{n} = y_{n}$ for each $n\in\mathbb{N}$.

Here is a proof. First, note that by this answer there is some $C > 0$ such that for every $z\in Y$ there is some $x\in X$ such that $\|x\| \leq C\|z\|$ and $Tx = z$. [In fact, for any $\varepsilon > 0$ we can take $C = \|T\| + \varepsilon$.] Note that this result makes use of the open mapping theorem.

Let $x\in X$ such that $Tx = y$. For each $n\in\mathbb{N}$ choose $w_{n}\in X$ such that $\|w_{n}\| \leq C\|y_{n} - y\|$ and $Tw_{n} = y_{n} - y$. Define $x_{n} := w_{n} + x$ for each $n\in\mathbb{N}$. We have \begin{equation} Tx_{n} = Tw_{n} + Tx = (y_{n} - y) + y = y_{n} \tag{1} \end{equation} and \begin{equation} \|x_{n} - x\| = \|w_{n}\| \leq C\|y_{n} - y\| \tag{2} \end{equation} for each $n\in\mathbb{N}$. Because $(y_{n})_{n\in\mathbb{N}}$ converges to $y$, it follows from $(2)$ that $(x_{n})_{n\in\mathbb{N}}$ converges to $x$ and from $(1)$ that $Tx_{n} = y_{n}$ for each $n\in\mathbb{N}$. This completes the proof.