Let $X$ and $Y$ be Banach Spaces. Let $T:X \rightarrow Y$ be a bounded linear map and T is onto. Let $(y_{n})$ be a sequence in $Y$ converging to $y$ in $Y$.Prove that there exists $x \in X$ and a sequence $(x_{n})$ in $X$ coverging to $x$ such that $T(x)=y$ and $T(x_{n})=y_{n}$ $\forall n \in \mathbb{N}$.
My attempt is the following:Let us call $Ker(T)=M$. Then we have an induced map $\hat{T}:X/M \rightarrow Y$ given by $\hat{T} \circ \pi=T$ where $\pi$ is the natural quotient map. With the given condition OPEN MAPPING THEOREM easily implies T is open. Hence $\hat{T}$ is continous and open and hence $\hat{T}$ is bounded below.
Now so, $||\hat{T}(x+M)|| \geq \delta ||x+M||$ $\implies$ $||T(x)|| \geq \delta ||x+M||$.
Now consider $y_{n}$. There exists $x_{n}$ such that $T(x_{n})=y_{n}$. Since $y_{n}$ is convergent we can use the bounded below condition to show that $x_{n} +M$ is cauchy in $X/M$. Hence $x_{n}+M \rightarrow x+M$. Then it is clear after some calculation and manipulation with $\hat{T}$ that $T(x)=y$. But what I'm unable to show is that $x_{n} \rightarrow x$ in $X$.
Any help will be appreciated. Thanks in advance!
You don't necessarily have $x_n \to x $. But we have $$ \inf_{m_n \in M} \|x - (x_n + m_n) || = \| x - x_n +M \| \to 0. $$ This allows you (how exactly?) to modify the sequence $(x_n)_n $ in such a way that the modified sequence satisfies $x_n ' \to x $ and still $T x_n' = y_n$ and $Tx = y $.