Creating convergent sequences in Banach spaces

[ScroogeMcDuck]

Sorry for the rather vague title!

Homework Statement

Given:

• Two Banach spaces $A$ and $B$, and a linear map $T: A\rightarrow B$
• The sequences $(x^n_i)$ in A. For each fixed n, $(x^n_i) \rightarrow 0$ for $i \rightarrow \infty$.
• The sequences $(Tx^n_i)$ in B. For each fixed n, $(Tx^n_i) \rightarrow y_n$ for $i \rightarrow \infty$.
• The sequence $(y_n)$ in B, with $y_n \rightarrow y$ for $n \rightarrow \infty$.

Problem:
I need to create a sequence $w_n$ in A, for which $w_n \rightarrow 0$ and $Tw_n \rightarrow y$ in B.

Homework Equations

So for the sequence $y_n$ of limits of $Tx_i^n$ we know:
$\forall \epsilon>0 \, \exists m \in \mathbb{N}$ such that $\forall n ≥ m: ||y_n - y||<\epsilon$
And for a fixed n, we know:
$\forall \epsilon>0 \, \exists m_1=m_1(n) \in \mathbb{N}$ such that $\forall n ≥ m_1: ||x_i^n||<\epsilon$
$\forall \epsilon>0 \, \exists m_2=m_2(n) \in \mathbb{N}$ such that $\forall n ≥ m_2: ||Tx_i^n - y_n||<\epsilon$.

Furthermore T is not necessarily continuous (it would be trivial if it were).

The Attempt at a Solution

I tried using the sequence $w_n = x_n^n$. Proving that $Tw_n \rightarrow y$ then required me to prove that $\forall \epsilon>0 \, \exists n' \in \mathbb{N} : \forall n≥n': ||Tx_n^n - y_n||< \epsilon/2$ (so that I could use the triangle inequality afterwards), but I couldn't manage this since y_n is not fixed.

I also tried using $w_n = x^n_{m_3(n)}$ where $m_3(n) = max\{m_1(n),m_2(n)\}$. The required convergence did work out, but then I realized that m_1(n) and m_2(n) depend on $\epsilon$ as well as n, so my sequence depends on $\epsilon$ which is of course not as it should.

Any suggestions, hints, ideas would be appreciated!

 

[gopher_p]

So your problem is that you essentially have three sequences, and you need to get some kind of uniform control of them. Passing to subsequences is obviously (?) the way to go, but you're running into trouble with your subsequences being dependent on an arbitrary $\epsilon$.

I suggest using the following fact about metric spaces to get a little better control over your subsequences:

If $u_n\rightarrow u$, then there is $n_1<n_2<...<n_j<...$ such that $d(u_{n_j},u)<j^{-1}$ for all $j=1,2,...$.

I.e. there is always a "nice" subsequence that converges in a controlled way. I think you can use that fact to remove your $\epsilon$ troubles. I'd also suggest making use of the trusty "WLOG" in order to make your notation manageable; since we know we can find a "nice" subsequence, we may as well assume WLOG that the original sequence was already "nice".