ScroogeMcDuck
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Sorry for the rather vague title!
Given:
Problem:
I need to create a sequence w_n in A, for which w_n \rightarrow 0 and Tw_n \rightarrow y in B.
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).
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!
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!