Equivalence of maps on l-infinity (involves limits, suprema and sums)

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SUMMARY

The discussion focuses on the convergence of the sequence of maps \( A_n: l^\infty \to \mathbb{R} \) defined by \( A_n(x) = \sup_{i \in \mathbb{N}} \frac{1}{n} \sum_{j=0}^{n-1} x_{i+j} \) for sequences \( x \in S \subset l^\infty \) consisting of convergent sequences. It is established that \( \lim_{n \to \infty} A_n(x) = \lim_{n \to \infty} x_n \) holds true. The sequence \( A_n(x) \) is shown to be monotone decreasing and bounded by \( \|x\|_\infty \), ensuring its convergence.

PREREQUISITES
  • Understanding of normed spaces, specifically \( l^\infty \) and its properties.
  • Familiarity with limits and convergence of sequences.
  • Knowledge of supremum and average calculations in mathematical analysis.
  • Experience with sequences and their behavior in functional analysis.
NEXT STEPS
  • Study the properties of \( l^\infty \) spaces and their subspaces.
  • Learn about monotonic sequences and their convergence criteria.
  • Explore the concept of suprema in the context of functional analysis.
  • Investigate the relationship between averages of sequences and their limits.
USEFUL FOR

Mathematicians, students of functional analysis, and anyone studying convergence in normed spaces will benefit from this discussion.

TaPaKaH
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Homework Statement


Normed space (l^\infty,\|\cdot\|_\infty) with subspace S\subset l^\infty consisting of convergent sequences x=(x_n)_{n\in\mathbb{N}}.

Given a sequence of maps A_n:l^\infty\to\mathbb{R} defined as $$A_n(x)=\sup_{i\in\mathbb{N}}\frac{1}{n}\sum_{j=0}^{n-1}x_{i+j}$$need to show that for any x\in S one has$$\lim_{n\to\infty}A_n(x)=\lim_{n\to\infty}x_n.$$

Homework Equations


Already shown that for any x\in l^\infty the sequence A_n(x) is monotone decreasing in n and is bounded by \|x\|_\infty therefore is convergent.

The Attempt at a Solution


It's more or less clear to me that if a sequence converges then a sequence of "averages" also converges, but I am struggling to find a way to write this out explicitly.
 
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nevermind, solved it myself via evaluations from both sides
 

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