Suppose I have an element $x\colon\mathbb{N}\to\mathbb{R}$ in $\ell^\infty$. Fix $i\in\mathbb N$ and $x_m(i)$ converges to some $x(i)\in\mathbb R$ Let $N\in\mathbb N$. When does $$\lim_{m\to\infty}\sup_{1\le i\le N}|x (i)-x_m(i)|=\sup_{1\le i\le N}\lim_{m\to\infty}|x (i)-x_m(i)|$$ hold?
When interchanging limit and supremum over a finite set is allowed
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real-analysis
sequences-and-series
functional-analysis
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0Does $(x_m(i))$ converge for all $1\le i \le N$? Then both sides are equal zero. – 2012-09-20
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0Since you just take a supremum over a _finite_ number of indexes, this is allowed true with the condition you mentionned. – 2012-09-20
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0@Davide Giraudo what condition do you mean? Does it follow from $x\in\ell^\infty$ or from the fact that $x_m(i)$ converges to $x(i)$ (for $i$ fixed)? – 2012-09-20
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0I meant the condition of convergence. – 2012-09-20
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0@Davide Giraudo : how does that follow from convergence? – 2012-09-20
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0Sorry, I didn't see that $i$ was fixed. First, we have to be sure that all the terms in the equality exist. – 2012-09-20