# When interchanging limit and supremum over a finite set is allowed

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?

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Does $(x_m(i))$ converge for all $1\le i \le N$? Then both sides are equal zero. – martini Sep 20 '12 at 12:50
Since you just take a supremum over a finite number of indexes, this is allowed true with the condition you mentionned. – Davide Giraudo Sep 20 '12 at 12:50
@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)? – gary Sep 20 '12 at 12:58
I meant the condition of convergence. – Davide Giraudo Sep 20 '12 at 13:12
@Davide Giraudo : how does that follow from convergence? – gary Sep 20 '12 at 16:50

Your right-hand side is equal to zero. For the left hand side, $$0\leq\lim_{m\to\infty}\sup_{1\leq i\leq N}|x(i)-x_m(i)| =\lim_{m\to\infty}\max_{1\leq i\leq N}|x(i)-x_m(i)| \leq\lim_{m\to\infty}\sum_{i=1}^N|x(i)-x_m(i)|\\ =\sum_{i=1}^N\lim_{m\to\infty}|x(i)-x_m(i)| =\sum_{i=1}^N0=0.$$ So the left-hand side is also zero.