Why do we denote the sequence space as $l^\infty$

Question

My question is really simple. I didn't understand the call the sequence space as $l^\infty$. I know this is from the $l^p$ spaces denotation but I didn't understand the connection between both.

$l^p$ space

Let $p\ge 1$ a fixed real number. By definition , each element in the space $l^p$ is a sequence $x=(x_j)=(x_1,x_2,\ldots)$ of numbers such that $|x_1|^p+|x_2|^p+\ldots$ converges and the metric is defined by

$$d(x,y)=\bigg(\sum_{j=1}^{\infty}|x_j-y_j|^p\bigg)^{1/p}$$

$l^{\infty}$ space

Let $X$ be a set of all bounded sequences of complex numbers; that is, every element of $X$ is a complex sequence $x=(x_1,x_2,\ldots)$ such for all $j=1,2\ldots$ we have $|x_j|\le c_x$, where $c_x$ is a real number. We choose the metric defined by

$$d(x,y)=sup_{j\in \mathbb N}|x_j-y_j|$$

Because when $p\to\infty$ the $p$-norm tends to the sup norm. — uniquesolution, Jul 10 '18 at 10:48
Possible duplicate of The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. — mechanodroid, Jul 10 '18 at 17:45

score -1 · Answer 1 · answered Jul 10 '18 at 11:00

-1

We denote it because $$\lim_{p\to\infty} \left(\sum_{j=1}^\infty |x_j-y_j|^p\right)^{\frac1p} = \sup_{j\in\mathbb N} |x_j-y_j|$$

answered Jul 10 '18 at 11:00

5xum

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As written, this is clearly false. You want to assume something on the sequences. – uniquesolution Jul 10 '18 at 11:13

Why do we denote the sequence space as $l^\infty$

$l^p$ space

$l^{\infty}$ space

1 Answers1