Is $\ell^{\infty} (\mathbb{R})$ A Banach Space?

Calculus Level 3

Let $V$ be a vector space over $\mathbb{R}$ . A norm on $V$ is a function $\|\cdot \| : V \to \mathbb{R}$ satisfying the following properties:

The norm is nonnegative: $\|v\| \ge 0$ for all $v\in V$ , with equality if and only if $v=0$ .
The norm scales with vectors: $\|cv\| = |c| \cdot \|v\|$ for all $v\in V$ and $c\in \mathbb{R}$ .
The norm satisfies the triangle inequality: $\|v+w\| \le \|v\| + \|w\|$ for all $v, w\in V$ .

A vector space equipped with a norm is called, unsurprisingly, a normed vector space .

Suppose $V$ is a normed vector space. For $v,w \in V$ , define a function $d: V\times V \to \mathbb{R}$ by $d(v,w) = \|v-w\|.$ One can verify that this function $d$ is a metric on $V$ , giving $V$ the structure of a metric space .

If the metric space structure on $V$ induced by the norm is complete (i.e., Cauchy sequences converge ), then $V$ is called a Banach space .

Consider the space $\ell^{\infty} (\mathbb{R})$ , consisting of bounded sequences of real numbers. This is a vector space over $\mathbb{R}$ , using coordinate-wise addition and scalar multiplication. One may define a norm on $\ell^{\infty} (\mathbb{R})$ by setting $\|(a_1, a_2, a_3, \cdots)\| = \sup_{i\ge 1} |a_i|.$ Under this norm, is $\ell^{\infty} (\mathbb{R})$ a Banach space?

Yes No

1 solution

Samrat Mukhopadhyay
Sep 22, 2016

Let $\{a(n)\}$ be a Cauchy sequence in the $l^{\infty}(\mathbb{R})$ space. Then, by definition, for any $\epsilon>0,\ \exists N\in \mathbb{Z}^{+}$ such that for any $m,n\ge N$ , $\|a(n)-a(m)\|<\epsilon\\\implies \sup_{i}|a_{i}(n)-a_{i}(m)|<\epsilon\\\implies |a_{i}(n)-a_{i}(m)|<\epsilon,\ \forall i\ge 1$ Hence, for every $i\ge 1$ ,the sequence $\{a_i(n)\}$ is Cauchy. Since $a_i(n)\in \mathbb{R},\ \forall i$ and since $\mathbb{R}$ is complete, $a_{i}(n)\to a_i,\ i\ge 1$ for some $a_i\in \mathbb{R}$ , as $n\to \infty$ . Thus, in the first condition, where it says $|a_{i}(n)-a_i(m)|<\epsilon$ , for all $m,n>N$ , take $m\to \infty$ , to get $|a_{i}(n)-a_i|<\epsilon$ for all $n>N$ , which implies that $\sup_{i}|a_i(n)-a_i|<\epsilon$ for all $n>N$ , which in turn implies that $a(n)\to a$ as $n\to \infty$ where $a=(a_1,a_2,\cdots)$ . Thus the space is complete and hence is a Banach space.

Nice proof to this Real Analysis problem, Samrat!

tom engelsman - 7 months, 2 weeks ago

Is ℓ ∞ ( R ) \ell^{\infty} (\mathbb{R}) ℓ ∞ ( R ) A Banach Space?

1 solution

0 pending reports

Is $\ell^{\infty} (\mathbb{R})$ A Banach Space?