08 - Completeness of Normed Vector Spaces for Bounded Functions

Completeness of Normed Vector Spaces for Bounded Functions

Let $X$ be a set then

$$B(x)\text{ is complete}$$

where $B(X)$ denotes the normed vector space of bounded functions $f:X\to \mathbb{R}$ with norm

$$||f|{|}_{\infty }=\underset{x\in X}{\sup }|f(x)|$$

Proof

Let $(f_n{)}_{n=1}^{\infty }$ be a Cauchy Sequence in $B(X)$ then
For each $x$, sequence $(f_n(x){)}_{n=1}^{\infty }$ is a Cauchy Sequence of Real Numbers

As $\mathbb{R}$ is complete then each sequence has a limit written as $f(x)$ so that

$$\underset{n\to \infty }{\lim }{f}_n(x)=f(x)$$

Showing that $f$ is bounded
Take $ϵ=1$ in Cauchy so there exists $N$ such that for $n,m\ge N$ then

$$\underset{x\in X}{\sup }|f_n(x)-f_m(x)|\le 1$$

Taking $m=N$ then

$$|f_N(x)-f_n(x)|\le 1\text{ for all }n\ge N\text{ and }x\in X$$

As $n\to \infty$ then we get

$$|f_N(x)-f(x)|\le 1\text{ for all }x\in X$$

As $f_N$ is a bounded function then so is $f$

Showing that $f_n\to f$ in the norm $||\cdot |{|}_{\infty }$ (have pointwise convergence currently)
Let $ϵ>0$ so there exists $N$ such that if $n,m\ge N$ then

$$|f_n(x)-f_m(x)|\le ϵ\text{ for all }x\in X$$

For fixed $n\ge N$ and $x\in X$ as $m\to \infty$ hence

$$|f_n(x)-f(x)|\le ϵ$$

Hence for al $n\ge N$ we have

$$||f_n-f|{|}_{\infty }\le ϵ$$

Hence $f_n\to f$ in the $||\cdot |{|}_{\infty }-$norm