6
$\begingroup$

X is a arbitrary non empty set , B(X) the set of bounded functions $f:X\rightarrow \mathbb{R}$ and $||f||_\infty = \sup_{x\in X} |f(x)|$

Completeness: Let $(f_n(x))_{n \in \mathbb{N}}$ be a cauchy sequence, then: $$||f_n-f_m||\le \frac{\epsilon}{2} \ \text{for n,m greater than some N}$$

the cauchy sequence $f_n$ will have a limit $f(x)$ for $x \in \mathbb{R}$, so there must be a $f_{n_k}$ with a $n_k > N$ such that : $|f_{n_k}(x)-f(x)|\le \frac{\epsilon}{2}$ so one can put:

$$|f_n(x)-f(x)| \le ||f_n(x)-f_{n_k}(x)||+ |f_{n_k}(x)-f(x)| \le \epsilon $$

And for every $x\in \mathbb{R}$: $$|f(x)|\le |f_{n_k}(x)|+|f_n(x)-f(x)|\le ||f_{n_k}(x)||+\epsilon < \infty$$

Is this sufficient to say that it was shown that $(B(X), ||.||_\infty )$ is a Banach space?

Is it also a Banach Algebra?

$\endgroup$
2
  • 3
    $\begingroup$ assuming that it is complete : I don't understand. $\endgroup$ Jan 19, 2013 at 17:35
  • $\begingroup$ It is just $\ell^{\infty}(X)$... $\endgroup$
    – shalop
    May 11, 2019 at 5:31

1 Answer 1

5
$\begingroup$

Let $\|f\|:= \sup_{x\in X} |f(x)|$.

For proving that your space $(B(X),\|\cdot \|)$ is complete I would begin with : Let $(f_n)_{n=0}^\infty$ be a Cauchy sequence in $B(X)$. Let $\epsilon >0 $. Then we find $N \in \mathbb N$ such that for all $n,m \geq N: \|f_n - f_m\| < \epsilon$. This implies that each $(f_n(x))_{n=0}^\infty$ converges pointwise because $\mathbb R$ is complete. Then we may define $$ f: X \rightarrow \mathbb R: x\mapsto \lim_{n \rightarrow \infty} f_n(x) $$ We now show that $f_n \rightarrow f$ uniform on $X$. Let $\epsilon > 0$. Then we find $N \in \mathbb N$ such that for all $n,m \geq N$ : $$ |f_n(x)-f_m(x)| < \epsilon, \forall x \in X $$ Taking the limit in $m \rightarrow \infty$ (this must be verified as legitimate step, too) leads to $$ \forall n \geq N: |f_n(x)-f(x)| \le \epsilon, \forall x \in X $$ which proves uniform convergence of the $f_n$ to $f$. Thus any Cauchy sequence in $B(X)$ converges which proves completeness of $(B(X),\|\cdot\|)$.

Further is boundedness preserved by uniform convergence.

What I mean by "thist must be verified": For a given sequence $(a_n)_{n=0}^\infty$ in $\mathbb R$ with limit $a$ we have for any metric d: $$ \lim_{n \rightarrow \infty} d(a_n,x) = d(a,x) $$ where $x$ is some point in your space.

$\endgroup$
3
  • $\begingroup$ Thanks, it does answer the question in the title. $\endgroup$ Jan 19, 2013 at 20:46
  • $\begingroup$ I am not familiar with "Banach Algebra" so sorry for that ^^ $\endgroup$
    – user42761
    Jan 20, 2013 at 9:37
  • $\begingroup$ There's a small error; in the last step, the inequality should not be strict, i.e., it should be $\forall n \geq N: |f_n(x)-f(x)| \leq \epsilon, \forall x \in X$. $\endgroup$
    – Yibo Yang
    Apr 18, 2023 at 18:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .