Prove that $(B(A),\|\cdot\|_\infty)$ is a Banach Space I need to prove that $(B(A),\|\cdot\|_\infty)$ is a Banach Space, where $B(A)$ is the set of bounded functios from $ A $ to $\mathbb R$. I have seen the proof but I don't understand it. This is what I have:
First, we take ${f_n}$ a Cauchy sucession. As it 'lives' in $\mathbb R$, because of the completeness of real numbers we know that this sucession is convergent. My question is: why the proof doesn't finish here? The proof that I'm following now proves that our sucession converges to another function with the infinite norm, but we know that, in $\mathbb R$, all norms are equivalent.
Can somebody help me?
 A: By a fenced function $f: A \rightarrow \mathbb{R}$, do you mean $|f(x)| \leq C$, for some $C < \infty$, $\forall x \in \mathbb{R}$ ? i.e., $B(A)$ is the set of all bounded functions.
Its true that $f_n(x) \rightarrow g(x)$ for some $g(x) \in \mathbb{R}$. what you now need to show is:

*

*You need to show $|f_n(x)-g(x)| \leq \epsilon$, forall $x \in \mathbb{R}$ (uniformly), for all $n \geq N'_{\epsilon}$. Note that $N'_{\epsilon}$ should not depend on $x$.


*Further you need to show that $|g(x)| \leq C'$ for some $C' < \infty$ for all $x \in \mathbb{R}$. Note that $C'$ should not depend on $x$. Inparticular as $x \rightarrow \infty$, $g(x) \rightarrow \infty$ should not happen.
a) Point 1):
If $f_n$ is cauchy, it means $||f_n(x) - f_m(x)|| \leq \epsilon$ for all $x \in \mathbb{R}$, for all $n,m \geq N_{\epsilon}$.
This implies $|f_{N_{\epsilon}}(x) - g(x)|  = |f_{N_{\epsilon}}(x) -f_n(x)+f_n(x)- g(x)| \leq |f_{N_{\epsilon}}(x) -f_n(x)|+|f_n(x)-g(x)|$.
By taking $n$ very large such that $|f_n(x)-g(x)| \leq \epsilon$, for all $n \geq \max(M(x),N_{\epsilon})$ ($M$ could be dependent on $x$. So i wrote it as $M(x)$).
we have
$|f_{N_{\epsilon}}(x) - g(x)| \leq |f_{N_{\epsilon}}(x) -f_n(x)|+|f_n(x)-g(x)| \leq |f_{N_{\epsilon}}(x) -f_n(x)|+\epsilon$, for all $n \geq \max(M(x),N_{\epsilon})$.
But we have that
$|f_{N_{\epsilon}}(x) -f_n(x)| \leq \epsilon$ for all $x$. Hence
$|f_{N_{\epsilon}}(x) - g(x)| \leq |f_{N_{\epsilon}}(x) -f_n(x)|+\epsilon \leq 2 \epsilon $, for all $n \geq \max(M(x),N_{\epsilon})$.
Hence we have,
$|f_{N_{\epsilon}}(x) - g(x)| \leq 2 \epsilon$ for all $x \in \mathbb{R}$ for all $n \geq \max(M(x),N_{\epsilon})$. But $n$ does not appear in the inequality. So we have that
$|f_{N_{\epsilon}}(x) - g(x)| \leq 2 \epsilon$ for all $x \in \mathbb{R}$.
Hence Point 1) is proved.
b) Point 2): From previous point, we have that, $||f_n(x)-g(x)|| \leq \epsilon$ for all $x \in \mathbb{R}$, for all $n \geq N'_{\epsilon}$.
Since $|f_{N_{\epsilon}}(x)| \leq C''$, for some $C'' < \infty$ for all $x \in \mathbb{R}$ and we have $||f_{N_{\epsilon}}(x)-g(x)|| \leq \epsilon$ for all $x \in \mathbb{R}$$\implies |g(x)| \leq |f_{N_{\epsilon}}(x)|+\epsilon \leq C''+\epsilon$ for all $x \in \mathbb{R}$ and $C'' < \infty$.
Hence Point 2) is proved.
A: Each sequence $\big(f_n(t)\big)_n \in \mathbb{R}^\mathbb{N}$ does converge to a certain real $f(t)$, however a priori this new function $f$ might not be the limit of the sequence of functions $(f_n)_n \in B(A)^\mathbb{N}$, nor is it a priori bounded (I'm guessing that what "fenced" means), however these are true statements, and that's what is to be proved here.
Take $f_n : t \in [0,1] \mapsto t^n$ for example.
The corresponding "limit" (which is called pointwise limit) $f$ would be the function:
$$f : t \mapsto \lim_{n \to \infty} f_n(t) = \begin{cases}0 \quad\text{if }t \neq 1\\ 1 \quad\text{if } t = 1\end{cases}$$
Yet, we have:
$$\|f_n - f\|_\infty = \sup_{t \in [0,1]} |f_n(t) - f(t)| = \max\left(|f_n(1) - f(1)|, \sup_{t \in [0,1)} |f_n(t)|\right) = 1 \underset{n \to \infty}{\not\to} 0$$
due to the fact that $f_n(t) \xrightarrow[t \to 1^-]{} 1$ for all $n$, and as such $(f_n)_n$ doesn't converge to $f$ in $B([0,1], \|\cdot\|_\infty)$.
This does not contradict the fact that norms are equivalent on $\mathbb{R}^d$, simply because you are trying to compare $(\mathbb{R}, |\cdot|)$ and $(B(A), \|\cdot\|_\infty)$, but $\mathbb{R}$ and $B(A)$ are two different spaces altogether!
