Distance of consecutive terms of a sequence goes to zero implies convergence in functional spaces.

Consider the vector space of real bounded functions $\Delta =$ {$f: \mathbb{R} \rightarrow \mathbb{R}: \text{ f is bounded}$}.

Then the following is a norm on $\Delta$:

$$\left\| f \right\|_{\infty}=\underset{x\in\mathbb{R}}{sup}|f(x)|$$

In fact, $(\Delta,\left\| \cdot \right\|_{\infty})$ is a Banach space (real normed complete vector space). Consider a sequence $f_n$ in $\Delta$ such that:

$$\left\| f_n-f_{n-1} \right\|_{\infty}\underset{n}{\longrightarrow }0$$

Then we know that {$f_n$} is not necessarily a Cauchy sequence, hence it does not necessarily converge.

The question is: is there a (functional) complete vector space $A \subseteq $ {$f:\mathbb{R} \rightarrow \mathbb{R}$} with a norm $|\cdot |$ such that if {$f_n$} is a sequence in $A$ them the following two conditions hold:

1. if $| f_n-f_{n-1}|\underset{n}{\longrightarrow }0$ then $f_n$ is Cauchy, that is, converges to a limit $f$.
2. if $f_n$ converges to $f$ in $(A, | \cdot |)$ then $f_n\to f$ point-wise (or at least almost everywhere convergence.).

Thank you