Proving that a set of functions is complete

Question

I need to prove that the set of functions of the form $\{g\colon [-1,1]\to \mathbb{R}\, |$ there exists a constant $c$ such that $g(x) = c\cdot x\}$ is complete. How shall I proceed? I am taking a Cauchy sequence and want to prove that it converges to a function in this set, but how can I guarantee the limit will have the same form? (that is linear)

Answer 1

Let $S$ be the set you defined. You can see $S$ inside $C([-1,1])$, the space of all continuous real functions defined on $[-1,1]$. In $C([-1,1])$ there is the supremum norm: $\| f \| = \sup_{x \in [-1,1]} |f(x)|$. With this norm, $C([-1,1])$ is a Banach space. So $S$ is complete with respect to the supremum norm if and only if $S$ is closed inside $C([-1,1])$, and this fact is easy to prove.

Answer 2

Set $F:[-1,1]\to\mathbb{R}$ given by $F(t)=t$. Define $$U=\{g:[-1,1]\to\mathbb{R}|g(x)=cx\mbox{ for some }c\in\mathbb{R}\}$$

Clearly $U=\mathrm{span}(F)$. Then $U$ has dimension $\mathrm{dim}\,U=1<\infty$. Thus, every norm in $U$ generates equivalent topologies. Hence, every topology generated for a norm has the same convergent sequences.

But also since $\mathrm{dim}\,U=1<\infty$, $U$ is complete in any norm.