Completion of the set of continous functions on $[0,1]$ with the metric $d(x(t),y(t)) = \sup_{t \in [0,1]} t^2|x(t)-y(t)|$.

Question

I am trying to find the completion of $C[0,1]$ with the metric $d(x(t),y(t)) = \sup_{t \in [0,1]} t^2|x(t)-y(t)|$. I understand that this metric space is not complete because the function $f(x) = 1 \text{ if } x = 0, 0 \text{ otherwise}$ can be approximated with $f_n(x) = (1-x)^n$. My guess for the completions is the set $A := \{ f : [0,1] \rightarrow \mathbb{R}| f \text{ is continous on } (0,1] \land \lim_{x \to 0^+} x^2f(x) = 0 \}$. I need to show two things: 1-) $(A,d)$ is complete. 2-) $(C[0,1],d)$ is dense in $(A,d)$. I couldn't show either.

Answer 1

Here's a possible strategy. Let's write $X$ for the space of continuous function endowed with your unusual metric, $d_X$, and let's write $C$ for the space of continuous functions on $[0,1]$ with the usual sup-norm metric, i.e. $$ d_C(x,y) = \sup_{t \in [0,1]} |x(t) -y(y)|. $$ It's well-known that $(C,d_C)$ is complete. Now consider the map $T : X \to C$ via $Tx(t) = t^2 x(t)$. This is well-defined because the product map $t \mapsto t^2 x(t)$ is continuous. Moreover, $$ d_X(x,y) = d_C(Tx,Ty), $$ so $T$ is an isometric embedding. This means we can isometrically identify $X$ and $TX \subseteq C$. From here we deduce that the completion of $X$, say $\bar{X}$, is isometric to the closure of $TX$ in $C$, i.e. $\bar{X} \sim \overline{TX}$. From here it's straightforward to get the the identification of $\bar{X}$ and $A$ and the two points you're trying to prove.