Completion of $C([0,1])$ with respect to a norm vanishing at $0$ Let $C([0,1])$ denote the collection of continuous function on the unit interval. Let $\psi$ be an element of $C_0((0,1])$, the continuous functions vanishing at $0$, which satisfies $0 < \psi \le 1$ and $\sup_{x \in [0,1]} \psi(x) = 1$.
Define a norm on $C([0,1])$ by 
$$
\|f\| = \sup_{x \in [0,1]} |\psi(x)f(x)|.
$$
What is the completion of $C([0,1])$ with respect to this norm?
Note: This norm is not equivalent to the sup norm on $C([0,1])$ since the sequence 
$$
f_n (x) = \begin{cases}
1 -nx & \text{if } x \in [0,1/n];\\
0 & \text{otherwise};
\end{cases}
$$
converges to zero in this norm.
Edit: It might be easier to consider the case where $\psi(x) = x$ in which case
$$
\|f\|=\sup_{x \in [0,1]}|xf(x)|.
$$
 A: When you complete $C([0,1])$ with that norm you pick up "all the functions that blow up at zero more slowly than $1/\psi$."
To make this more rigorous we will do the most basic thing possible, which is to embed our space in another complete space and figure out the closure there. We define the following:
$X =$ your space $C([0,1])$ with the $\psi$-norm, $||f||_\psi = \sup_{x \in [0,1]} |\psi(x)f(x)|$
$Y =$ the space $C([0,1])$ with the standard sup norm, $||f||_0 = \sup_{x \in [0,1]} |f(x)|$
$j : X \to Y, f \mapsto \psi f$
and
$C_0 \subset Y, C_0 = \{f\in Y | f(0) = 0\}$
(I'm just going to treat $\psi$ as a function defined on all of $[0,1]$ - it won't make any difference.)
By construction, $j$ is an (isometric) embdedding of $X$ into $Y$. As $Y$ has the standard $\sup$ norm $C_0$ is a closed subspace, and $j$ clearly maps $X$ into $C_0$. So we easily get that $\overline{j(X)} \subset C_0$, and I claim that $\overline{j(X)}$ is all of $C_0$.
So let's take an arbitrary $f \in C_0$ and $n \in \mathbb{N}$. We are going to find an $\epsilon > 0$ such that for all $x \in [0,\epsilon]$ we have both
$$
|f(x)| < 1/2n
$$
and
$$
\psi(x) < \psi(\epsilon)
$$
We do this by first finding $\epsilon_0$ so that the condition on $f$ holds, and then set 
$$ 
\epsilon = \operatorname*{arg\,max}_{x \in [0,\epsilon_0]} \psi(x)
$$
(For your simple example where $\psi(x) = x$, $\psi$ is strictly increasing near zero, we could find $\epsilon$ more easily, but we need this more complicated version to handle an arbitrary, possibly less well-behaved $\psi$ that could, for example, oscillate infinitely often between $x^2$ and $x^3$ as $x \to 0^+$)
Now define $f_n \in X$ as
$$
f_n(x) = \begin{cases}
f(\epsilon)/\psi(\epsilon) & x \in [0,\epsilon]\\
f(x)/\psi(x) & x \in [\epsilon,1]\\
\end{cases}
$$
Let's look at $j(f_n)$, i.e. $f_n$ mapped into $Y$. On most of $[0,1]$, away from zero, $j(f_n)$ wil be identical to $f$, while near zero $j(f_n)$ looks like a scaled version of $\psi$, but isn't too far from $f$. Explicitly, for $x \in [0,\epsilon]$,
$$
\begin{split}
|j(f_n)(x) - f(x)| & = |\psi(x)f(\epsilon)/\psi(\epsilon) - f(x)|  \\
& < |\psi(x)/\psi(\epsilon)| * |f(\epsilon)| + |f(x)|  \\
& < 1 * 1/2n + 1/2n = 1/n
\end{split}
$$
So $||j(f_n) - f||_0 < 1/n$, which gives us $j(f_n) \to f$ as $n \to \infty$ and thus $\overline{j(X)} = C_0$. QED
While this gives us a precise answer it'd be nice to see what this means for our original functions, without looking at them multiplied  by $\psi$. If we take @Jochen's idea of looking at them as living in $C( (0,1] )$ you can see that we only get the functions where $\lim\limits_{x\to 0} \psi(x)f(x)$ equals $0$, not the whole space of functions where the limit exists. In particular, the function $ f = 1/\psi$ is not in the completion. It's hard for me to see what's going on with the original functions in $C( (0,1] )$, but if you use  $j$ to map $1/\psi$  to my space $Y$ you get $\textbf{1}_{[0,1]}$, which can't be approximated (in the $\sup$ norm) by functions that vanish at zero.
(Thanks for posting this question. It was fun to think about.)
