TITLE: $C([0,1])$ with norm $\|f\| := \sum_{n=1}^\infty|\frac{f(q_n)}{2^n}|$ complete?
QUESTION [6 upvotes]: Let $(q_n)$ be a sequence of all rational numbers from $[0, 1]$. On $C([0,1])$ let's consider norm:
$$\|f\| := \sum_{n=1}^\infty|\frac{f(q_n)}{2^n}|$$
I want to check whether norm $\|f\|$ is complete.
My work so far
Let's take $(f_k) \subset C([0,1 ])$ Cauchy sequence. We want to prove or disaprove that $$\exists{f \in C([0, 1])}: \|f_k - f\|\rightarrow 0 $$
I wanted first to focus on more general problem:
$$\exists f: \|f_k - f\| \rightarrow 0$$
and after proving that, we can focus on showing that such $f$ has to be in $C([0, 1])$.
So - because $(f_k)$ is a Cauchy Sequence then:
$$\exists_l : \|f_k  - f_l\| = \sum_{n=1}^\infty \frac{|f_k(q_n) - f_l(q_n)|}{2^n} \le \epsilon$$
for any $\epsilon > 0$
$$\|f_k  - f\| = \sum_{n=1}^\infty \frac{|f_k(q_n) - f(q_n)|}{2^n}= \sum_{n=1}^\infty \frac{|f_k(q_n) - f_l(q_n) + f_l(q_n) - f(q_n)|}{2^n} \le$$
$$\le \sum_{n=1}^\infty \frac{|f_k(q_n) - f_l(q_n)|}{2^n} + \frac{|f_l(q_n) - f(q_n)|}{2^n} \le$$
$$\le \sum_{n=1}^\infty \frac{\epsilon}{2^n} + \frac{\sup\{|f_l(q_n) - f(q_n)| : n \in \mathbb{N} \}}{2^n} = \epsilon + \sup\{|f_l(q_n) - f(q_n)| : n \in \mathbb{N} \}$$.
And here I got stuck, because $\sup\{|f_l(q_n) - f(q_n)| : n \in \mathbb{N} \}$ is not arbitrary small. Do you know how to fix this/could you please give me a hint what is correct approach to this problem

REPLY [3 votes]: Let $E$ be the space of functions $f:[0,1]\cap \mathbb Q\to\mathbb R$ such that $\|f\| < +\infty$. We can quite easily see that $E$ is complete and $C([0,1])$ is (isomorphic to) a subspace of $E$. But we will show that it is not complete.
For example, let :
$$f_k(x) = \tanh\big(k(x-1/\sqrt{2})\big)$$
Then, we have :
$$\forall x \in \mathbb Q\cap[0,1], \lim_{k\to +\infty} f_k(x) = f(x)$$
where $f(x) = 1$ for $x>1/\sqrt{2}$ and $f(x)=-1$ for $x<1/\sqrt{2}$. Since $|f_k(x)|\leq 1$, the dominated convergence theorem gives :
$$\|f_k - f\| \to 0$$
Therefore, $f_k \to f$ in $E$. Therefore, $(f_k)$ is a Cauchy sequence in $E$, and also in $(C([0,1],\|\cdot\|)$, but has no limit in $C([0,1])$