Let $E=\mathcal{C}([a,b],\mathbb{R})$ and $w\in E$ do not vanish on $[a,b]$. define $d_w(f,g)=\sup_{t\in [a,b]} |w(t)(f(t)-g(t))|$
how to see if $(E,d_w)$ is complete or not ?
----> i take $(f_n)_n$ a Cauchy sequence from E, for any fixed $t\in [a,b]$ then $(f_n(t))$ converge to a real $f(t)$
but i don't know to continue.
We want to show that $(E, d_w)$ is complete, that is, show that every Cauchy sequence from $E$ converges in $E$.
Taking the 'normal' metric $\tilde d = \sup_{t\in [a,b]} |(f(t)-g(t))|$, what do we know? We know:
Suppose now, we have a Cauchy-sequence $(f_n)_n$ in $(E, d_w)$.
Step 1: We want to show it is also a Cauchy-sequence in $(E, \tilde d)$.
Proof: Let $\epsilon \gt 0$. $w \in E$ is continuous, so let $$\alpha := \max_{t\in [a,b]}{|w(t)|} \gt 0$$ and $$\beta := \min_{t\in [a,b]}{|w(t)|} \gt 0$$ (because $w$ doesn't vanish).
As $(f_n)_n$ is Cauchy-sequence in $(E, d_w)$, there exists $N \in \mathbb{N}$, so that $d_w(f_n, f_m) \lt \frac{\epsilon \cdot \beta}{2}$ $\forall n,m \geq N$. But then also for all $n,m \geq N$ and $t \in [a,b]$: $$ |f_n(t) - f_m(t)| = \frac{|w(t)|}{|w(t)|} \cdot |f_n(t) - f_m(t)| \leq \frac{|w(t)|}{\beta} \cdot |f_n(t) - f_m(t)| = \frac{1}{\beta} \cdot |w(t)(f_n(t)-f_m(t))|$$$$ \leq \frac{1}{\beta} \cdot d_w(f_n, f_m) \lt \frac{\epsilon \cdot \beta}{2\beta} = \frac{\epsilon}{2}$$
Building the maximum provides $\tilde d(f_n, f_m) \lt \epsilon$. $\square$
Step 2: We now know that $(f_n)_n$ is a Cauchy-sequence and thus converges to a $f \in E$ (in regard to metric $\tilde d$). We want to show that $(f_n)_n$ also converges to $f$ in regard to metric $d_w$. But this follows easily from seeing that $\forall n \in \mathbb{N}$ and $t \in [a,b]$: $$|w(t)(f_n(t)-f(t))| = |w(t)|\cdot|(f_n(t)-f(t))| \leq \alpha |f_n(t)-f(t)| \leq \alpha \cdot \tilde d(f_n, f)$$