I am taking an introductory course in functional calculus this semester, but we're right at the begining. I am trying to solve the following problem, which is actually part of an iff-relation. The other implication, I was able show following a similar approach.
Suppose for $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\} $ we have a sequence $(x_k)_k \subset \mathcal{C}([a,b], \mathbb{K})$ such that the limit $\lim_{k} x_k = x$ exists with respect to the weighted metric
$$d_g(x_k,x) := \max_{t\in[a,b]} \big\lvert g(t) \big(x_k(t)-x(t)\big)\big\rvert, $$
for a non-zero function $g \in \mathcal{C}([a,b])$.
I want to show that the assumption implies uniform convergence in the traditional, i.e. real or complex valued, sense. My approach is rewriting the assumption as
$$ \forall \varepsilon > 0 \ \exists k_0 \in \mathbb{N} \ \forall k \ge k_0 : \max_t \big\lvert g(t) \big(x_k(t)-x(t)\big)\big\rvert < \varepsilon.$$
Next, as $\max_t \lvert g(t)\rvert \neq 0$ exisits and is fixed, I would like to conclude something like $$ \forall \varepsilon > 0 \ \exists k_0 \in \mathbb{N} \ \forall k \ge k_0 : \max_t \lvert g(t)\rvert \cdot \max_t \lvert x_k(t)-x(t)\rvert < \varepsilon,$$
to go on argumenting via an appropriate choice of $\varepsilon$. However, this is neither correct nor elighting. Especially, since the lefthand side of the inequality has at worst even increased. I'm not sure what else to try here, as I cannot think of any other way to get rid of the $g(t)$ within the metric and also not increase the lefthand side. Thanks
Suppose $g>0$ without loosing generality. Then there exists $m:=\min_{[a,b]}g>0$. For any $\epsilon>0$, there exists $k_0\in \mathbb{N}$ s.t. for all $k\geq k_0$ s.t. $$m\cdot\max_t|x_k(t)-x(t)|\leq\max_t|g(t)(x_k(t)-x(t)|\leq m\epsilon$$ which implies uniform convergence.