Let $g: X \to \mathbb{R}_+$ be a continuous function where $X \subseteq \mathbb{R}^n$ is an open convex set. Denote the weighted supremum norm $\|\cdot \|_g$ of a real valued function $f: X \to \mathbb{R}$ by $$ \|f\|_g := \sup_{x\in X} \frac{|f(x)|}{g(x)}. $$
If $\lim_{n\to\infty}\|f_n - f^*\|_g = 0$, does $f_n$ converge to $f^*$ uniformly on compact sets?
My attempt: Let $\tilde X$ be an arbitrary compact subset of $X$. Since $\tilde X$ is compact, then $\underline g := \min_{x\in\tilde X} g(x) < \infty$. Therefore, on $\tilde X$ if $\lim_{n\to\infty}\|f_n - f^*\|_g = 0$ then $\lim_{n\to\infty}\|f_n - f^*\|_\infty = 0$ on $\tilde X$ where $\|\cdot \|_\infty$ is the supremum norm. I know that convergence in the supremum norm implies uniform convergence. Thus, since the choice of $\tilde X$ was arbitrary, we have uniform convergence on compact sets. Does my proof seem correct? Are there any flaws? Thanks.
Yes, you are correct. If $K$ is a compact set in $X$ then the continuous positive function $g$ has a positive maximum $M_K$ and $$\sup_{x\in K} |f_n(x)-f(x)|=M_k\sup_{x\in K} \frac{|f_n(x)-f(x)|}{M_K}\leq M_K\sup_{x\in K} \frac{|f_n(x)-f(x)|}{g(x)}\leq M_K\sup_{x\in X} \frac{|f_n(x)-f(x)|}{g(x)}.$$ Hence $\lim_{n\to\infty}\|f_n - f\|_g=0$ implies that $f_n$ converges uniformly to $f$ on the compact sets $K$. It is not necessary that $X$ is an open convex set.