Let's consider the Banach space $$X=\{u \in C([0,\infty),\mathbb{K})\; ; \;u(0)=0,\lim_{x \rightarrow \infty}u(x)=0\},$$ where, $\mathbb{K}$ is a field and $a:[0,\infty)\longrightarrow (0,\infty)$ is function such that $$\lim_{x \rightarrow \infty}\int_{0}^{x}\frac{1}{a(s)}ds=\infty,$$ with norm $||u||_{X}=\sup\{|u(x)|\; ; \; x \in [0,\infty)\}$. Set a linear operator $A:D(A)\subset X \longrightarrow X$, where $$D(A)=\{u \in X \; ; \; u\:\text{is differentiable and}\: au'\in X\}.$$ In this context, the formation of law operator $ A $ is irrelevant.
I want to show that $D (A) $ is dense in $X $. I tried to do this by using a corollary of the Hahn-Banach theorem-$2^{nd} $ geometric form. That is, I tried to show that $$\forall\:f \in X' \; ; \; f(u)=0,\: \forall\:u \in D(A) \Rightarrow f\equiv 0 ,$$ where, $X'=\{ g: X \longrightarrow \mathbb{K} \; ; \; g \; \text{is linear and bounded}\}$. But I could not finish this.