Density of a subspace in $C_0$

Let $$f\in C_0$$, the space of all continuous functions vanishing at infinity. Let $$A\subset \mathbb{R}$$ be such that $$\text{span}(f_x \mid x\in \bar{A})$$ is dense in $$(C_0,\|\cdot\|_{\infty})$$, where $$f_x:\mathbb{R}\to \mathbb{R}$$ defined by $$f_x(y)=f(xy)$$.

Is it true that $$\text{span}(f_x \mid x\in A)$$ also dense in $$(C_0,\|\cdot\|_{\infty})$$?

It suffices to show that for any $$x \in \bar{A}$$ there exists $$(h_n)_{n \in \mathbb{N}} \subset \text{span}(f_x \mid x \in A)$$ such that $$\|h_n-f_x\|_{\infty} \to 0$$ as $$n \to \infty$$.

To this end, we first observe that the denseness of $$\text{span}(f_x \mid x \in \bar{A})$$ in $$C_0$$ implies that $$f(0) \neq 0$$. Indeed, if $$f(0)$$ were zero, then $$h(0)=0$$ for any $$h \in \text{span}(f_x \mid x \in \bar{A})$$ which would mean that the family cannot be dense in $$C_0$$ (just pick some $$g \in C_0$$ with $$g(0) \neq 0$$).

Now let $$x \in \bar{A}$$ and $$\epsilon>0$$. Since $$f_0(y)=f(0) \notin C_0$$, we have $$x \neq 0$$. By definition, there exists $$(x_n)_{n \in \mathbb{N}} \subset A$$ such that $$x_n \to x$$. Moreover, $$f \in C_0$$ implies that we cann choose $$R>0$$ such that

$$|f(uy)| \leq \epsilon \quad \text{for all |u| \geq \frac{|x|}{2}, |y| \geq R}.\tag{1}$$

As $$f$$ is uniformly continuous we can also choose $$\delta>0$$ such that

$$|f(uy)-f(vy)| \leq \epsilon \quad \text{for all |u-v| \leq \delta, |y| \leq R}. \tag{2}$$

Since $$x_n \to x$$ and $$x \neq 0$$, it holds that $$|x_n| \geq |x|/2$$ and $$|x_n-x| \leq \delta$$ for $$n \geq N$$ sufficiently large. On the one hand, we have by $$(1)$$

$$|f_x(y)-f_{x_n}(y)| \leq |f(xy)|+ |f(x_ny)| \leq 2 \epsilon$$

for all $$|y| \geq R$$ and $$n \geq N$$; on the other hand, by $$(2)$$

$$|f_x(y)-f_{x_n}(y)| =|f(xy)-f(x_ny)| \leq \epsilon$$

for all $$|y| \leq R$$ and $$n \geq N$$. Hence,

$$\|f_x-f_{x_n}\|_{\infty} \leq 2 \epsilon, \qquad n \geq N.$$

As $$\epsilon>0$$ is arbitrary, this proves that we can find $$(h_n)_{n \in \mathbb{N}} \subset \text{span}(f_x \mid x \in A)$$ such that $$\|h_n-f_x\|_{\infty} \to 0$$ as $$n \to \infty$$.

• Thanks mate. I only abled to show that $h_n\to f_x$ pointwise and knew that I required an uniform convergent and uniformity comes from uniform continuity of $f$. – Mathemajician Jul 28 '19 at 14:09
• @Mathemajician You are welcome. – saz Jul 28 '19 at 14:10
• One humble request. Please complete equation (2) so that i can mark it as a answer – Mathemajician Jul 28 '19 at 14:20
• @Mathemajician What do you mean by "complete the equation"? Looks quite complete to me. – saz Jul 28 '19 at 15:21
• @equation (2) $|f(uy)-f(vy)|$ should be $\leq \epsilon$. Isn't it? – Mathemajician Jul 28 '19 at 16:25