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 $n \to \infty$.

  • $\begingroup$ 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 $f$. $\endgroup$ – Mathemajician Jul 28 '19 at 14:09
  • $\begingroup$ @Mathemajician You are welcome. $\endgroup$ – saz Jul 28 '19 at 14:10
  • $\begingroup$ One humble request. Please complete equation (2) so that i can mark it as a answer $\endgroup$ – Mathemajician Jul 28 '19 at 14:20
  • $\begingroup$ @Mathemajician What do you mean by "complete the equation"? Looks quite complete to me. $\endgroup$ – saz Jul 28 '19 at 15:21
  • $\begingroup$ @equation (2) $|f(uy)-f(vy)|$ should be $\leq \epsilon$. Isn't it? $\endgroup$ – Mathemajician Jul 28 '19 at 16:25

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