Uniform Continuous function f has limit zero at infinity Suppose $f(x)$ if uniformly continuous in $[0,+\infty)$. Prove that if $\{f(x+n)\}_{n}$ converges to $0$ $\forall x \in [0,1]$, then $\lim\limits_{x\rightarrow \infty} f(x) = 0$.
 A: The sequence $(g_n)$ in $C([0,1])$, where
$$g_n(x)= f(x+n)$$
is (uniformly) equicontinuous. By assumption, it converges to $0$ pointwise. On equicontinuous subsets of $C([0,1])$, the topologies of pointwise convergence and of uniform convergence coincide(1), hence the sequence $(g_n)$ converges to $0$ uniformly, thus for every $\varepsilon > 0$ there is an $n_\varepsilon$ with $\lvert g_n(x)\rvert < \varepsilon$ for all $x\in [0,1]$ and $n \geqslant n_\varepsilon$. That is, we have $\lvert f(x)\rvert < \varepsilon$ for $x \geqslant n_\varepsilon$, and since $\varepsilon > 0$ was arbitrary, $\lim\limits_{x\to\infty} f(x) = 0$ is shown.
(1) It is clear that the topology of uniform convergence is finer than the topology of pointwise convergence, so we need only show the converse. Let $H\subset C([0,1])$ be equicontinuous. For $f\in H$ and $\delta > 0$ we need to find an $\eta > 0$ and a finite subset $\mathscr{F}\subset [0,1]$ such that
$$\sup \left\{ \lvert f(x) - g(x)\rvert : x \in \mathscr{F}\right\} \leqslant \eta \implies \sup \left\{ \lvert f(x) - g(x)\rvert : x \in [0,1]\right\} \leqslant \delta.$$
Choosing $\eta = \delta/3$ works: By the equicontinuity of $H$, each $x\in [0,1]$ has an open neighbourhood $U_x$ such that
$$y\in U_x \implies \lvert h(y) - h(x)\rvert \leqslant \eta$$
for all $h\in H$. $\{ U_x : x \in [0,1]\}$ is an open cover of the compact space $[0,1]$, hence it has a finite subcover, i.e. there is a finite $\mathscr{F} \subset [0,1]$ with $$[0,1] = \bigcup_{x\in\mathscr{F}} U_x.$$
Then if $g\in H$ with $\sup \left\{ \lvert f(x) - g(x)\rvert : x \in \mathscr{F}\right\} \leqslant \eta$, for an arbitrary $y\in [0,1]$ we choose an $x_0\in\mathscr{F}$ with $y\in U_{x_0}$, and have
$$\begin{aligned}
\lvert g(y) - f(y)\rvert &= \lvert (g(y) - g(x)) + (g(x) - f(x)) + (f(x) - f(y))\rvert\\
&\leqslant \lvert g(y) - g(x)\rvert + \lvert g(x) - f(x)\rvert + \lvert f(x) - f(y)\rvert\\
&\leqslant \eta+\eta+\eta = \delta.
\end{aligned}$$
Thus on $H$, every neighbourhood of $f$ in the topology of uniform convergence is also a neighbourhood of $f$ in the topology of pointwise convergence, which means that on $H$ the topology of pointwise convergence is finer than the topology of uniform convergence.
