Let $f:\mathbb{R}\to \mathbb{R}$ be a function given by $f(x)=\sum_{k=1}^\infty \frac{g(x-k)}{2^k}$ where $g:\mathbb{R}\to \mathbb{R}$ is a uniformly continuous function such that the series converges for each $x$ belongs to $\mathbb{R}$. Then show that $f$ is uniformly continuous.
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Let $f:\mathbb{R}\to \mathbb{R}$ be a function given by $f(x)=\sum_{k=1}^\infty \frac{g(x-k)}{2^k}$ where $g:\mathbb{R}\to \mathbb{R}$ is a uniformly continuous function such that the series converges for each $x$ belongs to $\mathbb{R}$.
Let $\varepsilon >0$ since $g $ is uniformly continuous there is $\delta >0$ such that for every $x,y\in\mathbb R$ if $|x-y|<\delta$ then
$$ |g(x)-g(y)|< \varepsilon $$
Accordingly, if $ |y-x|<\delta$ then for every $k\ge 0$ $|(x-k)-(y-k)|= |x-y|< \delta $ that is $$|g(x-k) -g(y-k)|<\varepsilon $$ Therefore $$ |f(x) -f(y)|\le \sum_{k=1}^\infty \frac{|g(x-k) -g(y-k)|}{2^k} < \varepsilon \sum_{k=1}^\infty \frac{1}{2^k} = \varepsilon $$
If $g$ is uniformly continuous in $\mathbb{R}$ then given $\epsilon>0$ there exist $\delta>0$ such that if $|x-y|<\delta$ then $|g(x)-g(y)|<\epsilon$. Hence, if $|x-y|<\delta$, then $|(x-k)-(y-k)|=|x-y|<\delta$ and $$|f(x)-f(y)|\leq \sum_{k=1}^\infty \frac{|g(x-k)-g(y-k)|}{2^k}\leq \sum_{k=1}^\infty \frac{\epsilon}{2^k}=\epsilon,$$ that is $f$ is uniformly continuous in $\mathbb{R}$.