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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 $\mathbb{R}$. Then show that $f$ is uniformly continuous.

How I show this. Please help me to solve this.

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  • $\begingroup$ \infty gives you $\infty$, and the command for fraction is \frac. As for the question itself, what have you tried? $\endgroup$
    – Arthur
    Commented Sep 2, 2017 at 10:24
  • $\begingroup$ Try and use Rabee's Theorem @abcdmath $\endgroup$ Commented Sep 2, 2017 at 10:26
  • $\begingroup$ I tried to show that the series is uniformly convergent. But if the series is uniformly convergent , then can I say f is uniformly continuous? @Arthur $\endgroup$
    – abcdmath
    Commented Sep 2, 2017 at 10:27
  • $\begingroup$ I like this exercise. it looks weird but follows just by using definition. $\endgroup$
    – Guy Fsone
    Commented Oct 8, 2017 at 17:26

2 Answers 2

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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 $\mathbb{R}$.

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  • $\begingroup$ @abcdmath Any further doubt? $\endgroup$
    – Robert Z
    Commented Sep 2, 2017 at 15:36
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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 $$ |f(x) -f(y)|\le \sum_{k=1}^\infty \frac{|g(x-k) -g(y-k)|}{2^k} < \varepsilon \sum_{k=1}^\infty \frac{1}{2^k} = \varepsilon $$

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