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Let $f: \mathbb{R} \to \mathbb{R}$ uniformly continuous.We set $f_n(x)=f(x+\frac{1}{n})$.Show that $f_n \to f \text{ uniformly }$.

Let $\epsilon>0$.

Since $f: \mathbb{R} \to \mathbb{R}$ uniformly continuous, $\exists \delta>0 $ such that $\forall x,y \in \mathbb{R}$ with $|x-y|< \delta \Rightarrow |f(x)-f(y)|<\epsilon$.

For $y=x+\frac{1}{n}$,we have $|x-x-\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon \Rightarrow \frac{1}{n}< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon$

We pick $n_0$ such that $\frac{1}{n_0}< \delta$

$|f_n(x)-f(x)|=|f(x+\frac{1}{n})-f(x)| < \epsilon$

and taking the supremun,we get $\sup_{x \in \mathbb{R}} {|f_n(x)-f(x)|} < \epsilon$,so $f_n \to f \text{ uniformly }$.

Could you explain me why we pick $n_0$ such that $\frac{1}{n_0}< \delta$ ??

S L
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evinda
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3 Answers3

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Since you need $1/n<\delta$, you must choose any $n_0 \in \mathbb{N}$ such that $1/n_0<\delta$. As you write, $|x-y|=1/n$, and the choice of $n$ is dictated by the uniform continuity of $f$.

Siminore
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The part that says "$\frac{1}{n}\lt\delta\Rightarrow...$", is actually saying if
"$\frac{1}{n}\lt\delta$" this implies "..."

So to continue with the desired property we must pick an $n_0$ such that $\frac{1}{n_0}\lt\delta$

It may become more apparent if we look at a real life analogue:

If I pick up an apple, this implies that I can eat it.

Which means I must pick it up to eat it. But I am free to choose when I pick it up and with which hand. This is anologous to the choice of picking an $n_0$.

Ellya
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  • ellya! I understand!!! But..how can I formulate the answer at the beginning where I pick $n_0$?? Can I say it like that:

    $$\text{We choose } y=x+\frac{1}{n}$$ So,we have: $|x-x-\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon \text{ ,that implies } |\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|$.

    So,$\forall n \geq n_0:$

    $$|(x+\frac{1}{n})-x|< \delta \text{ and so we conclude that } |f(x+\frac{1}{n})-f(x)|< \epsilon$$

    Or do you have a better idea how I could formulate it?

     – evinda Jun 24 '14 at 15:43
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What does it mean for a sequence $f_n$ to be uniformly convergent to $f$? It just means that for every $\varepsilon>0$ you can choose a $n_0$ so that for all $n\geq n_0$: $sup_{x \in \mathbb{R}} {|f_n(x)-f(x)|} < \epsilon$. So what's been done is, that $n_0$ is chosen according to the premise, which is possible hence f is uniformly continous.

cQQkie
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  • Couldn't we take $\frac{1}{n_0}<\delta$ ? – evinda Apr 22 '14 at 13:35
  • Yes, you choose an $n_0$ so that $\frac{1}{n_0}<\delta$ – cQQkie Apr 22 '14 at 13:42
  • Could I take instead of $<$, $\leq$ ?So,that it is $\frac{1}{n_0} \leq \delta $ ? – evinda Apr 22 '14 at 14:20
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    If you admit $...\leq \delta$ in your formulation of uniform continuity... – cQQkie Apr 22 '14 at 15:54
  • how can I formulate the answer at the beginning where I pick $n_0$?? Can I say it like that:

    $$\text{We choose } y=x+\frac{1}{n}$$ So,we have: $|x-x-\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon \text{ ,that implies } |\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|$.

    So,$\forall n \geq n_0:$

    $$|(x+\frac{1}{n})-x|< \delta \text{ and so we conclude that } |f(x+\frac{1}{n})-f(x)|< \epsilon$$

    Or do you have a better idea how I could formulate it?

     – evinda Jun 24 '14 at 16:20