Uniform convergence of the sequence $f_n(x)=f(x+1/n)$ for uniformly continuous $f$

Question

Let $f$ be a uniformly continuous real-valued function on $(-\infty, +\infty)$, and for each $n\in I$ let $f_n(x)=f\left(x+\frac{1}{n}\right)$.

Prove that $\{f_n\}_{1}^{\infty}$ converges uniformly on $(-\infty, +\infty)$ to $f$.

I have no idea to prove this.

Answer 1

You have to show that given $\epsilon>0$ you can find an $k\in \Bbb{N}$ depending on $\epsilon$ only such that $$|f_n(x)-f(x)|=|f\left(x+\frac{1}{n}\right)-f(x)|<\epsilon \quad \forall n\ge k $$ Now use uniform continuity to show that above inequality is true.

Answer 2

$f$ is uniformly continuous on $\mathbb{R}$, so for $\forall \epsilon >0 \ \exists \delta >0; \ \forall x,\ y\in\mathbb{R} \ \text{if }\ |y-x|<\delta \Rightarrow|f(y)-f(x)|<\epsilon . \ \ (*)$

Now for this $\delta$, there exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\delta$. Hence for $\forall n\geq N$ and $\forall x\in\mathbb{R},\ y:=x+\frac{1}{n}$ we have $|y-x|=|(x+\frac{1}{n})-x|=\frac{1}{n}<\delta$. As a result by $(*)$, we have $|f(y)-f(x)|=|f(x+\frac{1}{n})-f(x)|<\epsilon$.

This prove your problem. Since, with my notations, for $\forall \epsilon>0 \ \exists N\in \mathbb{N}$ such that for $\forall x\in\mathbb{R} \ $ and $\forall n\geq N$, we have:

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