Convergence of $(f_n)$

Question

Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be an uniformly continuous function. Suppose $f_n(x)=f\left(x+\cfrac{1}{n}\right)$ for all $n\in\mathbb{N}$ and for all $x\in\mathbb{R}$. How can I prove that $(f_n)$ converges uniformly to a continuous function?

I just need some hint to begin please. Any help is welcome!

Answer 1

Let $\epsilon > 0$ be given.
Since $f$ is uniformly continuous, there exists $\delta > 0$ such that $$|f(x) - f(y)| < \epsilon,$$ whenever $|x - y| < \delta.$

Now choose $N \in \Bbb N$ such that $1/N < \delta$. Then, for all $n \ge N$, we have $$\left|f\left(x + \dfrac{1}{n}\right) - f(x)\right| < \epsilon$$ for all $x \in \Bbb R$.

Using the defintion $f_n$, we get that $$|f_n(x) - f(x)| < \epsilon,$$ for all $n \ge N$ and $x \in \Bbb R$.

Answer 2

Hint: We have $$|f_n(x)-f(x)|=\left|f\left(x+\frac1n\right)-f(x)\right|.$$ Now use uniform continuity of $f$.