Prove that $f_n$ converges uniformly on $\mathbb{R}$

Question

Suppose that $f$ is uniformly continuous on $\mathbb{R}$. If $\displaystyle \lim_{n\to\infty} y_n =0$ and $f_n(x) := f(x + y_n)$ for $x \in \mathbb{R}$, prove that $f_n$ converges uniformly on $\mathbb{R}$.

Answer 1

Let $\epsilon > 0$. Since $f$ is uniformly continuous, there exists a $\delta > 0$ such that $$|f(x) - f(y)| < \epsilon$$ whenever $x,y \in \mathbb{R}$ with $|x-y| < \delta$. Since $y_n \to 0$, there exists a $N > 0$ such that $|y_n| < \delta$ for all $n > N$. Thus for any given $x \in \mathbb{R}$ and $n > N$ we have $|x + y_n - x| < \delta$ which implies $$|f_n(x) - f(x)| = |f(x+y_n) - f(x)| < \epsilon.$$ This means that that $f_n \to f$ uniformly.

Answer 2

$$\|f_n-f\|_\sup=\sup_{x}\,|f_n(x)-f(x)|=\sup_{x}\,|f(x+y_n)-f(x)|\,.$$