## Winter 1995 #4

Problem Statement: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be uniformly continuous and let $g_k(x)=f(x+\dfrac{1}{k})$ for every $k\in\mathbb{N}$. Show $g_k$ converges uniformly.

Proof: First I claim that $g_k(x)$ converges pointwise to $f(x)$.

Let $x\in\mathbb{R}$. Then $\lim\limits_{k\rightarrow\infty}g_k(x)=\lim\limits_{k\rightarrow\infty}f(x+\dfrac{1}{k}$. Since $f$ is continuous and $\{x+\dfrac{1}{k}\}$ converges to $x$ it follows that $\lim\limits_{k\rightarrow\infty}f(x+\dfrac{1}{k}=f(x)$. This holds for all $x\in\mathbb{R}$ and so $g_k$ converges to $f$ pointwise.

Now to show that this convergence is uniform. Let $\varepsilon>0$. Since $f$ is uniformly continuous there exists a $\delta>0$ such that for every $|x-y|<\delta$ it follows that $|f(x)-f(y)|<\varepsilon$.

Let $x\in\mathbb{R}$, then there exists a $N\in\mathbb{N}$ such that for every $k\geq N$ it follows that $|\dfrac{1}{k}|<\delta$ and so $x+\dfrac{1}{k}\in (x-\delta,x+\delta)$.

For all $k\geq N$ and for every $x\in\mathbb{R}$ it follows that $|(x+\dfrac{1}{k}-x|=|\dfrac{1}{k}|<\delta$. Uniform continuity implies that $|f(x+\dfrac{1}{k})-f(x)|<\varepsilon$ for all $k\geq N$ and for all $x\in\mathbb{R}$. But $g_k(x)=f(x+\dfrac{1}{k})$, thus, $|g_k(x)-f(x)|<\varepsilon$ for every $k\geq N$ and for all $x\in\mathbb{R}$. Therefore, $g_k$ converges to $f$ uniformly on $\mathbb{R}$.

$\Box$

Reflection: The main thing in this proof is that we know $f$ is uniformly continuous. It makes sense that the convergence has to be uniform. It convergence was only pointwise then your function could be doing crazy things and so $f$ wouldn’t be uniformly continuous.