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.

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This entry was posted in Analysis, Sequence, Sequence of Functions, Uniform Continuity, Uniform Convergence. Bookmark the permalink.

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