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Let $C_{b}(\mathbb{R},\mathbb{R}):=\{f:\mathbb{R}→\mathbb{R} \,\,;f \style{font-family:inherit}{\text{ is continuous and }}||f||_{\infty}<\infty\}$. Let $\operatorname{Lip}(\mathbb{R}):=\{f:\mathbb{R}→\mathbb{R}\,\,: f \style{font-family:inherit}{\text{ is Lipschitz continuous}}\}$.

Problem: Show that $ \operatorname{Lip}(\mathbb{R})\cap C_{b}(\mathbb{R},\mathbb{R}) $ is not dense in $C_{b}(\mathbb{R},\mathbb{R})$ relative to metric $||\cdot||_{\infty}$.

Attempt. My idea is to choose a $\epsilon>0$ and a $f\in C_{b}(\mathbb{R},\mathbb{R})$ such that for all $g\in \operatorname{Lip}(\mathbb{R})\cap C_{b}(\mathbb{R},\mathbb{R})$, we have that $||f-g||_{\infty}\geq \epsilon$. But I can't think of how to start this.

Gabriel Ribeiro
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2 Answers2

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As hinted in the comments, Lipschitz implies uniformly continuous, and uniform continuity survives the $\|\cdot\|_{\infty}$ closure (basic epsilon-delta exercise, e.g. check here). So, if you had density, then all bounded continuous maps on $\Bbb R$ would be uniformly continuous.

So, it is enough to find one bounded continuous map on $\Bbb R$ which is not uniformly continuous, for example $f(x) = \sin(x^2)$.

User
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Let $f_n$ be a function whose graph is a triangular hat of height $1$ centered at $x=n$ and which has width $1/n$. Then $f = \sum_{n=1}^\infty f_n$ is in $C_b(\mathbb{R})$. However, if for every $\varepsilon>0$ there existed Lipschitz $g$ such that $\|f-g\|_\infty <\varepsilon$, then $f$ would be uniformly continuous since by choosing an appropriate $g$ we get $$|f(x)-f(y)|\leq |f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|\leq 2 \varepsilon + L|x-y|$$

Which can be made small in a way depending only on $|x-y|$. However, $f$ is not uniformly continuous due to the shrinking support of the $f_n$ as $n\to\infty$, a contradiction.

kieransquared
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