Density of Lipschitz functions on $C^0_b(\mathbb{R},\mathbb{R})$

Question

My problem:

Suppose $$V=C^0_b(\mathbb{R},\mathbb{R})= \{ f :\mathbb{R} \to \mathbb{R} | f \mbox{ is continuous and bounded}\}$$ with the usual norm $\|.\|_{\infty}$. I am asking if $$W= \{ f \in V | f \mbox{ is Lipschitz continuous }\}$$ is dense or not in $V$.

Attempt:

I tried using convolution but it is NOT true that: $$\lim\limits_{\varepsilon \to 0} \| \rho_{\varepsilon}*f - f \|_{\infty} = 0$$

because it is true only if $f \in L^p$ where $p<+\infty$.

Answer 1

I will leave it to you to prove each of the following statements:

1. Every Lipschitz continuous function is uniformly continuous.
2. The uniform limit of uniformly contiuous functions is again uniformly continuous.
3. Thus, a function that is not uniformly continuous cannot be uniformly approximated by Lipschitz functions.

Thus, if there exists a bounded, continuous function which is not uniformly continuous, then the set of Lipschitz functions is not dense.

Such a function indeed exists, e.g. for $$ f : \quad \Bbb{R} \to \Bbb{R}, \quad x \mapsto \sin(2 \pi \cdot x^2) $$ and arbitrary $n \in \Bbb{N}$ we have that $f(n) = \sin(2 \pi n^2) = 0$, but $$ f \Big( n + \frac{1}{8 n} \Big) = \sin \bigg( 2 \pi \Big(n^2 + \frac{1}{4} + \frac{1}{64 n^2} \Big) \bigg) = \sin \Big( \frac{\pi}{2} + \frac{\pi}{32 n^2} \Big) \to 1 \quad \text{as} \quad n \to \infty. $$ This easily shows that $f$ is not uniformly continuous.