Construction of a certain function. Sorry for the vague title but here is the question. Let $F := C_b([0,1],\mathbb R)$ and 
$$
 U_n := \left \{f \in F: \forall x \in [0,1] \exists y \in [0,1]: \left \lvert \frac{f(x)-f(y)}{x-y} \right \rvert > n \right \}
$$
I want to prove that $U_n$ lies dense in $F$ given the $\sup$-norm. Heuristically given $\epsilon > 0$ and $f \in F$ we can make a function $g:[0,1] \rightarrow \mathbb R$ as follows: 
$f$ is uniform continuous since $[0,1]$ is compact (standard metric). So we can find a $\delta > 0$ s.t. for all $x,y \in [0,1]: |x-y|< \delta \Rightarrow |f(x)-f(y)|<\epsilon$. Now we can find points $0=x_1<x_2<\cdots<x_n=1$ s.t. $|x_i-x_{i+1}|<\delta$ for each $i$. Define $g(x_i) := f(x_i)$ .
Now I run into trouble. We can make $g$ piecewise linear and having some kind of a "saw-tooth" shape in order to have $g \in U_n$. Further it should be possible to get this shape and still having $\|f-g\|_\infty < \epsilon$. How can I formalize this ?
 A: The trick, as you say, is to superimpose a saw-tooth function--or something similar--on $f$. As you've also suggested, uniform continuity of $f$ means that for all $\epsilon>$ we can find a piecewise-linear function $\tilde{f}$ such that $\|f-\tilde{f}\|_\infty<\epsilon$. It would therefore suffice to approximate piecewise-linear functions by elements of $U_n$.
We define $\Lambda_{\epsilon,r}:[0,1]\to\mathbb{R}$ as the saw-tooth function with amplitude $\epsilon$ and slope $\geq r$ such that $\Lambda_{\epsilon,r}(0)=\Lambda_{\epsilon,r}(1)=0$. Note that, naturally, $\|\Lambda_{\epsilon,r}\|_\infty\leq\epsilon$ and $\Lambda_{\epsilon,n}\in U_n$.
If $f(x) = ax+b$ ($a\neq 0$) is defined on $[c,d]\subseteq[0,1]$, we let $g_{\epsilon,n}(x)=f(x) + \Lambda_{\epsilon,(d-c)(n+a)}(\frac{x-c}{d-c}))$, and note
$$\frac{|g_{\epsilon,n}(x)-g_{\epsilon,n}(y)|}{|x-y|} = \frac{|a(x-y) + \Lambda_{\epsilon,(d-c)(n+a)}(\frac{x-c}{d-c})-\Lambda_{\epsilon,(d-c)(n+a)}(\frac{y-c}{d-c})|}{|x-y|}\geq\\
\geq \frac{|\Lambda_{\epsilon,(d-c)(n+a)}(\frac{x-c}{d-c})-\Lambda_{\epsilon,(d-c)(n+a)}(\frac{y-c}{d-c})|}{|x-y|}-a$$
and as we know for all $x\in[c,d]$ we can find $y\in[c,d]$ such that
$$\frac{|\Lambda_{\epsilon,(d-c)(n+a)}(\frac{x-c}{d-c})-\Lambda_{\epsilon,(d-c)(n+a)}(\frac{y-c}{d-c})|}{|x-y|} =\\
= (d-c)^{-1}\frac{|\Lambda_{\epsilon,(d-c)(n+a)}(\frac{x-c}{d-c})-\Lambda_{\epsilon,(d-c)(n+a)}(\frac{y-c}{d-c})|}{|\frac{x-c}{d-c}-\frac{y-c}{d-c}|} = (n+a),$$
so that
$$\frac{|g_{\epsilon,n}(x)-g_{\epsilon,n}(y)|}{|x-y|}\geq (n+a)-a = n.$$
Also, $g_{\epsilon,n}(c)=f(c)$; $g_{\epsilon,n}(d)=f(d)$.
Finally, if $f:[0,1]\to\mathbb{R}$ is piecewise-linear, a combination of approximations on each segment of linearity will clearly finish the exercise (and continuity is also preserved).
