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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 ?

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What is $C_b$? ${}{}$ – copper.hat Sep 21 '13 at 22:20
Continuous and bounded but since $[0,1]$ is compact, bounded is redundant I now recognize. – Epsilon Sep 21 '13 at 22:22
Redundancy is good. Like I say, redundancy is good. – copper.hat Sep 21 '13 at 22:25
up vote 0 down vote accepted

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).

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Could you explain why $g \in U_n?$ I cant figure it out. – Epsilon Sep 22 '13 at 9:54
@André, it's not, of course. Thanks, I'll fix that. – Jonathan Y. Sep 22 '13 at 11:37
You made a misktake while manipulationg the numerator. It is $(d-c)^{-1} \cdot $ the fraction. Changing the slope $\frac{n+a}{d-c}$ to $(n+a)(d-c)$ fixes this problem. – Epsilon Sep 23 '13 at 10:23
@André, many thanks. – Jonathan Y. Sep 23 '13 at 11:50

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