Fix $L>0$. Show that the set $$ A=\{f\in \mathcal{C}([0,1]): \exists t\in [0,1] s.t. |f(t)-f(s)|\leq L|t-s| \forall s\in [0,1] \} $$ has empty interior in $\mathcal{C}([0,1])$.
Hint: Show that the complement of $A$ is dense by approximating an arbitrary continuous function by "sawtooth" functions with arbitrarily large slope.
Anyone have a hint for how to construct sawtooth functions the way the hint mentioned? I tried, but it ended looking really ugly. Is that inevitable?
Just define the triangle function $$g(x) = \begin{cases} x & \text{ if } x \in [0,1/2] \\ 1-x & \text{ if } x \in (1/2,1].\end{cases}$$ and set $g(x) =0$ for $x \notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows: $$g_n(x) := \sum_{k=1}^{n} g(nx-k).$$ In any point $x \in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| \quad \text{for} \quad x,y \in (k/n,(k+1/2)/n).$$ For an $f \in \mathcal{A}$, we can define $h_\varepsilon := f+ \varepsilon g_n$, where $n$ is taken so large that $\varepsilon n \ge 3L$. Then $$\|f-h_\varepsilon\| = \varepsilon \|g_n\| \le \varepsilon,$$ but for $x \in (k/n,(k+1/2)/n)$ $$|h_\varepsilon(x)-h_\varepsilon(y)| \ge \varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$ i.e. $h_\varepsilon \notin \mathcal{A}$. We conclude that $\mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $\mathcal{A}$ is closed: Take a sequence $(f_n)_{n \in \mathbb{N}} \subset \mathcal{A}$ with $f_n \rightarrow f$ uniformly. Then there exists a $t_n \in [0,1]$ with $$\tag{1}|f_n(t_n)-f_n(s)| \le L|t_n-s| \quad \text{for all} \quad s \in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t \in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n \rightarrow \infty$ in (1) shows that $f \in \mathcal{A}$.