# (existence of) not Lipschitz function for every bound closed interval in $[0,1]$ with real values

Background: this question was asked in the course "Introduction to topology and metric spaces"

Prove that there exists a continuous function $f\colon[0,1]\to\mathbb R$ such that that for every $0 \leq r < s \leq 1$, $f$ is not Lipschitz on $[s,t]$.

Since the question was not asked in calculus or any kind of analytical course, I'm pretty sure that they don't expect a constructive proof.

Since all continuous functions $f\colon[0,1]\to\mathbb R$ are bound, then it follows that the metric space of these functions (with the supremum metric function) is a complete metric space. So far my main direction was using Cantor's intersection theorem to find the desired function in the intersection of a series of closed sets. The main problem is that the set of all such functions which are not Lipschitz on a specific $[s,t]$ is not a closed set. Another problem would be ordering these sets so that they are well-ordered and contained in one another. Both these problems make me think that even if Cantor's theorem is the right tool, I'm going about it the wrong way.

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Non Constructive proof involving the Baire theorem : (I don't know what you mean by Cantor's theorem)

Let $C_{n,k,m}$ be the set of all continuous functions that are m-Lipschitz on $[\frac{k}{n}, \frac{k+1}{n}]$. Each of those are closed, with empty interior for the sup topology (because there are very uniformly small but non lipschitz continuous functions).

Since the set of all continuous functions on $[0,1]$ with the sup topology is complete, it belongs to the Baire category and the countable union $\bigcup_{(n,k,m) \in \mathbb{N}^3 ; k+1 \leq n} C_{n,k,m}$ has empty interior, and therefore there are elements that do not belong to any of the $C_{n,k,m}$, QED.

Note : it should be possible to construct manually such a function : the usual construction of a continuous function which is nowhere differentiable works

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C_n,k,m are not closed (think of (x^(1/2))/n), but the interior of their closure is empty, which is enough for belonging to the Baire category. Thanks, good answer –  eladidan Dec 3 '11 at 17:17
p.s.: by Cantor's Intersection Theorem I meant: en.wikipedia.org/wiki/Cantor's_intersection_theorem often referred to as Cantor's Lemma –  eladidan Dec 3 '11 at 17:23

A concrete example is not much more difficult to give than a nonconstructive existence proof. For example, consider the Takagi function, defined by the uniformly convergent series
$$f(x) = \sum_{n=0}^\infty \operatorname{dist}(x,2^{-n}\mathbb Z)$$ To show the failure of Lipschitz condition, note that every subinterval contains a dyadic rational, i.e., a number $t=k/2^m$ with $k$ odd. Let $f_m(x)=\sum_{n=0}^m \operatorname{dist}(x,2^{-n}\mathbb Z)$ and $g_m=f-f_m$. For $N>m$ we have $$g_m(t+2^{-N}) = \sum_{n=m+1}^{N-1} 2^{-N} = (N-m-1) 2^{-N}$$ Since $f_m$ is $(m+1)$-Lipschitz, it follows that $$\begin{split} f(t+2^{-N})-f(t) &\ge g_m(t+2^{-N})-g_m(t) -(m+1)2^{-N} \\ &= (N-m-1) 2^{-N}-0 -(m+1)2^{-N} \end{split}$$ hence $$\frac{f(t+2^{-N})-f(t)}{2^{-N}} \ge N-2(m+1)$$ which tends to infinity as $N\to\infty$.

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