Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to prove the following: given a family $\mathcal F$ of Lipschitz functions $f: [0,1] \rightarrow \mathrm R^2$, with a common Lipschitz constant, such that $\{f(0): f \in \mathcal F\}$ is bounded, there exists a continuous function $g: [0,1] \rightarrow \mathrm R^2$ whose graph intersects each of the graphs of the functions in $\mathcal F$.

Since there are at most $\mathfrak c$ functions in $\mathcal F$, we can conclude that there exists an almost everywhere continuous function $g$ that solves the problem (for example, defining $g$ in Cantor set and extending linearly). However, a continuous approximation of such a function may not solve the problem.

I would appreciate other suggestions to solve this problem.

share|cite|improve this question
Can you clarify what it means for the graph of two functions $[0, 1] \to \mathbb{R}^2$ to intersect? – Srivatsan Jan 25 '12 at 16:11
With "graph" I meant the subset of $\mathrm R^3$, that is, for every $f \in \mathcal F$, there is $t \in [0,1]$ such that $f(t) = g(t)$. – Daniel Jan 25 '12 at 16:38

There is a square containing all the graphs of the family $\mathcal{F}$. Now take $g$ to be your favourite square filling continuous curve.

share|cite|improve this answer

Consider the family of constant (therefore Lipschitz) functions $f:[0,1] \to [0,1]^2$. Then the problem is to find a $g:[0,1] \to [0,1]^2$ continuous surjection. This is possible, see e.g. Hilbert curve. This also solves the general case.

Let $|f(0)|<C$ and the common Lipschitz constant is $L$ then $|f(x)|<C+|x|L \le C+L$

Therefore all possible values of all $f \in \cal F$ are contained in a disk of radius $C+L$. This is homeomorphic to the square $[0,1]^2$. Now let $g$ be the composition of the Hilbert curve and such a homeomorphism. Then the range of $g$ covers the disk of radius $C+L$ around the origin.

edit: I guess the above proof works in more than two dimensions, so using the fact that the graph of a Lipschitz function is itself Lipschitz, the $\bf R^3$ version is answered too.

share|cite|improve this answer
Since a cube-filling curve is not the graph of a function, I don't think this will work – Daniel Jan 25 '12 at 17:18

It might not solve, but it's an idea... Have you tried the Arzelà–Ascoli Theorem? You've got an equicontinuous bounded sequence of functions defined in a compact subset of $R$. Then there exists a subsequence $f_{n_i}$ which converges uniformly to $f$ and $f$ is continuous. Now, perturb $f$ a little bit in a way that it will intersect all $f_{n_i}$ if $i\geq N$. There are only finite $f_{n_i}$ left so you can extend $f$ in a way that it passes over all points $f_{n_i}(0)$. We didn't find an $f$ that intersects all of the $f_n$'s but it crosses a whole subsequence of it. Maybe this idea can give some new directions... That's the best a could do.

share|cite|improve this answer
Why is $\mathcal{F}$ a sequence of functions? (In fact, the OP clearly states that there are at most $\mathfrak{c}$ functions in it...) – Srivatsan Jan 25 '12 at 16:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.