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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.

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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.

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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.

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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.

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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

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