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Let $Q=[-1,1]^2$ denote the unit square and let $f:Q\to Q$ be a Lipschitz function such that for any ball $B(a,r)\subset Q$ with radius $r$, the width of the image $f(B(a,r))$ is at least $cr$ for some absolute constant $c>0$. (The width of a compact set is defined (as in here) to be the smallest distance between two parallel lines containing the set within them).

Can such a function have a nowhere dense image? In other words is there such a function $f$ for which $f(Q)$ contains no open ball?

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    $\begingroup$ What if you look at the image of a small ball around a point at which $f$ is differentiable? $\endgroup$ – Bill Johnson Sep 2 '14 at 21:28
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This answer builds on Bill Johnson's comment. This is not a full answer since a step is missing, but too long for a comment.

A Lipschitz function is almost everywhere differentiable, so let $a\in Q$ be a point where $f$ is differentiable. For simplicity, we can take $a$ to be an interior point. There is a $2\times2$ matrix $A$ and a continuous function $R:Q\to\mathbb R^2$ so that $$ f(a+x)=f(a)+Ax+|x|R(x) $$ and $R(0)=0$.

(This paragraph is not necessary, but may give better insight to the next one.) If $A=0$, then $|f(a+x)-f(a)|\leq|x||R(x)|$ and so $\text{diam}(f(B(a,r)))\leq2r\sup_{B(0,r)}|R|$. Since the width of $f(B(a,r))$ is at least $cr$, its diameter also has this lower bound. Thus $c\leq2\sup_{B(0,r)}|R|$ for all $r>0$. But this is impossible since the supremum tends to zero as $r\to0$. Therefore $A\neq0$.

Suppose $A$ does not have full rank and let $v$ be a unit vector perpendicular to its image. Now $$ v\cdot(f(x+a)-f(a)) = |x|v\cdot R(x). $$ The width of $f(B(a,r))$ is bounded from above by its width in the direction of $v$, which in turn is bounded by $$ \sup_{y\in B(a,r)}|v\cdot(f(y)-f(a))| = \sup_{x\in B(0,r)}|x||R(x)|. $$ As this is bounded from below by $cr$, we get $$ c \leq \sup_{x\in B(0,r)}|R(x)| $$ for all $r>0$. This is a contradiction since the supremum tends to zero as $r\to0$ by continuity of $R$. Thus $A$ has full rank.

The mapping $x\mapsto f(x)-Ax$ is Lipschitz. Suppose one can show that the Lipschitz constant can be made arbitrarily small by restricting it to $B(a,r)$ for $r$ small enough. (This would be easy if $f$ was continuously differentiable.) If this Lipschitz constant is less than $\frac12\|A^{-1}\|^{-1}$, then $f(x)=f(y)$ implies $$ |x-y| = |A^{-1}(f(x)-Ax-f(y)+Ay)| \leq \frac12|x-y| $$ and thus $x=y$. Thus $f$ is injective in a small neighborhood of $a$. Then it follows from invariance of domain that $f(a)$ is an interior point of $f(Q)$.

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  • $\begingroup$ After Bill's comment I arrived at a similar proof, but was too lazy to write it here. This answer gives good amount of details so I will go ahead and accept it (and I learned about invariance of domain!). $\endgroup$ – Joel Moreira Sep 16 '14 at 21:00

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