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Let $F:[0,1]\times[0,1]\to \mathbb {R}^2$ be a smooth function. Given $x\in [0,1]$, let $\ell_x:=\{x\}\times [0,1]$, and given $y\in [0,1]$, let $\ell_y:=[0,1]\times \{y\}$.

My question feels intuitive, by I have failed for a long time to find a rigorous argument. The most naive version of it is the follows:

Is there a constant $C>0$ s.t for all closed $A\subseteq [0,1]\times[0,1]$, $$\mathrm{Area}(F[A])\leq C\cdot \max_{x\in[0,1]}\mathrm{Length}(F[\ell_x\cap A])\cdot \max_{y\in[0,1]}\mathrm{Length}(F[\ell_y]) ?$$

Any drawing I make makes me feel like this should be true, but I cannot prove it. Moreover, it is of interest to me to understand what happens if the image of $F$ is in fact a manifold of bounded curvature?

Any help, guidance, or reference would be appreciated!

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    $\begingroup$ Erm... How about the identity map and the $\varepsilon$-neighborhood of the diagonal? $\endgroup$
    – fedja
    Commented Jan 7 at 6:55
  • $\begingroup$ It seems to work: the area is $2\epsilon\cdot \sqrt2 $, and for all $x$ $\mathrm{Length}(F[\ell_x\cap A])=2\sqrt2 \epsilon$, and for all $y$ $\mathrm{Length}(F[\ell_y])=1$. $\endgroup$
    – JustSomeGuy
    Commented Jan 7 at 17:25
  • $\begingroup$ Ah, yes. I misinterpreted the question. You are right. Let me think a bit then :-) $\endgroup$
    – fedja
    Commented Jan 7 at 17:28
  • $\begingroup$ Note the constant $C$ cannot be chosen independently from $F$. If $C=C(F)$, then it seems to be easy to prove. $\endgroup$
    – Anton Petrunin
    Commented Jan 7 at 18:33
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    $\begingroup$ If $F$ is $L$-Lipschitz, then $Area(F(A))\le L\int_{[0,1]}\text{length}(F(\ell_x\cap A))\,dx$, so if $\max_y\text{length}(F(\ell_y))>0$, you have your $C(F)$, but if the latter maximum is $0$, then the whole image of the square has $0$ area. So, if the constant is allowed to depend on $F$, there is no problem. Unfortunately, as Anton said already, a universal constant is out of question. $\endgroup$
    – fedja
    Commented Jan 7 at 20:13

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