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Let $f :\mathbb{R}^2\rightarrow \mathbb{R}^2 $ be everywhere differentiable such that the Jacobian is not singular at any point in $\mathbb{R}^2$. Assume $|f|\leq 1$ whenever $|x|=1$. Prove that $|f|\leq 1$ whenever $|x|\leq 1$.

I think this is straight forward if we apply maximum modulus principle. But how may I prove it without using it? I tried to use Implicit function theorem by defining $g:\mathbb{R}^2\rightarrow \mathbb{R}$ by $g(x)=|f(x)|^2$, but no success.

At least a hint is appreciated.

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Consider $g(x) = |f(x)|^2$. We note by the chain rule that $$ Dg(x) = 2[f(x)]^T Df(x) $$ In order for $g$ to achieve a maximum at $x$ on the interior of the unit ball, it must have a critical point at that same $x$, i.e. a point such that $Dg(x) = 0$. Because the Jacobian $Df$ is non-singular at all points, this would imply that $f(x) = 0$.

So, $g(x) = 0$ at any potential maximum on the interior. So, $g$ has no local maximum on the interior of the unit ball. The conclusion follows.

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  • $\begingroup$ I also got upto the line before the last line of your answer. How would it imply $g$ cannot attain max outside the boundary? Can you please elaborate it if you don't mind? $\endgroup$ – Extremal Dec 28 '15 at 18:40
  • $\begingroup$ @EpsilonDelta I showed that if $g$ attains a max at $x$ (on the interior, which is to say outside the boundary), then $f(x) = 0$. If $f(x) = 0$, then $g(x) = |f(x)|^2 = 0$. $\endgroup$ – Omnomnomnom Dec 28 '15 at 18:43
  • $\begingroup$ I've rephrased that last bit. $\endgroup$ – Omnomnomnom Dec 28 '15 at 18:44
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Let $x_0$ be a point in the disk $D=\{|x|\le 1\}$ where $|f|$ has an absolute maximum. Suppose by contradiction that $|f(x_0)|>1$. Then $x_0$ is an internal point of $D$ (because on the boundary $|f|=1$. Hence, by assumption, $Df_{x_0}$ is invertible, and by the local invertibility Theorem the function $f$ is invertible in a neighbourhood $U$ of $x_0$ all contained in $D$. Hence $f(D) \supset f(U) \supset V$ where $V$ is a neighborhood of $f(x_0)$. But this is a contradiction, because in $V$ there are points $y$ with $|y|$ larger than $|f(x_0)|$.

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The set $C=\{x | \|x\| \le 1 \}$ is compact, hence there is a maximiser $x_0 \in C$ of of $|f|$.

If $|f(x_0)| \le 1$ then there is nothing to show, so suppose $|f(x_0)| >1$, in which case we must have $x_0 \in \{x | \|x\| < 1 \}$ and hence $\phi(x)=|f(x)|$ has a local maximiser at $x_0$. Since $\phi$ is differentiable at $x_0$, we have $D \phi(x_0) = {f(x_0) \over |f(x_0)| } D f(x_0) = 0$ we see that $Df(x_0) = 0$ which is a contradiction.

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