TITLE: Vanishing eigenvalues of Jacobian
QUESTION [13 upvotes]: Let $f: \mathbb{R^2}\to \mathbb{R^2}$ be a Schwartz  function. If the eigenvalues of $Df$ vanish everywhere, must $f$ be constant? Does an analogous result hold when we replace $2$ by $n$?
Any properties of $f$ would be of interest. For example, clearly $\nabla \cdot f = 0$.

REPLY [11 votes]: For some reason, the 'edit' button didn't appear for my earlier answer, maybe because it was already accepted.  Thus, I'm adding the general $n$ argument as a separate answer.
In fact, there is a stronger result:  Suppose that $f:\mathbb{R}^n\to\mathbb{R}^n$ is smooth and has the properties that (i) the Jacobian $Df(x)$ has no negative real eigenvalues for any $x\in \mathbb{R}^n$ and (ii) there exist $\epsilon>0$
and a constant $K$ such that
$$
\bigl| f(x)\bigr| \le \frac{K}{\bigl(1+|x|\bigr)^\epsilon}
\qquad \forall x\in\mathbb{R}^n.
$$
Then $f\equiv0$.
Here is the argument:  For $t\ge 0$, 
consider the smooth mapping $\Phi_t:\mathbb{R}^n\to\mathbb{R}^n$ defined by
$$
\Phi_t(x) = x + tf(x).
$$
Of course, $\Phi_0(x) = x$.  Moreover,
$$
\det(D\Phi_t(x)) = \det\bigl(I_n + t Df(x)\bigr) > 0
$$
by the hypothesis that $Df(x)$ has no negative real eigenvalues for any $x$.
Thus, $\Phi_t:\mathbb{R}^n\to\mathbb{R}^n$ 
is a local diffeomorphism for all $t$.  In particular, all values of $\Phi_t$ are regular values.  Moreover, when $|x|=R>0$, 
$$
 \bigl|\Phi_t(x)-x\bigr|  \le \frac{tK}{(1+R)^\epsilon},
$$
and, by taking $R$ very large, 
it follows easily that $\Phi_t$ must be surjective 
(since it barely moves the sphere $|x|=R$ for $R>>0$).  Then, 
by degree theory (and the fact that $\Phi_t$ is an orientation-preserving 
local diffeomorphism), it follows that $\Phi_t$ must be injective as well.  I.e., $\Phi_t$ is a diffeomorphism of $\mathbb{R}^n$ with itself for all $t\ge 0$.
Suppose now that $f$ were not identically zero.  By translation, I can assume,
without loss of generality, that $f(0)\not=0$, 
say $\bigl|f(0)\bigr| = M > 0$.   Now, choose an $R>>0$ satisfying
$$
\frac{K}{(1+R)^\epsilon} < M,
$$
which is possible because $\epsilon>0$, and then choose $t>>0$ so that
$$
t\left(M-\frac{K}{(1+R)^\epsilon}\right) > R.
$$
Then, for all $x\in\mathbb{R}^n$ with $|x|=R$,
$$
\left|\Phi_t(x)\right| =  \left|x + tf(x)\right|
\le R + \frac{tK}{(1+R)^\epsilon} < t M.
$$
In particular, $\Phi_t$ carries the sphere $|x|=R$ 
into the ball $|x|\le R'$ for some $R' \lt tM$. 
Thus, because $\Phi_t$ is a diffeomorphism of $\mathbb{R}^n$ with itself,
it must carry the ball $|x|\le R$ into the ball $|x| \le R'$. 
On the other hand, $\left|\Phi_t(0)\right| = \left|tf(0)\right| = tM > R'$,
and this is a contradiction.
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