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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 $\nabla \cdot f = 0$.

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No, the function $f(x, y) = (y, 0)$ is a counter-example to your conjecture. – Matthias Ludewig Sep 1 '14 at 14:40
@Kofi That function isn't Schwartz. – Benji Sep 1 '14 at 14:44
@AndrásBátkai: that function is still not Schwartz – Willie Wong Sep 1 '14 at 15:10
@AndrásBátkai To expand on Willie Wong's comment, the function doesn't decay in the 'x-direction.' – Benji Sep 1 '14 at 15:12
The constant is 0. As you noted, $f$ is a divergence free vector field which is rapidly decreasing. For $n=2$ it is a Hamiltonian vector field, thus $f=\text{grad}^{\omega}(g)$ for a function $g$ on $\mathbb R^2$. Moreover, $g$ falls rapidly to a constant, so we may assume that $g$ is Schwartz also. – Peter Michor Sep 1 '14 at 15:51
up vote 19 down vote accepted

Actually, I made a sign mistake in my original calculation of $\det(Df)$, so my original argument was not right. Sorry. Here is (I hope and believe) a correct one.

In fact, when $n=2$, such an $f$ must be zero. (If it's constant, it must be zero, as Peter noted.)

As Peter pointed out, $f = (-u_y, u_x)$ for some function $u:\mathbb{R}^2\to\mathbb{R}$. The condition that the eigenvalues of $Df$ vanish requires, in addition, that $$ \det(Df) = u_{xx}u_{yy}-u_{xy}^2=0, $$ This implies that the graph $\bigl(x,y,u(x,y)\bigr)$ is a complete flat surface in $\mathbb{R}^3$ (i.e., its Gauss curvature vanishes). In particular, by the classic theorem on complete flat surfaces, this graph is ruled by parallel straight lines. The projection of this ruling onto the domain plane $\mathbb{R}^2$ then gives a ruling of the plane by parallel straight lines. By a rotation, we can assume that the rulings project to the parallels to the $x$-axis. The parallel ruling condition on the graph then implies that $u_x$ is a constant $a$ and so $u$ must be of the form $u(x,y) = ax + h(y)$ for some function $h$ and some constant $a$. However, this gives $f = (-h'(y),a)$, and so the requirement that $f$ be Schwartz implies that $a = h'(y) = 0$, so $f$ is zero.

N.B.: Note that the condition of $f$ being Schwartz is not used until the very end, and, in fact, it is far stronger than necessary. Obviously, it would be enough to know merely that $\|f(x,y)\|\rightarrow0$ as $\|(x,y)\|\rightarrow\infty$ in order to conclude that $f\equiv0$.

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Any guesses on the case $n > 2$? This proof is highly dependent on $n = 2$ (both the first step where $f = *du$ and the classification of surfaces). But I wonder whether the basic idea can be lifted: the condition $\det(Df) = 0$ is suggestive (up to problems with the kernel having higher dimension) of there being a vector field $v$ such that $v\cdot Df = 0$. And a potential proof would then rest on showing that the integral curves of $v$ extends to infinity. (In your proof the integral curves are precisely the parallel straight lines.) – Willie Wong Sep 2 '14 at 7:31
Indeed, it's true for all $n$, and with considerably weaker hypotheses on the decay of $f$. I'll put in a proof today when I have time, probably this afternoon. – Robert Bryant Sep 2 '14 at 11:04

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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Very nice proof. – Peter Michor Sep 3 '14 at 9:46
Doesn't your proof show that, as in the $n=2$ case, it is enough to assume that $f$ vanishes at infinity, i.e., $f =\mathcal{O}(|x|^0).$ – Benji Sep 3 '14 at 10:02
Sure, one can weaken the decay assumptions on $f$ even further and still get the vanishing conclusion, but I wasn't interested in that so much as in stating a simple decay assumption along the lines of Schwartz that would make the idea of the proof clear. It would suffice to suppose that, for every $\epsilon>0$, there is an $R>0$ such that $|f(x)|\leq\epsilon$ when $|x|\ge R$. – Robert Bryant Sep 3 '14 at 10:32
A side note: My freshman calculus exam consisted in developing Fourier theory for functions with exactly these decay conditions. – Vít Tuček Sep 3 '14 at 15:03
It seems that in fact you only need all eigenvalues to be of the same sign for in the case when $Df(x)$ has no positive eigenvalues you can use $\Phi_t(x) = x - tf(x)$. – Vít Tuček Sep 3 '14 at 15:18

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