I'm looking for a proof or counterexample of the following statement:

Conjecture: Let $\phi :[0,1]^n \to [0,1]^n $ be smooth. Suppose that each component of $\phi$ is convex, nondecreasing (in the sense that $\phi_i (x) \leq \phi_i (y)$ whenever $x_k \leq y_k$ for all $1\leq k \leq n$). Let $\mathbf{1}_n$ be the vector with all entries equal to $1$ and suppose $\phi(\mathbf{1}_n) = \mathbf{1}_n$. Assume further that the Jacobian $\mathcal{J}_\phi (\mathbf{1}_n)$ at that point has an eigenvalue $\lambda$ strictly greater than $1$. Further, suppose that $\phi(e_i)$ is a strictly positive vector for every unit vector $e_i$. Then $\phi$ has a fixed point in its domain other than $\mathbf{1}_n$.

The $n=1$ case is rather trivial, and I'm looking to generalize it. If the above doesn't work, can we add more 'niceness' assumptions to $\phi$ in order to make it work (like being a diffeomorphism)?

Any help is greatly appreciated.


Here is an analysis of your problem with simplified hypotheses on derivatives of $\phi$, but without convexity.

Assume $\phi: [0, 1]^n \to [0, 1]^n$ is $C^1$, $\dfrac{\partial \phi_i}{\partial \phi_j}(1_n) > \dfrac{1}{n}$ for any $(i, j)$, $\phi$ is non-decreasing (as defined in your post), and $\phi(1_n) = 1_n$.

Because of the second hypothesis, there is for any $i$ some $\epsilon_i > 0$ such that $\phi_i((1 - \epsilon_i) 1_n) \leq 1 - \epsilon_i$.


Assume $\forall \epsilon > 0$, $\phi_i((1 - \epsilon_i) 1_n) > 1 - \epsilon$

Using $\phi(1_n) = 1_n$, you get $\dfrac{\phi_i(1_n) - \phi_i((1-\epsilon) 1_n)}{\epsilon} < 1$

Since $\dfrac{\phi_i(1_n) - \phi_i((1-\epsilon) 1_n)}{\epsilon} \to \sum_j \dfrac{\partial \phi_i}{\partial x_j}(1_n)$ as $\epsilon \to 0$, you have $\sum_j \dfrac{\partial \phi_i}{\partial x_j}(1_n) \leq 1$, which is a contradiction.

So, $\exists \epsilon_i > 0$ such that $\phi_i((1 - \epsilon_i) 1_n) \leq 1 - \epsilon_i$

Take $\epsilon = \min_i \epsilon_i$, then for any $i$, $\phi_i((1 - \epsilon) 1_n \leq 1 - \epsilon$

Since $\phi$ is non-decreasing, then $\phi$ maps $[0, 1-\epsilon]^n$ to $[0, 1-\epsilon]^n$ (which does not contain $1_n$).

As $\phi$ is continuous, you get existence of a fixed point in $[0, 1-\epsilon]^n$ (convex compact) by the Brouwer fixed-point theorem.


The $\dfrac{1}{n}$ in the second hypothesis is surprising for me, so if someone finds a mistake, please tell me.

  • $\begingroup$ Thanks for answering! It seems like using Brouwer is a good idea, I hadn't thought of that. When you take $\epsilon = \min_i \epsilon_i$, how does this imply that $\phi_i((1-\epsilon)1_n) \leq 1-\epsilon$? $\endgroup$ Oct 31 '20 at 9:11
  • $\begingroup$ Well yes, in fact with more hypotheses (convexity of each $\phi_i$ and $\sum_j \dfrac{\partial \phi_i}{\partial x_j} > 1+\delta$, with $\delta > 0$ uniform in $i$), using a first order Taylor expansion with Lagrange remainder and the definition of derivative, it seems that you can prove that $\phi_i((1-\epsilon)1_n) \leq 1- \epsilon$. $\endgroup$
    – FredV
    Oct 31 '20 at 10:15

Inspired by the previous answer, here's what I managed to come up with myself:

Let's assume, on top of what is stated in the question, that $\mathcal{J}_\phi(1_n)$ is a positive matrix and that the supposed eigenvalue $\lambda>1$ is also the dominant one. Then, we know that the corresponding eigenvector $v_\lambda$ has positive components, by the Perron-Frobenius theorem.

Since $\phi$ is smooth, we have that $$\phi(1_n+tv) = \phi(1_n)+\mathcal{J}_\phi(1_n)(tv)+o(t^2) = 1_n +\lambda t v + o(t^2).$$

Since $\lambda >1$ and $v$ is positive, we know that $\lambda tv < t v$ for $t < 0$. For $t$ negative but sufficiently close to zero, we hence obtain $$\phi(1_n + tv)<1_n+tv. $$ Hence, by monotonicity of $\phi$, the box with opposite vertices $0_n$ and $1_n + tv$ gets mapped into itself. Then we can apply the Brouwer fixed point theorem to obtain what we wanted.

Remark: I really only need that $v$ is nonnegative (and nonzero), so assuming non-negative $\mathcal{J}_\phi(1_n)$ and using the analog of the Perron-Frobenius theorem for non-negative matrices also works.

I also think that convexity of the components of $\phi$ can guarantee uniqueness of the fixed point.

  • $\begingroup$ If anyone could comment or answer on the uniqueness of the fixed point, that would also be great! $\endgroup$ Oct 31 '20 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.