Fixed-point theorem for a component-wise convex function.

Question

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.

Answer 1

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 $\phi_i((1 - \epsilon_i) 1_n) \leq 1 - \epsilon_i$.

Proof:

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.

Remark:

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

Answer 2

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.