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

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

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.

• 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 $\phi_i((1-\epsilon)1_n) \leq 1-\epsilon$? Oct 31 '20 at 9:11
• 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 $\phi_i((1-\epsilon)1_n) \leq 1- \epsilon$. 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.

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