Bending of $F$ into $G$ (section 5.2)

Let $C_s$ be the distance to the origin of the second most distant stable ss of $F$ (it is possible for more than 1 such ss to be situated at distance $C_s$ from the origin), let $C$ be a real number strictly superior to $C_s$, and $\epsilon > 0$. We will add a "pull-back" term to $F$ at points whose norm is greater than $C$. Since the function $x\mapsto <x,F(x)>$ is continuous, there is a constant $M \in \Re_{+}^{*}$ such that:

$$\forall x \in D \mathrm{st} \parallel x \parallel \le \sqrt{n} (C+\epsilon), <x,F(x) > < M$$

In the following, we will use a smooth function $\Xi$ such that:

• $\forall x \in \Re \mathrm{st} x \le C, \Xi (x)=0$
• $\forall x \in \Re \mathrm{st} x \ge C+\epsilon, \Xi (x)=1$
• $\forall x \in \Re \Xi '(x) \ge 0$

We will consider the new vector field $G$ defined by

$$\forall x\in D, \forall i \in I, G_i(x)=F_i(x) - \alpha \Xi (x_i) x_i,$$

where $\alpha \in \Re_{+}^{*}$ will be defined later on. Function $\Xi$ is defined in such a way that $G$ shares at least two stable ss with $F$. Because of our assumption about expression leakage, we have:

$$\forall i \in I, \forall x \in D \mathrm{st} x_i=0, F_i(x)>0$$

This is also verified by $G$:

$$\forall i \in I, \forall x \in D \mathrm{st} x_i=0, G_i(x)>0$$

We have:

$$\forall x \in D, <x,G(x)> = <x,F(x)> - \alpha \sum_{i\in I} \Xi(x_i)x_{i}^{2}$$

Let us study the behaviour of the "pull-back" term we added to $F$:

$$\sum_{i \in I}x_{i}^{2}= \sum_{i \in J_{C+\epsilon}(x)}x_{i}^{2}+\sum_{i \in I \backslash J_{C+\epsilon}(x)}x_{i}^{2},$$

where $J_{c}(x)=\{ i \in I \mathrm{st} x_i < c \}$. It is obvious that:

$$\sum_{i \in J_{C+\epsilon}(x)}x_{i}^{2}< n(C+\epsilon)^2,$$

and by definition of $\Xi$ we have:

$$\forall x \in D, \sum_{i \in I}\Xi(x_i)x_{i}^{2} \ge \sum_{i \in I \backslash J_{C+\epsilon}(x)}x_{i}^{2}$$

We thus conclude that

$$\forall t \mathrm{st} t> \sqrt{n}(C+\epsilon), \mathrm{Min}_{\parallel x \parallel =t}\sum_{i \in I} \Xi(x_i)x_{i}^{2}>0$$

Therefore, we can choose $\alpha \in \Re_{+}^{*}$ such that:

$$\forall x \in D \mathrm{st} \parallel x \parallel = \sqrt{n}(C+\epsilon), \alpha \sum_{i \in I} \Xi(x_i)x_{i}^{2}> M$$

Finally, we derive:

$$\forall x \in D \mathrm{st} \parallel x \parallel = \sqrt{n}(C+\epsilon), <x,G(x)> < 0$$

$G$ is thus inward-pointing on the sphere of centre 0 and radius $\sqrt{n}(C+\epsilon)$ intersected with $D$. $G$ has at least two stable ss inside this portion of sphere, because $G$ coincides with $F$ within the sphere of centre 0 and radius $C$ intersected with $D$. The area of space which we will consider is the intersection $\Delta$ of this portion of sphere with the cone $\{ x\in D \mathrm{st} \forall i \in I, x_i>0\}$. Because of the hypotheses we made, $G$ is inward-pointing on $\Delta$; the problem with $\Delta$ is that it is not smooth, but it is possible to "round the angles" into a surface $S$ close enough to $\Delta$ for $G$ to be inward-pointing on $S$ (the demonstration will be detailed elsewhere). Finally we can apply corollary 2 to surface $S$: $G$ has a positive circuit at some point of $S$; since we added no positive circuits (the terms we added yield supplementary negative diagonal terms in the Jacobian matrix), $F$ has the same positive circuit. By letting $\epsilon$ be arbitrarily small, and $C$ arbitrarily close to $C_s$, we see that this positive circuit is at most at distance $\sqrt{n}C_s$ from the origin.