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Consider a nonlinear continuous function $f$ that maps, the 4-simplex \begin{equation} S=\left\lbrace (y_{1},y_{2},y_{3},y_{4})\in \mathbb{R}^{4}\vert\quad y_{i}\geq 0,\quad y_{1}+y_{2}+y_{3}+y_{4}=1\right\rbrace \end{equation} into itself. The function $f$ has components $f_{1}, f_{2}, f_{3}, f_{4}$ given next: \begin{eqnarray} f_{1}(y_{0},y_{1},y_{2},y_{3})&=&(1-C_{1})(1-C_{2})y_{0}+p_{1}y_{1}+p_{2}y_{2}+p_{3}y_{3}\\ f_{2}(y_{0},y_{1},y_{2},y_{3})&=&C_{1}(1-C_{2})y_{0}+(1-p_{1})(1-C_{2})y_{1}\\ f_{3}(y_{0},y_{1},y_{2},y_{3})&=&(1-C_{1})C_{2}y_{0}+(1-p_{2})(1-C_{1})y_{2}\\ f_{4}(y_{0},y_{1},y_{2},y_{3})&=&C_{1}C_{2}y_{0}+(1-p_{1})C_{2}y_{1}+(1-p_{2})C_{1}y_{2}+(1-p_{3})y_{3} \end{eqnarray} where $p_{1}, p_{2}, p_{3}$ are constants $p_{i}\in (0,1),\;$ $i=1, 2, 3$ and \begin{eqnarray} C_{1}=1-\exp(-(y_{1}+y_{3})),\\ C_{2}=1-\exp(-(y_{2}+y_{3})). \end{eqnarray} I prove that my function $f$ has a fixed point at $E_{0}=(1,0,0,0)$ and I "believe" and I want to prove that $f$ has only another fixed point in $S$. In order to do that, I want to use the Banach fixed point theorem in $S$. However, the theorem is not true in $S$ since I already have $E_{0}$ as a fixed point and I need another fixed point in the interior of $S$. My idea is to find a subset of $S$ that does not contain $E_{0}$ where I can use Banach's theorem. Nevertheless, for example, the set $S-E_{0}$ is not compact and I cannot apply the theorem. In summary, what I want is to find a compact subset of $S$ that not contains $E_{0}$, in which I can apply Banach's theorem. Thanks in advance.

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    $\begingroup$ You need Brouwer's fixed-point theorem, not Banach. But also, without certain other guarantees (is $f$ even continuous?), we really can't say anything. $\endgroup$ – Michael Lee Mar 2 '18 at 22:57
  • $\begingroup$ Brower's fixed-point theorem tells me that there is at least one fixed point, but I already know that because I have $E_{0}$. $\endgroup$ – Destiny Mar 2 '18 at 23:04
  • $\begingroup$ Warning $(1,0,0,0)\ne 1$ $\endgroup$ – Piquito Mar 2 '18 at 23:07
  • $\begingroup$ Right, so if you want to prove the existence of another, you either need to find one by solving $f(x) = x$ (using Newton's method, perhaps?), or you need to find a convex, compact subregion of $S$ that gets mapped into itself by $f$ (again, assuming that $f$ is continuous, which you haven't explicitly stated). $\endgroup$ – Michael Lee Mar 2 '18 at 23:08
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    $\begingroup$ Right, but $f$ maps the simplex to itself, not the simplex to $\mathbb{R}$. $\endgroup$ – Michael Lee Mar 3 '18 at 1:12
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As you point out, the fixed point theorem ensures the existence of a fixed point. It does not ensure unicity or multiplicity (think of the constant mapping or the identity mapping). You should provide more information about the function $f$ you are applying.

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