Exercise 8.29

Question

Exercise 29: 
    Prove that every continuous mapping $f$ of $\bar D$ into $\bar D$ has a fixed point in $\bar D$.

analambanomenos · Accepted Answer

\def\ints{\mathbb{Z}}

Following the hint, assume $f(z)
e z$ for all $z\in\bar D$. For $z\in\bar D$ let $g(z)\in T$ be the point which lies on the ray starting at $f(z)$ (but not including $f(z)$) and passes through $z$.
    Note that if $z\in T$, then $g(z)=z$. The points on that ray are the points $$f(z)+r\big(z-f(z)\big),\quad r\in(0,\infty),$$ so the $r$ defining $g(z)$ satisfies 
    \begin{align*}
        1 &= \big|f(z)+r\big(z-f(z)\big)\big|^2 \
        1 &= \Big(f(z)+r\big(z-f(z)\big)\Big)\Big(\overline{f(z)}+r\big(\overline{z}-\overline{f(z)}\big)\Big) \
        0 &= \big|z-f(z)|^2r^2+2\Re\big(f(z)\overline{z}-|f(z)|^2\big)r+\big(|f(z)|^2-1\big)
    \end{align*}
    This is a quadratic equation in $r$ with coefficients which are continuous functions of $z$ and $f(z)$. From the geometry of the ray, for all $z\in\bar{D}$ there will always be only one positive
    solution $r(z)$, given by the familar quadratic formula, and so $r(z)$ and $g(z)$ are continuous functions of $z$. Then $g(z)$, a continuous function which maps $\bar{D}$ into $T$, satisfies the
    conditions of Exercise 28, so there must be a $z\in T$ such that $g(z)=-z$, contradicting $g(z)=z$ for all $z\in T$.

Amit Awasthi · Answer

\begin{proof}
      Assume, to get a contradiction, that $f(z) 
e z$ for every $z \in \overline{D}$. Now associate to each $z \in \overline{D}$ the point $g(z) \in T$ which lies on the ray that starts at $f(z)$ and passes through $z$. Then, $g$ is continuous because
      \begin{equation*}
        g(z) = z - s(z) \{ f(z) - z \}
      \end{equation*}
      where $s(z)$ is the nonnegative root of $-1 + |z|^2 + 2t(z(f(z)-z)) + |f(z) - z|^2t^2$, which corresponds to the value of $t$ at which the ray starting from $z$ intersects $T$. $g$ is continuous because because $f$ is, and therefore also $s(z)$, since its denominator $2|f(z) - z|^2$ is never 0. By construction, $g(z) = z$ for all $z \in T$. Then, by Exercise ef{8.28} there exists a $z$ such that $g(z) = -z$; however, this contradicts that $g(z) = z$ if $z \in T$. Thus, $f(z) = z$ for at least 1 value of $z \in \overline{D}$.
    \end{proof}

Exercise 8.29

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Exercise 8.29

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