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Let $f: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $g:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous and compact valued.

Consider the function $F: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n \times \mathbb{R}^n$ defined as $$ F(x,y) := \left[ \begin{matrix} f(x,y) \\ f(x,y) + g(y) + y \end{matrix} \right]. $$

Find conditions on $g$ such that $F$ has a fixed point.

Comments. I tried assuming that the function $y \mapsto g(y) + y$ has a fixed point $\bar{y}$, i.e., $\bar{y} = g( \bar y) + \bar y \Leftrightarrow g(\bar y) = 0$. But it does not seem to work.

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  • $\begingroup$ what does compact valued mean? $\endgroup$ – user66081 Feb 5 '15 at 15:32
  • $\begingroup$ It means that the range of $f$ is a compact set. $\endgroup$ – user693 Feb 5 '15 at 16:57
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At a fixed point of $F$ we must have $$\begin{bmatrix}x\\y\end{bmatrix}=F(x,y)=\begin{bmatrix}f(x,y)\\f(x,y)+g(y)+y\end{bmatrix}$$

i.e. $f(x,y)=x$ and $f(x,y)+g(y)+y=y$.

Therefore the system $$\begin{cases}f(x,y)&=x\\g(y)&=-x\end{cases}$$ must have solution.

In other words $f(-g(y),y)=-g(y)$ must have a solution. Conversely, if $f(-g(y),y)=-g(y)$ has a solution then $\begin{bmatrix}-g(y)\\y\end{bmatrix}$ is afixed point of $F$.

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  • $\begingroup$ In the second statement, do you mean "if $f(-g(y),y)=-g(y)$" has solution? $\endgroup$ – user693 Feb 5 '15 at 16:47
  • $\begingroup$ @Adam Yes, the same equation. $\endgroup$ – Pp.. Feb 5 '15 at 16:51
  • $\begingroup$ Thanks. So your statement is: $F$ has a fixed point if and only if $y \mapsto f( -g(y), y ) + g(y)$ has a zero. $\endgroup$ – user693 Feb 5 '15 at 16:54
  • $\begingroup$ @Adam Yes, you can write it that way. $\endgroup$ – Pp.. Feb 5 '15 at 16:55
  • $\begingroup$ A follow up problem on the existence of such zero is in [math.stackexchange.com/questions/1142019/existence-of-a-zero]. Thanks! $\endgroup$ – user693 Feb 10 '15 at 12:58
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The condition $F(x, y) = (x, y)$ implies $f(x, y) = x$, so $x = -g(y)$, so $f(-g(y), y) = -g(y)$.

Suppose $g$ is invertible with a continuous inverse, such that the range of $-f$ is contained in the range of $g$.

Consider $\psi : y \mapsto g^{-1}(-f(-g(y), y))$. This is a well defined, continuous function.

I speculate that one can to argue that $\psi(y) = y$ has a solution using the intermediate value theorem and boundedness of $f$. Then take $x := -g(y)$.

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  • $\begingroup$ I guess $\psi(y) = y$ has solution because $\psi$ is continuous and compact valued, therefore it has a fixed point. $\endgroup$ – user693 Feb 5 '15 at 16:46
  • $\begingroup$ @Adam: well is it?, I don't know $\endgroup$ – user66081 Feb 5 '15 at 17:13

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