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let $K$ be a closed convex subset of a normed space $V$. For any $f: K \to K$ define the fixed-point set of $f$ as follows: $fix(f)=\{x$ belongs to $K$ $|f(x)=x \}$. I have to show that a nonempty subset of $K$ is a fixed point set of some continuous function $f:K \to K$ iff it is closed subset of $K$?

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I'm thinking of the point, if F is a closed subset of K then u(x)= inf norm x-y with y belongs to K.I want to show this to be continuous function from K to [0, infinity) which is zero only on F. But i'm stuck with this thing, could anyone of you help me out? –  Athar Raheel Ahmad Oct 24 '12 at 20:47

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First of all, the fact that the fixed point set is always closed follows directly from continuity. For the other direction your idea in the comment is a good start. Given a closed non-empty set $F\subseteq K$, fix some point $x_0 \in F$, and let $u(x)=\inf_{y\in F} \|x-y\|$ be the distance to $F$. Since $F$ is closed, $u(x)=0$ iff $x\in F$. Now define the map $f(x) = e^{-u(x)}x + (1-e^{-u(x)})x_0$. Since $K$ is convex, and $f(x)$ is a convex combination of $x$ and $x_0$, we get $f(K) \subseteq K$. For any $x\in F$ we have $e^{-u(x)}=1$, so $f(x)=x$. For any $x\in K \setminus F$ we have $f(x)-x = (1-e^{-u(x)}) (x-x_0) \ne 0$ since $x\ne x_0$ and $e^{-u(x)} \ne 1$. Together this shows that the fixed point set of $f$ in $K$ is exactly $F$.

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Thanks Lukas for your help,I hope now I can figure out the problem. –  Athar Raheel Ahmad Oct 25 '12 at 14:55

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