# Three questions about fixed points

Pick out the true statements.

1. Let $f : [0, 2] \to [0, 1]$ be a continuous function. Then, there always exists $x \in [0, 1]$ such that $f(x) = x$.
2. Let $f : [0, 1] \to [0, 1]$ be a continuous function which is continuously differentiable in $]0, 1[$ and such that $|f'(x)| \le \frac12$ for all $x\in ]0, 1[$. Then, there exists a unique $x \in [0, 1]$ such that $f(x) = x$.
3. Let $S = \{p = (x, y) \in R^2 : x^2 + y^2 = 1\}$. Let $f : S \to S$ be a continuous function. Then, there always exists $p\in S$ such that $f(p) = p$.

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There are a lot of fixed point theorems. When you say "the fixed point theorem", what exactly does it say? –  MJD Sep 2 '12 at 2:38
Also, there is now a tutorial about how to write math formulas on this site. –  MJD Sep 2 '12 at 2:45
for 3, consider rotations of the circle. for 2 consider, for instance, a constant function. for 1 you can apply some version of the brouwer fixed-point theorm –  yoyo Sep 2 '12 at 2:56
Homework, anyone? –  user22805 Sep 2 '12 at 4:17
poton, you are a member for 28 days now. If this is enough for zombies to rule London, you should be able to familiarize yourself with the proper way of asking questions in that time. –  Michael Greinecker Sep 2 '12 at 17:36

1. Consider the restriction of $f$ to $[0,1]$. This setup should look really close to a fixed-point theorem on an interval (1-D fixed-point theorem).
2. Brian Scott's solution covers this nicely.
3. Consider a function that rotates $S$ a "little".
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For (2) a constant function has the specified property. –  Brian M. Scott Sep 2 '12 at 7:18
Of course. Thank you for catching that. Your answer covers (2) quite nicely. –  shoda Sep 2 '12 at 15:54

I believe (a) is true and (c) is false. For (a), consider $f(x) - \frac{1}{2}x$ and use intermediate value theorem. For (c), a counterexample is the function $f: S \to S$ such that $f(x, y) = (-x, -y)$. For (b), I would try and make a translated, scaled, etc. cosine function as a counterexample.

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By $]a,b[$ the writer probably means the open interval $(a,b)$. Some people use this awful notation to avoid confusion with the ordered pair $(a,b)$. –  MJD Sep 2 '12 at 2:41
@MJD Why awful? –  Did Sep 2 '12 at 7:42
@did: I don’t know about Mark, but I find it extraordinarily counterintuitive to have the brackets turned the wrong way: even after years of occasional exposure, I stumble over it a little when I encounter it. –  Brian M. Scott Sep 2 '12 at 18:48
@Brian Funny that you mention intuition, the explanation I heard for [ and ] as beginning/end of open/closed intervals is that you automatically know whether the limit point is in or out by placing it in your imagination on the bracket. For example, ]a,b] does not contain a because when you place a dot at mid-height on ], it falls on the left of the vertical bar, that is, outside of the interval. (On the other hand, I learned ( and ) while reading papers in English but never found any intuition in them.) –  Did Sep 2 '12 at 19:12
@did: Oh, I understand the intuition; it just doesn’t work for me. This might simply be because I didn’t encounter the notation until grad school, by which time the more traditional version was already deeply ingrained. –  Brian M. Scott Sep 2 '12 at 19:15

HINT: (b) is true. To prove that there is at least one point $x\in[0,1]$ such that $f(x)=x$, use the intermediate value theorem. To prove that there is at most one, use the mean value theorem. (Other answers have covered (a) and (c).)

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Hint for 1: $f$ is a retraction. Consider the image of the set $[0,1]$ under $f$.

Hint for 2: Always look for the simplest possible counterexample.

Hint for 3: What shape is $S$?

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