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Thread: Check 1 more Real

  1. #1
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    Check 1 more Real

    Suppose we have a continuous function f that is on the closed interval [0,1] and has a range that's contained in [0,1] as well. Prove f has to have a fixed point (ie: show $\displaystyle f(x) = x$ for at least 1 value of $\displaystyle x \in [0,1]$.

    SOLUTION:

    We are done if $\displaystyle f(0) = 0$ or $\displaystyle f(1) = 1$

    What about if $\displaystyle f(0) > 0$ and $\displaystyle f(1) < 1$.

    Let $\displaystyle g(x) = x - f(x)$, then we know that $\displaystyle g$ is continuous and further $\displaystyle g(0) < 0, g(1) > 0$. By the Intermediate Value Theorem (IVP), there exists a $\displaystyle c \in (0,1)$ with $\displaystyle g(c) = 0$. That is, $\displaystyle f(c) = c$. And hence the proof is complete.
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  2. #2
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    Yes, that proof works nicely.
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