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Math Help - Proof Using Continuity

  1. #1
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    Proof Using Continuity

    This is a problem that my professor has in his notes. I could really use some guidance. I don't understand how to prove problems such as these.

    Let f:[a,b]\rightarrow R be continuous. If there is a c<1, such that for any  x\in [a,b], \exists y \in [a,b], |f(y)|\leq c|f(x)|, then prove that there is a r\in [a,b] such that f(r)=0. [Hint: Consider g(x)=|f(x)|]
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  2. #2
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    Suppose the statement is false; no such r exists.
    Then the function g being continuous has absolute minimum of [a,b].
    Call it g(d),~d\in [a,b]. Now it should be clear that g(d)>0.
    Could we have y\in [a,b] such that g(y)\le c\cdot g(d)~?.
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    Well, if we know that g(d) is the absolute minimum of [a,b] and y\in [a,b]. Then, any element <g(d)will not be \in [a,b]. Then, since c<1, c*g(d)<g(d). Hence, c*g(d) cannot be contained in [a,b]. Thus, there is no y\in [a,b] such that g(y)\leq c*g(d)????
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  4. #4
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    Correct. So what does that prove?
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  5. #5
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    Well, since we showed that when g(d)>0, no y\in [a,b] exists such that g(y)<c*g(d) then, |f(y)|\leq c|f(x)| can only be true when |f(x)|<0 or |f(x)|=0 for some x\in [a,b]. However,since we have |f(x)| must be greater then zero, we know that |f(x)| must equal zero for some x\in [a,b]. We can call this element r. Hence, there is an r\in [a,b] such that f(r)=0???
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  6. #6
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    Good. The contradiction means that the assumption is false.
    So r must exist.
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