Use Extreme Value Theorem

  • Thread starter cxc001
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  • #1
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Suppose f:[0,1]->R is continuous, f(0)>0, f(1)=0.
Prove that there is a X0 in (0,1] such that f(Xo)=0 & f(X) >0 for 0<=X<Xo (there is a smallest point in the interval [0,1] which f attains 0)

Since f is continuous, then there exist a sequence Xn converges to X0, and f(Xn) converges to f(Xo).
Since 0<=(Xo-1/n)<Xo
Can I just let Xn=Xo-1/n so that 0<=Xn<Xo
So when Xn->Xo, f(Xn)->f(Xo)

I wasn't convinced enough this is the right approach...
 

Answers and Replies

  • #2
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No, this won't work, because you begin by assuming the existence of the number [tex]x_0[/tex], which existence you are required to prove.

I suggest two other approaches, either of which will work.

1. What does the set [tex]f^{-1}(\{0\})[/tex] look like, topologically?

2. What would happen if the set of points [tex]x[/tex] such that [tex]f(x) = 0[/tex] had no smallest element in [tex][0,1][/tex]?
 

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