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- Apr 13, 2013

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I am given the following exercise:

$f:[a,b] \to \mathbb{R}$ continuous and $\forall x$ there is a $y$ such that $|f(y)| < \frac{|f(x)|}{2}$ .Show that there is a $z$ such that $f(z)=0$.

That's what I have tried:

Suppose that there is not a $z$ such that $f(z)=0$.Then $f(x)>0 , \forall x$ or $f(x)<0 , \forall x$.

If $f(x)>0 , \forall x$ then from the relation $|f(y)| <\frac{|f(x)|}{2}$ for $x=y$ we find $\frac{-f(x)}{2}>0$,that can't be true,as we have supposed that $f$ is positive $\forall$ x.

Now,suppose that $f(x)<0 , \forall x$,from the relation $|f(y)| < \frac{|f(x)|}{2}$ for $x=y$ we find $|f(x)|<0$,that also can't be true.

So,we conclude that it can't be true that $f$ doesn't change sign,so there has to be a $z$ such that $f(z)=0$.

Could you tell me if it is right?