1. ## 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 $\displaystyle f:[a,b]\rightarrow R$ be continuous. If there is a $\displaystyle c<1$, such that for any$\displaystyle x\in [a,b], \exists y \in [a,b], |f(y)|\leq c|f(x)|$, then prove that there is a $\displaystyle r\in [a,b]$ such that $\displaystyle f(r)=0$. [Hint: Consider g(x)=|f(x)|]

2. Suppose the statement is false; no such $\displaystyle r$ exists.
Then the function $\displaystyle g$ being continuous has absolute minimum of $\displaystyle [a,b].$
Call it $\displaystyle g(d),~d\in [a,b]$. Now it should be clear that $\displaystyle g(d)>0$.
Could we have $\displaystyle y\in [a,b]$ such that $\displaystyle g(y)\le c\cdot g(d)~?$.

3. Well, if we know that $\displaystyle g(d)$ is the absolute minimum of $\displaystyle [a,b]$ and $\displaystyle y\in [a,b]$. Then, any element $\displaystyle <g(d)$will not be $\displaystyle \in [a,b]$. Then, since $\displaystyle c<1, c*g(d)<g(d)$. Hence, $\displaystyle c*g(d)$ cannot be contained in $\displaystyle [a,b]$. Thus, there is no $\displaystyle y\in [a,b]$ such that $\displaystyle g(y)\leq c*g(d)$????

4. Correct. So what does that prove?

5. Well, since we showed that when $\displaystyle g(d)>0$, no $\displaystyle y\in [a,b]$ exists such that $\displaystyle g(y)<c*g(d)$ then, $\displaystyle |f(y)|\leq c|f(x)|$ can only be true when $\displaystyle |f(x)|<0$ or $\displaystyle |f(x)|=0$ for some $\displaystyle x\in [a,b]$. However,since we have $\displaystyle |f(x)|$ must be greater then zero, we know that $\displaystyle |f(x)|$ must equal zero for some $\displaystyle x\in [a,b]$. We can call this element $\displaystyle r$. Hence, there is an $\displaystyle r\in [a,b]$ such that $\displaystyle f(r)=0$???

6. Good. The contradiction means that the assumption is false.
So r must exist.