Thread: Proof Using Continuity

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 $f:[a,b]\rightarrow R$ 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)|]

2. 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 $g(y)\le c\cdot g(d)~?$.

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

4. Correct. So what does that prove?

5. Well, since we showed that when $g(d)>0$, no $y\in [a,b]$ exists such that $g(y) 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 $f(r)=0$???

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