Problem on Continuity Let $I = [a,b]$,  $f: I → R$ be continuous function on $I$ such  that for each $x$ in $I$ there exists y in $I$ such that, 
$|f(y)| ≤ (1/2)|f(x)|$
Then prove that there exists $c$ in $I$ such that $f(c) = 0$
How to start up? 
I know Intermediate value theorem but how can I use it? 
 A: As $I$ is compact, $f$ attains its minimum and maximum there (extreme value theorem). Therefore there exists $c \in I$ be such that $\lvert f(c) \rvert \leq \lvert f(x) \rvert$ for all $x \in I$. By hypothesis, there exists $y \in I$ such that $\lvert f(y) \rvert \leq \lvert f(c) \lvert/2 $. As we then have $$ 2\lvert f(y) \rvert \leq \lvert f(c) \lvert \leq \lvert f(y) \rvert,$$
the only possibility is $\lvert f(y) \rvert = 0$, i.e. $f(y) = 0$.
A: If you proved it using the intermediate value theorem you would need a positive and negative value. This is not at all required with your requisites. A way to prove it is shown below.
First off all, note that $f\left(\left[a, b\right]\right)$ is an interval because of continuity. If you create a sequence that converges in this set, its limit will also be a part of this set.
You can create a sequence that converges to zero by picking $x_1$ at random and then letting every subsequent $x$ be such that 
$$|f\left(x_{n+1}\right)|\le \left(\frac 12\right)|f\left(x_{n}\right)|\,.$$ 
This converges to zero, so $0 \in f\left(\left[a, b\right]\right)$.
