let $f:[a,b]\to \mathbb{R}$ be a continuous function which has the property let $f:[a,b]\to \mathbb{R}$ be a continuous function which has the property that for each $x\in[a,b]$ there is $y\in[a,b]$ such that $|f(y)|\le\frac{1}{2}|f(x)$
(a) prove that there is sequence $\{x_n\} \subset[a,b]$ such that $f(x_n) \to 0$
(b) prove that there is point $c\in [a,b]$ such that $f(c)=0$
i have no idea to how to prove this can any help
 A: Hint: Let $x_0$ be in $[a,b]$, suppose defined $x_n$, consider $x_{n+1}$ such that $|f(x_{n+1})|\leq {{|f(x_n)|}\over 2}$, show recursively that $|f(x_{n})|\leq {{|f(x_0)|}\over 2^n}$.
Since $[a,b]$ is compact, you can extract a subsequence $x_{n_k}$ of $x_n$ which converges towards $c$, show that $f(c)=0$.
A: You can approach the problem like this: fix $x_0 \in [a,b]$. By hypothesis you can find $x_1 \in [a,b]$ such that $|f(x_1)| \leq \frac{1}{2}|f(x_0)|$. You can go on and construct $(x_n)_{n \in \mathbb{N}}$ such that
$$|f(x_n)| \leq \frac{|f(x_{n-1})|}{2} \leq \dots \frac{|f(x_0)|}{2^n}$$
and then $0 \leq \lim_n |f(x_n)| \leq \lim_n \frac{|f(x_0)|}{2^n} = 0$. To prove the existence of $c$ observe that $[a,b]$ is compact, so for any sequence you can get a convergent subsequence. So from $(x_n)_n$ you can get a subsequence $(x_{n_k})_k$ such that $x_{n_k} \rightarrow x \in [a,b].$ $x$ plays the role of your $c$. By continuity of $f$:
$$0 = \lim_n f(x_n) = \lim_k f(x_{n_k}) = f(\lim_k x_{n_k}) = f(x).$$
