Applications of intermediate value theorem Let $f: [a, b] \rightarrow \mbox{R}$ be a continuous function such that for each $x \in [a, b]$ there exists $y \in [a, b]$ such that $|f(y)| \leq \frac{1}{2} |f(x)|.$ Prove that there exists $c \in [a, b]$ such that $f(c) = 0.$
I am trying to use intermediate value theorem for continuous function. Here $|f(x)|$ is maximum with respect to $|f(y)|.$ Hence $$|f(y)| \leq 1/2(|f(x)|+ |f(y)|) \leq |f(x)|.$$ Now I can say $f(c) = 1/2(|f(x)|+ |f(y)|$ for some $c \in (a, b).$ But how to show $f(c) = 0?$
 A: Let $x_0 \in [a,b]$. Then there is  $x_1 \in [a,b]$ such that $|f(x_1)| \leq \frac{1}{2} |f(x_0)|.$
Now we get inductively a sequence $(x_n)$ in $[a,b]$ with
(*) $|f(x_n)| \leq \frac{1}{2^n} |f(x_0)|.$
$(x_n)$  contains a convergent subsequence  $(x_{n_k})$ with limit $c \in [a,b]$.
$f$ is continuous, thus  $f(x_{n_k}) \to f(c)$ .
From (*) we get: $f(x_{n_k}) \to 0$.
Consequence: $f(c)=0$
A: There is $c \in [a,b]$ such that $|f(c)|= \min\{|f(x)|: x \in [ab]\}$
We get an $ y \in [a,b]$ with $|f(y)| \leq \frac{1}{2} |f(c)|$.
Therfore: $|f(c)| \leq \frac{1}{2} |f(c)|$. This gives $f(c)=0$.
A: First, suppose the range of $f$ was all positive. Then, using the extreme value theorem, let $f(x_0) = m$ be the minimum achieved by $f$. Then there exists $y\in [a,b]$ such that $\vert f(y) \vert \leq \frac{1}{2}m < m$. But then $f(y) < m$, a contradiction. A similar issue arises if the range of $f$ is all negative. Thus, either $f$ has a zero (in which case, we're done), or else there exists $x,y\in [a,b]$ such that $f(x) < 0 < f(y)$. In the second case, the intermediate value theorem then guarantees the existence of some $c$ between $x$ and $y$ so that $f(c)=0$.
A: For any $x_0\in[a,b]$ there could be constructed sequence $(x_k)_{k=0}^{\infty}$ such that $f(x_{k+1})=1/2*f(x_k)$. Then $f(x_{k+1})\rightarrow0$, but $[a,b]$ is compact, $f$ is continuous, so $f([a,b])$ is compact too. By definition of compactness $0\in f([a,b])$.
