Existence of a root for a specific function

Question

Let $f$ be a continuous function on the compact interval $I$. Suppose that for each $x$ $\in I$, there exists a $y$ $\in I$ such that, $$|f(y)| \leq \frac{1}{2}|f(x)|$$

Then show that there exists a point $\eta$ $\in I$ such that $f(\eta) = 0$. I am also given a hint which says- Note that $f^2$ is continous on $I$(why?)

Attempt: I squared both the sides of the given eqn. to get $$f(x)^2\geq4f(y)^2$$ but I don’t see how this tells me that $f^2$ is continous on $I$ let alone how to use this to solve the question. Can someone please provide the solution.

Edit : I see that since $f$ is continous $f^2$ will always be continous, but I still don’t see how that helps mo solve the problem.

Answer 1

Attempt:

$f^2(y) \le (1/4)f^2(x).$

$f$ continuous on compactum $I$ implies $f^2$ continuous on $I$.

$f^2 \ge 0$ attains its minimum on $I$, i e. there is a $x_m$ \in $I$ s.t.

$f^2(y)\ge f^2(x_m)$, $y \in I$ .

By assumption there is a $y \in I$ s.t.

$0\le f^2(y) \le (1/4) f^2(x_m) \le (1/4)f^2(y)$.

$\Rightarrow$ $f^2(y)=0=f^2(x_m)$.

Hence $f(y)=0$, for a $y \in I$.

Answer 2

Let $\eta:I\rightarrow I$ be such that $2^{n}|f(\eta^{(n)}(x))|\leq|f(x)|$, where $\eta^{(n)}=\eta\circ\cdots\circ\eta$ for $n$-times. Fix an $x_{0}\in I$, then the sequence $(\eta^{(n)}(x_{0}))$ is a bounded sequence in $I$, so there exists some subsequence $(n_{k})$ (depends on $x_{0}$, but this doesn't matter) such that $\eta^{(n_{k})}(x_{0})\rightarrow L(x_{0})$. We also have \begin{align*} |f(\eta^{(n_{k})}(x_{0}))|\leq\dfrac{1}{2^{n_{k}}}|f(x_{0})|. \end{align*} Taking $k\rightarrow\infty$, continuity of $f$ gives \begin{align*} |f(L(x_{0}))|\leq 0, \end{align*} so $f(L(x_{0}))=0$.