Last problem of MATH202 tutorial note #14

Problem. Consider a continuous function $f$ on $[a,b]$, suppose for any $a\leq x\leq b$, there is $a\leq y\leq b$, such that $|f(y)|\leq \frac{1}{2}|f(x)|$. Prove that there is a $c\in[a,b]$ such that $f(c)=0$.

My way.
It is often easy to prove the converse to be impossible for this kind of problems. We on the contrary suppose that $f(x)\neq 0,\forall x\in[a,b]$, then either $f(x)>0$ or $f(x)<0$ for all $x\in[a,b]$.

We assume that $f(x)>0$, due to continuity there exists $x_0$ such that
$(*):\qquad f(x)\ge f(x_0)>0, \forall x\in[a,b].$

However, we have the following observation.
$\exists y_1\in [a,b], f(y_1)\leq \frac{1}{2}f(y_0)$;
$\exists y_2\in [a,b], f(y_2)\leq \frac{1}{2}f(y_1)$;
$\qquad \vdots$
$\exists y_n\in [a,b], f(y_n)\leq \frac{1}{2}f(y_{n-1})$.

thus $\displaystyle f(y_n)\leq \frac{1}{2}f(y_{n-1})\leq \left(\frac{1}{2}\right)^2f(y_{n-2})\leq \cdots \leq \left(\frac{1}{2}\right)^nf(y_0)$. Since right hand side tends to 0 as $n\to\infty$, there exists an $N$ such that $n>N\implies f(y_n)<\epsilon < f(x_0)$, a contradiction with (*). The case that $f(x)<0$ is essentially the same, hence there exists $c\in[a,b]$ such that $f(c)=0$.

Alternatively, if we know that any bounded sequence have a convergent subsequence, then this problem can be solved in a much neater way. As above we show that $$|f(y_n)|\leq \frac{1}{2}|f(y_{n-1})|\leq \left(\frac{1}{2}\right)^2|f(y_{n-2})|\leq \cdots \leq \left(\frac{1}{2}\right)^n|f(y_0)|,\forall n \in \mathbb{N}\cup\{0\},$$ then since there exists a convergent subsequence $\{y_{n_k}\}$ with $\lim_{k\to\infty}y_{n_k}=c\in[a,b]$, thus taking limit on the proved inequality, we get $$0\leq \lim_{k\to\infty}|f(y_{n_k})|=\left|f\left(\lim_{k\to\infty}y_{n_k}\right)\right| =|f(c)|\leq\lim_{k\to\infty}\left(\frac{1}{2}\right)^{n_k}|f(y_0)| = 0,$$ here we used the fact that a composite of continuous function is still continuous, we are done.