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.
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