Let $f:[a,b]\to \mathbb{R}$ be a function such that for every $x\in[a,b]$ there exists a $y \in [a,b]$ such that $|f(y)|<\frac{1}{2}|f(x)|$. What is the infimum of $|f(x)|$ on $[a,b]$?
My idea:
Suppose $\inf\limits_{x \in [a,b]}|f(x)|$ exists, then there exists $x_0 \in [a,b]$ such that $$|f(x_0)|\le |f(x)|, \hskip{0.2cm} \forall x\in [a,b]$$ Then by hypothesis, there exists a $y \in [a,b]$ such that $$|f(y)|\le \frac{1}{2}|f(x_0)|\le \frac{1}{2}|f(x)|$$
Thus, $|f(y)|\le 2|f(y)|<|f(x_0)|$, a contradiction to the infimum attained.
Are my arguments valid?
Let $\theta$ be the infimum of $|f|$ on the given interval $I$. Obviously $\theta \in [0,|f(a)|].$
Our $\theta$ must be zero, for if we suppose $\theta>0$, there must be $t \in I$ satisfying $|f(t)|<2\theta$.
Hence there exists $s \in I$ such that
$$|f(s)|\,\,<\,\,\frac{1}{2}\,|f(t)|\,\,<\,\,\theta.$$
Thus $\theta$ is not the infimum after all.