Let $f : [a,b] \to\mathbb R$ and for every $x \in [a,b]$ there exists $y \in [a,b]$ such that $|f(y)| < \frac12|f(x)|$.
Find $\inf\{|f (x)| : x \in [a, b]\}$, and show that $f$ is not continuous on $[a, b]$.
I am fairly new to this type of calculus so I am not being able to start an approach regarding this, I think infimum will be a function out of the range of $f$ as according to the definition there will always be another term less than half of it.
Part One.
Let $I=\{|f(x)|: x\in [a,b]\}.$ Let $D=\inf I.$
Then $D=0.$ Proof by contradiction: Suppose $D>0.$ By def'n of $D$ we have $$(i).\quad I\subset [D,\infty).$$ Now if $E>D$ then $$(ii).\quad I\cap [D,E]\ne \emptyset.$$ Otherwise, by $(i)$ we would have $I\subset [E,\infty),$ implying $ D=\inf I\ge \inf \,E>D.$
So in $(ii)$ take $E=3D/2.$ Then there exists $x\in [a,b]$ with $|f(x)|\in [D,E]=[D,3D/2].$ But then there exists $y\in [a,b]$ with $|f(y)|<|f(x)|/2\le 3D/4<D,$ contrary to $(i).$ This is the desired contradiction that follows from assuming $D>0$.
Part two.
$0=D=\inf I\not \in I.$ Otherwise for some $x\in [a,b]$ we would have $|f(x)|=0$ but then there would exist $y\in [a,b]$ with $|f(y)|<|f(x)|/2=0.$