If $\|f\|\leq 1,$ then is it true that $\{x\in\mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in\mathbb{R}: |f(x)|\geq a\}?$

Question

Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a continuous function such that $$\|f\|_\infty:=\sup_{x\in\mathbb{R}}|f(x)|\leq 1.$$

Question: Assume that $a>0.$ Is it true that $$\{x\in\mathbb{R}: |1-f(x)|\leq 1-a\} = \{x\in\mathbb{R}: |f(x)|\geq a\}?$$

Observe that $1-f$ is the reflection of $f$ about the line $y=\frac 12.$

My earlier question shows that the answer is negative if we do not assume that $\|f\|_\infty\leq 1.$

How abou we impose the bounded assumption? Is the conjecture true?

Answer 1

Take $$ f(x) = \begin{cases} 1, &\text{if $x \geq 1$}, \\ -1 ,&\text{if $ x \leq -1 $}, \\ x , &\text{otherwise}, \end{cases} $$ and set $a=1$. Then

$$ \{ x: |1-f(x) | \leq 1-a \} = \{ x: |1-f(x) | \leq 0 \} = \{ x: f(x) = 1 \} = [1, \infty), $$ while $$ \{x: |f(x)| \geq a\} = (-\infty, -1] \cup [1,\infty). $$

Answer 2

Assume $|1-f(x)|\le 1-a$ with $a>0$. Then $f(x)>0$ (or the left hand side is $\ge1$); together with $|f(x)|\le 1$, this gives us $|1-f(x)|=1-f(x)$ and so $f(x)\ge a$.

Conversely, assume $f(x)\ge a$. With the additional bounds $1\ge f(x)\ge a>0$, we immediately get $|1-f(x)|=1-f(x)\le 1-a$.

Therfore, for $a>0$ and $\|f\|_\infty\le 1$, $$\{\,x\in\mathbb{R}: |1-f(x)|\leq 1-a\,\} = \{\,x\in\mathbb{R}: f(x)\geq a\,\}$$ Note that in general $$\{\,x\in\mathbb{R}: f(x)\geq a\,\}\ne \{\,x\in\mathbb{R}: |f(x)|\geq a\,\}$$