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 on $\mathbb{R}.$ Let $a$ be a positive constant.

Question: 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 $y=\frac{1}{2}.$ I use desmos to check the two sets and it seems that they coincide.

I tried to prove the equality. Let $x$ be a real number such that $$|1-f(x)|\leq 1-a.$$ By reverse triangle inequality, $$1-|f(x)|\leq |1-f(x)|\leq 1-a.$$ So we have $$|f(x)|\ge a.$$ However, if $x$ is such that $|f(x)|\geq a,$ then I am not able to obtain $$|1-f(x)|\leq 1-a.$$ It seems that I need $$|1-f(x)|\leq 1-|f(x)|$$ but the inequality above may not be true.

Answer 1

Let's look at a simple case:

$f(x) = x$. And let's take $a = 0$.

Then you're asking if the set $$ \{ x \in R : |1-x| \le 1\} $$ is the same as the set $$ \{x : |x| \ge 0\} $$

Well...$x = 3$ is in the second set, but not the first, so the conjecture is false.