Is $\sup_{\{x \in S : |f(x)|\leq |\inf_S f|\}} |f(x)| = |\inf_S f|$?

Question

Let $f: \Bbb R \to \Bbb R$ be a bounded function and let $S \subset \Bbb R$. Suppose that the set $$A=\{x \in S : |f(x)|\leq |\inf_S f|\}$$ is non empty.

Question: Is $\sup_A |f(x)| = |\inf_S f|$?

I was just able to prove that $\sup_A |f(x)|\leq |\inf_S f|$, since $|f(x)|\leq |\inf_S f|$ for all $x \in A$, but I can't get the reverse inequality.

Answer 1

Yes.

If $\inf_S f = \min_S f$ then you also have $\sup_A |f| = \max_A |f| = |\min_S f|$, just by considering a point $x^*\in S$ such that $f(x^*)=\min_S f$.

Otherwise for all $x$ in $S$, you have $f(x) > \inf_S f$. Then for $A$ to be non-empty, you must have $\inf_S f < 0$.

Therefore there exists a negative decreasing sequence $f(x_n)$ converging to $\inf_S f$. It follows that $x_n \in A$ and $|f(x_n)|$ converges to $|\inf_S f|$.

Thus $\sup_A |f| = |\inf_A f|$.

Answer 2

Note that $\inf_S f$ is a real number.

• Case 1: $\inf_S f<0$.

Let $x_n\in S$ be such that $f(x_n)$ decreases to $\inf_S f$. This exists by definition of infimum. As $f(x_n)<0$ eventually, $|f(x_n)|$ increases to $|\inf_S f|$. Thus $x_n\in A$, and $$\sup_A |f|\ge \lim |f(x_n)| = |\inf_S f|.$$

• Case 2: $\inf_S f\ge 0$.

Then $x\in S$ implies $f(x)\ge 0$ so the definition of $A$ simplifies slightly $$ A = \{ x\in S : f(x) \le \inf_S f\}.$$ Now $x\in A$ implies $x\in S$ and $f(x)\le \inf_S f$, therefore $f(x)=\inf_S f$. Thus $$\sup_A |f| = \sup_A f = \inf_S f = \inf_S |f|.$$