Let $f\colon X \rightarrow \mathbb{R}$ be a function such that $\max_{x\in X} f(x) + \min_{x\in X} f(x) = 0$. Does it then follow that $\max_{x\in X} f(x) = \max_{x\in X} |f(x)|$?
I'm quite sure it does but I don't know how to prove it. Here's what I tried: First of all, $\max_{x\in X} f(x)$ and $\min_{x\in X} f(x)$ cannot be both negative, and since $\max_{x\in X} f(x) \geq \min_{x\in X} f(x)$, we have $\max_{x\in X} f(x) \geq 0$.
Case 1: $\max_{x\in X} f(x) = 0$. Then $\min_{x\in X} f(x) = 0$, so $f(x) = 0$ for all $x$ and thus $\max_{x\in X} f(x) = \max_{x\in X} |f(x)|$.
Case 2: $\max_{x\in X} f(x) > 0$. Then $\min_{x\in X} f(x) < 0$, so $-\min_{x\in X} f(x) > 0$ and thus $\max_{x\in X} (-f(x)) > 0$.
How could I proceed? Thanks.
EDIT: Following Michael Burr's advice, I'll try to prove the three steps:
Step 3: $\max_{x\in X} f(x) + \min_{x\in X} f(x) = 0$ implies $\max_{x\in X} f(x) = -\min_{x\in X} f(x)$, and with step 2 we get $\max_{x\in X} f(x) = \max_{x\in X} -f(x)$.
Step 2: Let $x'\in X$ be such that $-f(x') = \max_{x\in X} -f(x)$. Then \begin{align*} & &-f(x') &\geq -f(x) &\quad \forall x\in X \\ &\Rightarrow &f(x') &\leq f(x) &\quad \forall x\in X \\ &\Rightarrow &\min_{x\in X} f(x) &= f(x') \\ &\Rightarrow &-\min_{x\in X} f(x) &= -f(x') = \max_{x\in X} -f(x) \\ \end{align*}
Okay so far? I have no clue how to prove step 1.
Sketch:
Step 1: Show that, in general, $\max\limits_{x\in X}|f(x)|=\max\left\{\max\limits_{x\in X}f(x),\max\limits_{x\in X}-f(x)\right\}$
Step 2: Show that, in general, $\max\limits_{x\in X}-f(x)=-\min\limits_{x\in X}f(x)$.
Step 3: Show that, in the present case, $\max\limits_{x\in X}f(x)=\max\limits_{x\in X}-f(x)$.