True or false: $\left|\max\limits_{x\in D}f(x) - \max\limits_{x\in D}g(x)\right| \leq \max\limits_{x\in D}\left|f(x)-g(x)\right|$

Question

Is this true or false?

$$\left|\max\limits_{x\in D}f(x) - \max\limits_{x\in D}g(x)\right| \leq \max\limits_{x\in D}\left|f(x)-g(x)\right|$$

I know it is an easy question but I forget all analysis. Thank you for your help

Answer 1

Wlog. $\max f(x)\ge \max g(x)$. If $f(x_0)=\max f(x)$ then $$\max|f(x)-g(x)|\ge f(x_0)-g(x_0)= \max f(x)-g(x_0)\ge \max f(x)-\max g(x)$$

Answer 2

Yes, it is true, but I rather prefer to replace $\max$ with $\sup$.

Since $\sup\limits_{t\in D}g(t)\geq g(x)$ for all $x\in D$, it follows that $$f(x)-\sup\limits_{t\in D}g(t)\leq f(x)-g(x)\leq |f(x)-g(x)|$$ which implies $$\sup\limits_{x\in D}f(x)-\sup\limits_{t\in D}g(t)\leq \sup\limits_{x\in D}|f(x)-g(x)|.$$ Similarly, by symmetry, $$\sup\limits_{x\in D}g(x)-\sup\limits_{t\in D}f(t)\leq \sup\limits_{x\in D}|f(x)-g(x)|.$$ Hence $$\left|\sup\limits_{x\in D}f(x)-\sup\limits_{x\in D}g(x)\right|\leq \sup\limits_{x\in D}|f(x)-g(x)|.$$