Absolute value of supremums vs Supremum of absolute value

Question

This question is related to this. Let $f,g$ be real continuous bounded functions. Does $$\lvert \sup_x f(x)-\sup_x g(x)\rvert\leq \sup_x\lvert f(x)-g(x)\rvert?$$

By visualizing some plots, it makes sense. But how to prove it?

Answer 1

This fact is true. To see this, write $|f(x)|=|f(x)-g(x)+g(x)|\leq |f(x)-g(x)|+|g(x)|$. Taking supremum of both sides, we have $$ \sup_x |f(x)|\leq\sup_x(|f(x)-g(x)|+|g(x)|)\leq \sup_x|f(x)-g(x)|+\sup_x|g(x)| $$ so $$ \sup_x |f(x)|-\sup_x|g(x)|\leq \sup_x|f(x)-g(x)|. $$ Repeat the same argument but interchange $f$ and $g$.

EDIT: It occurred to me that I only showed that it's true when $f=|f|$ and $g=|g|$, so when both functions are nonnegative. I wonder if the claim is true when $f$ and $g$ have mixed signs.