Inequality for supremums of functions

Question

We have the reverse triangle inequality:

$|x-y| \geq ||x| - |y||$

Let $f, g \in \mathbb{R}^I$ for some interval $I$. Fix $x \in I$.

Do we have that sup$\{f(x)-g(x)\} \geq$ sup$\{f(x)\}$ - sup$\{g(x)\}$ ?

P.S. I'm asking this because I believe I need that to proof that $(X,$$d_{\infty,\infty})$ is a metric space.

Answer 1

Yes. For every $x\in I$, we have $$ f(x) = \big(f(x)-g(x)\big) + g(x) \leq \sup (f-g)(I)+\sup g(I).$$ Taking the supremum over $x\in I$ thus gives $$ \sup f(I) \leq \sup (f-g)(I)+\sup g(I).$$ Now re-arrange to get $$\sup (f-g)(I) \geq \sup f(I) - \sup g(I),$$ as requested.

Answer 2

We have $$ \sup_{\Omega} f(x) - \sup_{\Omega} g(x) \le \sup\limits_{\Omega}(f(x)-g(x)) \le \sup_{\Omega} f(x) - \inf_{\Omega} g(x). $$ This assumes that the domain of both functions is $\Omega$.