As we know that: If $f$ and $g$ are bounded functions on some arbitrary set, then $ |\sup f - \sup g | \leq \sup |f-g| $.

Indeed, $\sup f = \sup (f -g + g) \leq \sup (f-g) + \sup g \leq \sup |f-g| + \sup g$. Then, it implies $\sup f - \sup g \leq \sup |f-g|$. By changing the roles of $f$ with $g$, we finally obtain $ |\sup f - \sup g | \leq \sup |f-g| $.

However, I am wondering that if it is even possible to extend $f$ and $g$ be any unbounded functions and above property still holds?

Thanks so much in advance!

  • $\begingroup$ If neither $f,g$ is bounded above, then $\sup f - \sup g = \infty - \infty$, which doesn't make much sense. So maybe only look at cases when at least one of $f,g$ is bounded above. Also, what are your definitions for arithmetic in $\overline{\mathbb{R}} = [-\infty, + \infty]$? Most authors define $a \pm \infty = \pm \infty + a = \pm \infty$ if $a$ is a real number, as well as $\pm \infty \pm \infty = \pm \infty$ if they make definitions at all for arithmetic in $\overline{\mathbb{R}}$ $\endgroup$ – joeb Feb 22 '17 at 1:05
  • $\begingroup$ Yeah, the extended real number is defined by $\bar{\mathbb{R}} = [-\infty, +\infty]$. Indeed, $\infty - \infty$ is indeterminant. $\endgroup$ – Paradiesvogel Feb 22 '17 at 1:20
  • $\begingroup$ If you allow things like $\infty-1 = \infty$, then letting $f,g: \mathbb{R} \to \mathbb{R}$ be respectively defined by $f(x) = x$ and $g(x) = 1$, then you would have $h = |f-g|: x \to |x - 1|$, in which case $|\sup f - \sup g| = | \infty - 1 | = | \infty | = \infty = \sup h$, provided you define $| \pm \infty | = \infty$. $\endgroup$ – joeb Feb 22 '17 at 1:31

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