$sup_x [ f(x)+sup_x g(x)]= sup_x[f(x)+g(x)]$?

Question

$sup_x [ f(x)+sup_x g(x)]= sup_x[f(x)+g(x)]$ is the statement correct?

Can I prove like this:

$sup_x [ f(x)+sup_x g(x)] = sup_x[f(x)] + sup_x[g(x)] = sup[f(x)+g(x)]$.

Answer 1

Take $f(x) = x = -g(x)$ in $[-1,1]$.

Answer 2

This is false. Take $f(x)=1_{\left\lbrace x=1\right\rbrace}$ and $g(x)=1_{\left\lbrace x=2\right\rbrace}$. Then :

$$sup_x [ f(x)+sup_x g(x)]=sup_x [ f(x)+1]=2$$ But : $$sup_x[f(x)+g(x)]=1$$

Answer 3

The equality is NOT true. Let $f(x) = x$ and $g(x) = 1 - x,$ and use $[0,1]$ for their domains.

Then $\;\sup_x[f(x) + \sup_xg(x)] = 2\;$ and $\;\sup_x[f(x) + g(x)] = 0.$