Property of supremum : do we always have $\sup(f+g)\leq \sup f+\sup g$? I was wondering something : Let $f,g:\mathbb R\to \mathbb R$ be two functions. Do we always have that $\sup(f+g)\leq \sup(f)+\sup(g)$ ? I can prove this for $f$ and $g$ non-negative, but is it still true for functions that do not preserve their sign? Same question for $\sup(fg)\leq \sup(f)\sup(g)$. Is it always true? I know that for $f$ and $g$ non-negative it is true. But is it also true for $f$ and $g$ that don't preserve their sign?
 A: The first is true in general. If you have a sequence of points $x_i$ with $f(x_i)+g(x_i)\to\sup(f+g)$, then there is some subsequence on which $f(x_i)\to y$ for some $y\in[-\infty,\infty]$. Similarly there is a subsequence of this subsequence for which also $g(x_i)\to z$ for some $z$. Now $\sup(f+g)=y+z$, but $\sup(f)\geq y$ and $\sup(g)\geq z$.
The second is not true: simply take some function $f$ which is everywhere negative but not constant. Then $\sup(f(x)^2)=(\inf(f))^2>(\sup(f))^2$.
A: The first statement is certainly true. For all $x\in\Bbb{R}$ we have
$$(f+g)(x)=f(x)+g(x)\le\sup(f)+\sup(g)$$
so $\sup(f+g)\le\sup(f)+\sup(g)$.
For the second, consider $f(x)=-1$ for all $x\in\Bbb{R}$, and $$g(x)=\begin{cases}-1&x\neq0,\\0&x=0\end{cases}$$
Then $\sup(fg)=1$, and $\sup(f)\sup(g)=0$.
A: If both $\sup f$ and $\sup g$ is finite, then $\sup(f+g)$ is finite and $$\sup(f+g)\leq \sup f + \sup g$$
while
$$\sup(fg)\leq \sup f \sup g$$
is only true if $f,g$ are nonnegative.

If one of the two supremums is infinite and the other is not negative infinite, then the expression becomes $\sup(f+g)\leq \infty$ which is trivially true.
A: Let $\sup f=F$ and $\sup g=G.$  Then $$\forall x \;(f(x)+g(x)\leq F+g(x)\leq F+G).$$ So $F+G$ is an upper bound for all $f(x)+g(x).$ So the LEAST upper bound for all $f(x)+g(x)$  cannot exceed $F+G.$
