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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?

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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 $\sup(f(x)^2)=(\inf(f))^2>(\sup(f))^2$.

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    No. We can find a sequence of points approaching the supremum (possibly a constant sequence if the supremum is attained) just by definition of "supremum". Then we can pass to convergent subsequences by Bolzano-Weierstrass. – 2017-01-09
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    I don't understand your comment. If $\sup(f+g)$ is attained somewhere, then you just look at the values of f and g at the point where it is attained. These are at most $\sup(f)$ and $\sup(g)$ respectively. – 2017-01-09
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    If it is not attained, you need to look at a sequence $x_i$ such that $f(x_i)+g(x_i)$ approaches the supremum. This exists by definition. The problem is that $f(x_i)$ and $g(x_i)$ need not have limits. You can take a subsequence for which $f(x_i)$ has a limit. Then this subsequence in turn has a subsequence where $g(x_i)$ has a limit. So on this sub-sub-sequence, both have limits. – 2017-01-09
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    Condensed comments: My concern was about the sub-sub-sequence. For instance $f(x)=floor(x)/x$ and $g=floor(2-x)/(2-x)$ near $1$. sup is equal to $2$ and reached in $x=1$. But a convergent subsequence for $f$ has $x_i\ge 1$ while a convergent subsequence for $g$ has $x_i\le 1$ so the only common sub-sub-sequence is the constant sequence $(1)_n$, this is a bit troubling, but I understood your argument (separating the cases reached / not reached). Still what about $y,z$ are both $\infty$ but different sign, $y+z$ has no sense ? My feeling is that your answer raises a lot questions. – 2017-01-09
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    Yes, I should have separated out the (trivial) case where $\sup(f)$ (or g) is infinite first, to avoid $\infty-\infty$. – 2017-01-09
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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 $\sup(f)\sup(g)=0$.

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    Good answer. As a side note, I don't see any problem with infinite supremums. If one or both of them equal $+\infty$ then the right hand side of $$\sup(f+g)\le \sup f + \sup g$$ equals $\infty$ and so the inequality is trivially true. – 2017-01-09
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    Ah yes that's certainly true. Thanks for pointing that out – 2017-01-09
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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.

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    Is it even possible that $\sup f=-\infty$ ? – 2017-01-09
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    @zwim Oh yeah, I was way too into measure theory. You're right. – 2017-01-09
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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 $F+G.$