TITLE: Sum of the supremum and supremum of a sum
QUESTION [19 upvotes]: Consider two real-valued functions of $\theta$, $f(\cdot): \Theta \subset\mathbb{R}\rightarrow \mathbb{R}$ and $g(\cdot):\Theta \subset \mathbb{R}\rightarrow \mathbb{R}$. 
Is there any relation between
(1) $\sup_{\theta \in \Theta} (f(\theta)+g(\theta))$
and 
(2) $\sup_{\theta \in \Theta} f(\theta)+\sup_{\theta \in \Theta} g(\theta)$
?
Could you provide some informal proof or intuition behind your answer?

REPLY [6 votes]: For $\alpha = \sup f \ge f(x)$ for all x, and $\beta = \sup g \ge f(x)$ for all x then $\alpha + \beta \ge f(x) + f(y)$ for all x.  So $\sup f + \sup g = \alpha + \beta \ge \sup (f + g)$.
That's what others are calling "intuitive".  I like to think of it as degree of limitation.  We  have more degree of freedom for f and g operating independently than together as a fixed sum (f + g).  So the sum of the sups is greater (or equal) than the sup of the sums because we simply have more options.   
But that's a pretty vague definition and if it isn't clear when I first say it, it'll just be confusing.
To show that equality might not hold, simply imagine f and g "get big" at different places.  Imagine $f(x) < 1$ for all $x \ne 1$ but $f(1) = 1$ (example $f(x) = 1 - (x-1)^2$) $\sup f = 1$.  Imagine $g(x) < 1$ for all $x \ne 0$ but $f(0) = 1$ (Ex. $g(x) = 1 - x^x$) $\sup g = 1$.  But $f + g$ is always significantly less than 2.  (In our examples $f + g = 1 - 2(x^2 - x) \le 3/2$)
Then $\sup f + \sup g > \sup (f + g)$.  In our example $2 = \sup f + \sup g > \sup (f + g) = 3/2$
Or better yet let $g(x) = -f(x)$ where $f$ is bounded above and below but includes both positive and negative values.  $f(x) + g(x) = 0$ so $\sup(f + g) = 0$ but $\sup f > 0$ and $\sup g = - \inf f > 0$.