Supremum of sum of functions strictly less than the sum of the functions' suprema (sup f+g < sup f + sup g)

Question

Given the bounded functions $f, g: X \rightarrow \mathbb{R}$, I understant that $\sup (f+g) \leq \sup f + \sup g$. However I can't think of an example where $\sup (f+g) < \sup f + \sup g$, strictly.

(In case the notation is not clear, $\sup f = \sup \{f(x): x \in X\}$ and $\sup f+g = \sup \{f(x)+g(x): x \in X\}$.)

The textbook I'm using cites $f, g: [0, 1] \rightarrow \mathbb{R}$, $f(x)=x$, $g(x)=-x$ as such a case, but from what I understood $\sup f+g = \sup [-1,1]= 1$ and $\sup f + \sup g = \sup [0,1] + \sup [-1,0] = 1+0 = 1$.

To be clear, I would like an explanation of an example where the strictly inequality holds or what I'm missing in the textbook example.

Answer 1

Let consider as another example $f(x)=\cos x$ and $g(x)=\sin x$ with

$$\sup (f+g) =\sqrt 2< \sup f + \sup g=2$$

Answer 2

Choose $f = -g$ with $f=\sin(x)$.

Answer 3

If you think of $f$ and $g$ as sound waves (ignoring the need to be periodic, etc.), you can think of their sum $f+g$ as exhibiting some amount of constructive or destructive interference at each point $x\in X$. For instance, if $f(a)$ is positive while $g(a)$ is negative (or vice versa), they exhibit destructive interference at $x=a$, because $g(a)$ brings the values of $f(a)$ down (and $f(a)$ brings the value of $g(a)$ up). If both $f(a)$ and $g(a)$ are positive, there is some amount of constructive interference: the value of both $f(a)$ and $g(a)$ is amplified by adding them together. Perhaps this constructive interference at $x=a$ isn't as large as possible, as $g$ and $f$ both have larger values away from $x=a$.

If things synch up just right, the maximum value (or ramp of values if the supremum is attained by a limit) from $f$ and $g$ occur simultaneously, allowing them to demonstrate their maximum possible constructive interference. In this case, $sup(f+g)=sup(f)+sup(g)$. If $f$ and $g$ aren't in this perfect state of synchronization, we get a strict inequality. The examples given so far in other answers rely on offsetting $f$ and $g$ so that they don't synch up to give their maximum possible constructive interference. This is possible without having any destructive interference too: both functions can be positive. Take for instance $f$ and $g$ so that $f(0)=2$, $f(x)=1$ for $x\neq0$, $g(1)=2$, and $g(x)=1$ for $x\neq1$. Even though summing $f(x)+g(x)$ amplifies both values at any input $x$, the supremum of $f+g$ is 3, whereas if they were synched up better their supremum could be 4.

Answer 4

I'm adding another answer, because it's good to have an example where both $ f $ and $ g $ take only positive values; it's not just that they can cancel each other out, with one positive and one negative. Some of the posted answers already do this when restricted to certain intervals, but here's one where both functions are positive on the entire real line: $$ \eqalign { f ( x ) & = \frac 1 { x ^ 2 + 1 } \text ; \\ g ( x ) & = \frac 1 { ( x - 1 ) ^ 2 + 1 } \text . } $$ Then $ \sup f + \sup g = 2 $, but $ \sup { ( f + g ) } = 8 / 5 $.

ETA: Oh, I guess that the answer by @blark does have an example with positive functions at the end. But I'll still leave this up, in case you want an example with continuous (even smooth) positive functions.