in general I know that the supremum of a sum of functions is always less or equal than the sum of suprema. But how does this look like in the following case?
When I have two functions, $f(x)$ and $g(y)$ for $x,y \in [0,1]$ and each function only depends on one of the two parameters, does this equality hold?
$\sup \limits_{x,y \in [0,1] \times [0,1]} (f(x) + g(y)) = \sup \limits_{x \in [0,1]} f(x) + \sup \limits_{y \in [0,1]} g(y)$
Yes, we have $\sup \limits_{x,y \in [0,1] \times [0,1]} (f(x) + g(y)) = \sup \limits_{x \in [0,1]} f(x) + \sup \limits_{y \in [0,1]} g(y).$
Proof: let $s_1:= \sup \limits_{x \in [0,1]} f(x)$ and $s_2:= \sup \limits_{y \in [0,1]} g(y).$
Then we have $f(x)+g(y) \le s_1+s_2$ for all $(x,y) \in [0,1] \times [0,1]$.
Now let $ \epsilon >0$. Then there are $x_1,y_1 \in [0,1]$ such that
$$f(x_1) > s_1 -\epsilon/2$$
and
$$g(y_1) > s_2 -\epsilon/2.$$
This gives
$$f(x_1)+g(y_1) > s_1+s_2 - \epsilon.$$