Proof that if $ f_n \rightrightarrows f $ then $ \sup_{x \in D} f_n \rightarrow \sup_{x \in D} f $ 
Proof that if $ f_n \rightrightarrows f $ then $ \sup_{x \in D} f_n \rightarrow  \sup_{x \in D} f$ both in case when $\sup_{x\in D} f = \infty$ 

Thesis is equal to
$$ \forall \epsilon >0 \exists n_0 \forall n>n_0  |\sup_{x \in D} f_n (x) -  \sup_{x \in D} f(x) | < \epsilon $$
But from
$$ f_n \rightrightarrows f  $$ 
we know that:
$$\forall \epsilon > 0 \exists n_0 \forall_{x\in X} \forall_{n > n_0}
| f(x) - f_n (x) | < \epsilon $$
I don't know how to start this task. I have some doubts to thesis, because what if $sup$ of $f_n$ and $f$ will be before $n_0$? 
 A: Don't get confused. You are not taking the supremum by $n$, you take it by $x$. $\sup_{x\in D} f_n$ is the supremum of the set $\{f_n(x): x\in D\}$ for a specific $n\in\mathbb{N}$, and this is a real number. You need to show that the sequence of real numbers $\sup_{x\in D} f_n$ converges to $\sup_{x\in D} f$. I'll show how to do it in the case where all the functions are bounded, and I'll leave the case where the supremum is infinite for you. 
So let $\epsilon>0$. As we know:
$\exists n_0\in\mathbb{N}\forall n\geq n_0\forall x\in D    |f_n(x)-f(x)|<\epsilon$.
Now let $n\geq n_0$. For all $x\in D$ we have $f_n(x)<f(x)+\epsilon\leq \sup_{x\in D} f+\epsilon$. If it is true for all $x\in D$ then it is true for the supremum as well, so $\sup_{x\in D} f_n\leq \sup_{x\in D} f+\epsilon$. 
Similarly, we also have $\sup_{x\in D} f_n\geq f_n(x)>f(x)-\epsilon$ for all $x\in D$. In other words $f(x)\leq \sup_{x\in D}f_n+\epsilon$ for all $x\in D$. If it true for all $x$ then it is true for the supremum, so $\sup_{x\in D}f\leq \sup_{x\in D}f_n+\epsilon$ which implies $\sup_{x\in D}f-\epsilon\leq\sup_{x\in D}f_n$. 
So we showed that if $n\geq n_0$ then $\sup_{x\in D}f-\epsilon\leq\sup_{x\in D}f_n\leq\sup_{x\in D}f+\epsilon$ which means $|\sup_{x\in D}f_n-\sup_{x\in D}f|\leq\epsilon$, just as we wanted. 
