For some $c \geq 0$ $\text{sup} \ \{ c \cdot f(x): \text{some domain of $x$} \}$ = $c \cdot \text{sup} \ \{f(x): \text{same domain of $x$} \}$ How would you formally justify this? Or is it just notationally obvious? (As opposed to 'conceptually' obvious, which is never an excuse in mathematics.)
Edit: For some $c \geq 0$
$\text{sup} \ \{ c \cdot f(x): \text{some domain of $x$} \}$ = $c \cdot \text{sup} \ \{f(x):  \text{same domain of $x$} \}$
To illustrate what I mean, consider $\text{sup}\  \{ 1, 5\} = 5$ which however conceptually obvious can still be proven. 

Suppose otherwise, if $\text{sup}\  \{ 1, 5\} < 5$, then the supremum
  is less than an element of  set. A contradiction. If $\text{sup}\  \{1, 5\} >  5$, then there exists the number 5 less than the supremum
  which nonetheless is an upper bound of the set. A contradiction.

But I'm not sure how I would 'prove' my original assertion. 
 A: Note that you need the condition $c>0$ in order to prove your point. 
The proof needs to address two things. 
First you show that upon multiplying by $c$ the upper bound property is preserved 
Then you have to show that you have the least upper bound.  
Both statements are done by definitions and the fact that multiplying by the positive number $c$ preserve the orientation of inequalities. 
A: If $g:\mathbb{R}\to \mathbb{R}$ is increasing and continuous, then $g(\sup A)=\sup \{g(x): x\in A\}$.  Here $g(\infty)=\lim_{x\to \infty}g(x)$ and $g(-\infty)= \lim_{x\to -\infty} g(x)$. 
Fix $x\in A$ and note that, since $x\leqslant \sup A$, $g(x)\leqslant g(\sup A)$. Therefore $g(\sup A)$ is an upper bound for $\{g(x): x\in A\}.$ In order to finish, we need to show that $g(\sup A)$ is the least upper bound, which means we need to show that values in $\{g(x): x\in A\}$ can be arbitrarily close to $g(\sup A)$. By definition of $\sup A$, we can find  sequence $(x_n)_{n=1}^\infty$ in $A$ with $\lim_n x_n=\sup A$. By continuity, $\lim_n g(x_n)=g(\sup A)$, so among the points $g(x_n)$, we can indeed get arbitrarily close to $g(\sup A)$.  
Now apply this general fact to the specific case in which $g(x)=cx$ for some $c\geqslant 0$ and $A=\{f(x): x\in D\}$, where $D$ is your domain of $x$. 
