Function Spaces: How do I show that $\|x(t)\|=\sup\{|x(t)|:t \text{ contained in }I\} $ satisfies the three conditions of Absolsute Value Thm? I am currently taking Real Analysis 2 (a.k.a Advanced Calc 2) and have been assigned the following question:
Let I be any set, and let B(I)  be the set of all bounded real-valued functions on I. An absolute value can be defined on B(I) by the formula
\begin{align*}
\|x\|=\sup\{|x(t)|\ :\ t\in I\}, 
\end{align*}
and then a distance by the formula $d(x,y)=\|x-y\|$.
Show that the absolute value just defined satisfies:
a.) $\|x\|\geq0$, and $\|x\|=0$ only if $x=0$.
b.) $\|a\cdot x\|=|a|\cdot \|x\|$ if $a$ is a real number
c.) $\|x+y\|\leq\|x\|+\|y\|$.
There is another part to this question having to do with the specified d(x,y) satisfying some other rules, but at this point I am just looking for some direction rather than anybody answering the question for me:)
My thoughts on this: I guess I am thrown of a bit now because I am dealing with functions. I looked back at the definition of supremum and found the following axiom:
"Every non-empty set of real numbers has a least upper bound." With this axiom alone I conclude that $\|x\|\geq0$, and $\|x\|=0$ only if $x=0$. So a is satisfied and there is nothing left to show. Do you agree?
I should mention that I am still looking at $\|x\|$ as a real-valued vector and not so much as a function (I guess). I am interpreting the difference between $|x|$ and $\|x\|$ to simply be a way of distinguishing two different norms, one based on $\langle x,x\rangle$ and the other on the Supremum. 
Part B: This seems so trivial that I can't think of a way to show that this is true. 
Just a bit lost so any help would be appreciated. 
Andrew
 A: a) No. The axiom, you mentioned, just guarantees that the supremum of the set $\{|x(t)|\ :\ t\in I\}$ exists. It does not show that $||x||\geq0$ or anything else.
Ok. We have to show that $||x||\geq0$. Note that $|x(t)|\geq0$ for each $t\in I$, since the absolute value is nonnegative. Thus, the supremum over all those values must be nonnegative as well, which shows $||x||\geq0$. 
Now, let $||x||=0$ for some bounded function $x$. By definition, this means that $\sup\{|x(t)|\ :\ t\in I\}=0$ (so the maximal value of the function $|x|$ is 0). This implies $|x(t)|=0$ for each $t\in I$, and we conclude that $x(t)=0$ for each $t\in I$. Note that $|\cdot|$ is a norm on the set of real numbers, so it satisfies $|a|=0\ \Leftrightarrow a=0$ for $a\in\mathbb R$.
b) This is indeed not difficult. Just use that $|a\cdot x(t)|=|a||x(t)|$ for each $t\in I$, and that $\sup\{ |a b|\ : b\in B\}=|a|\sup\{|b|\ : b\in B\}$, whenever $B\subset\mathbb R$.
c) Try this on your own first and don't hesitate to ask if you get stuck somewhere.
