Proving something is a norm Let $a<b$ be real numbers and $X=C[a,b]$ be the space of continuous functions $f:[a,b] \rightarrow \mathbb R$
Prove that $||f||_1 = \sup _{t \in [a,b]} |f(t)|$ indeed defines a norm on $X$.
The general definition is:
Let $X$ be a vector space over $\mathbb R$. A norm on $X$ is a map $||.||: X \rightarrow \mathbb R$ such that@
($1$) $||x|| \geq 0$ for all $x \in X$ and $||x||=0 \iff x=0$
($2$) $||\alpha x ||= |\alpha| ||x||$ for all $\alpha \in \mathbb R$, $x \in X$
($3$) $||x+y|| \leq ||x||+||y||$ for all $x,y \in \mathbb R$
I cant even do the first condition because i dont know how it works with the "$x$" being inside the norm symbol. I feel that if i see how the first condition checks, i would be able to do the rest. Please help.
 A: The first condition says $||x||\geq0$ for all $x\in X$. In your case $X=C[a,b]$, such that elements of $X$ are continuous functions $f:[a,b]\to\mathbb{R}$.
In other words, you have to prove $||f||\geq0$ for all $f\in X$.
Since $|f(t)|\geq0$ for any $t\in[a,b]$, you have $\sup_{t\in[a,b]}|f(t)|\geq0$, proving $||f||\geq0$.
I think you'll be able to prove the second statement of the first condition, as well as the other two conditions. Good luck!
A: The $x$ is an element that you can take norm of, and $0$ is an neutral element under addition (ie such that $0+f = f$). In your example it would translate to:


*

*$||f||_1\ge 0$

*$||f||_1 = 0$ iff $f(t)=0$ for all $t\in[a,b]$

*$||\alpha f||_1 = |\alpha|\cdot||f||_1$

*$||f + g||_1 \le ||f||_1+||g||_1$


So for example for the first we note that since $|f(t)|\ge0$ we would have that $||f||_1=\sup_{t\in[a,b]}|f(t)|\ge 0$. 
I think you can do the rest now. The first three should be quite straight forward, but the last requires a little bit more work (but should hopefully be not to hard).
