Prove that $\| f \| := \max_{x \in [0,1]} |f(x)|$ is a norm on the vector space $\mathcal{C}[0,1]$. 
Let $\mathcal{C}[0,1]$ be the set of all continuous functions $f: [0,1] \to \mathbb{R}$.
  Prove that $\| f \| := \max_{x \in [0,1]} |f(x)|$ is a norm on the vector space $\mathcal{C}[0,1]$.

I know a norm has to satisfy the three conditions


*

*$\| x \| = 0 \iff x = 0$

*$\| a x \| = |a| \| x \| \ \forall a \in \mathbb{R}, x \in \mathbb{R}^m$

*The triangle inequality


For this function, I am confused as to how to satisfy conditions 1 and 3
For condition 1 , how do I prove max |f(x)| = 0 iff the function equals zero?
For condition 3, I think we start with $\| f_1 + f_2 \| \le \| f_1 \| + \| f_2 \|$ but don't know where to go from there.

For condition $(1)$ it is clear that $\|f\| \geq 0$ for all $f$. And: $$\|f\| = \sup_{x \in [0,1]}|f(x)| = 0 \implies |f(x)| = 0,\quad\forall\,x\in[0,1]\implies f(x) = 0,\quad\forall\,x\in[0,1].$$ For $(3)$, you take the supremum in the right order: $$|f(x)+g(x)|\leq |f(x)|+|g(x)| \leq \|f\|+\|g\|,\quad \forall\,x\in[0,1]\implies \|f+g\|\leq\|f\|+\|g\|.$$
