How do I prove that $\Vert f \Vert_{L^p} := \left(\int^b_a\vert f(x) \vert^p dx \right)^{\frac{1}{p}}$ defines a norm on $C^0([a,b])$?

Question

In my textbook I have to prove that $$\Vert f \Vert_{L^p} := \left(\int^b_a\vert f(x) \vert^p dx \right)^{\frac{1}{p}}$$

defines a norm on $C^0([a,b])$ for any $1\leq p\lt\infty$.

Now I don't even get what $C^0$ means(specifically the $0$ confuses me. Is it the space of continuos functions which vanish at $\infty$?

Also, I tried using a result in my textbook for the triangle inequality part of my proof where the result is as follows(it's a lemma btw):

Suppose that $N : X → [0, ∞)$ satisfies $1$ and $2$ of the definition of a norm (basically the parts which are not the triangle inequality) and in addition that the set $B:= \{x : N(x) ≤ 1\}$ is convex.
Then $N$ satisfies the triangle inequality $N(x + y) ≤ N(x) + N(y)$ and so defines a norm on $X$

But then I'm not sure whether to even start using this as I'm not sure about what $C^0$ means.

Also, I think that before proving the triangle inequality for this I have to show that $\Vert f \Vert_{L^p}=0$ implies $f=0$, right?

Answer 1

Triangle inequlity is just Minksowski's Inequality. See https://en.wikipedia.org/wiki/Minkowski_inequality

If $f(x) \neq 0$ for some $x$ then (by continuity) there exist $t,s >0$ such that $|f(y)| >s$ for $|y-x| <t$. This makes $\|f\|_p >s (2t)^{1/p} >0$ Hence $\|f\|_p=0$ implies $f(x)=0$ for all $x$.