Question: I have an exercise in my notebook and I don't understand one thing in the answer to it:

Exercise: Show that $$||f||\_{L^p}:=\biggl(\int^b\_a|f(x)|^p|dx\biggr)^{1/p}$$

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

So the bit from the answer that concerns me is an explanation of why would $f(x)>0$ if $f(a)$ or $f(b)$. Here is the part of the answer:

Suppose that $f\in C([a,b])$ with $\int^b\_a|f(x)|^pdx=0$

but $f\neq 0$. Then $f(c)\neq 0$ for some $c\in (a,b)$. [If $f(a)>0$ then $f(x)>0$ for some $x>c$, since $f$ is continuous: likewise if $f(b)>0$ then $f(x)>0$ for some $c<b$.]

Here I don't understand why would $f(a)>0 \Rightarrow f(x)>0$ since $f$ is continuous. Is there some property of continuous functions I don't know? My reasoning for this, although not rigorous, is that if $f(a)>0$, then it should be that in the small area after $f(a)$, there are some values of $f$ that take values bigger than $0$, otherwise the function would have a "bump" or "dot" at $f(a)$ which would make it discontinuous.

Answer: 
Because $f$ is continuous, and so if at some point $x\_0\in[a,b]$ we have that $f(x\_0)=c>0$ then there is a neighborhood $(x\_0-\delta ,x\_0+\delta )$ of $x\_0$ such that $f(x)>c/2$ for all $x\in(x\_0-\delta ,x\_0+\delta ) \cap [a,b]$.
That is not a Mathematical reasoning so the question of correctness does not arise.

There exists $\delta >0$ such that $|f(x)-f(a)| < f(a)$ for $a<x<a+\delta$. So $f(x)\geq f(a)-|f(x)-f(a)| >0$ for any $x$ with $a<x<a+\delta$. Similar argument works for points near $b$.