How do I prove that the following integral is a norm? 
Show that
  $$ \|f\| := \int_{a}^{b} |f(x)|\,\mathrm{d}x $$
  is a norm on $C[a,b]$, the space of continuous functions on the interval $[a,b]$.

Specifically I'm stuck on the property N1, which requires that a norm be positive definite.
How do I show that the definite integral of the absolute value of some generic function $f(x)$ is always positive over the interval $[a,b]$?
Do I have to split it up into three cases $a$, $b$ positive; $a$ negative, $b$ positive; and $a$, $b$ negative?   And if so how do I show that each case is positive?
 A: Consider the following result about integration: if $f,g \in C[a,b]$ are such that $f(x) \geq g(x)$ for all $x \in [a,b]$, then 
$$\int_a^b f(x) \ dx \geq \int_a^b g(x) \ dx.$$
Now take any function $f \in C[a,b]$ and note that $|f(x)| \geq 0$ for all $x \in [a,b]$. Using the previous result we have that
$$\|f\| = \int_a^b |f(x)| \ dx \geq \int_a^b 0 \ dx = 0.$$
A: The Setting
To show that $\|\cdot\|$ is a norm on $C[a,b]$, we have to show three things:


*

*(Positive Definiteness) $\|f\| \ge 0$ for all $f\in C[a,b]$, with equality if and only if $f(x) = 0$ for all $x\in [a,b]$,

*(Triangle Inequality) $\|f + g\| \le \|f\| + \|g\|$ for all $f,g\in[a,b]$, and

*(Homogeneity) $\| \lambda f\| = |\lambda| \|f\|$ for all $\lambda\in\mathbb{R}$ and $f\in C[a,b]$.


Positive Definiteness
Integral showed part of this in another answer, but I'll repeat it for completeness.  Recall that if $f,g\in C[a,b]$ with $f(x) \ge g(x)$ for all $x\in [a,b]$, then
$$ \int_{a}^{b} f(x)\, \mathrm{d}x \ge \int_{a}^{b} g(x)\, \mathrm{d}x. $$
Then for any $f\in C[a,b]$, we have $|f(x)| \ge 0$ for all $x\in [a,b]$, so by the above result, with $g \equiv 0$, we get
$$ \| f \|
 = \int_{a}^{b} |f(x)|\,\mathrm{d} x
 \ge \int_{a}^{b} 0\,\mathrm{d}x
 = 0,$$
which gives positivity.
For definiteness, suppose that $f\in C[a,b]$ with $f\not\equiv 0$.  It then follows that there is some $\varepsilon > 0$ and some $c\in [a,b]$ such that $|f(c)| = \varepsilon$.  By the assumption that $f$ is continuous on $[a,b]$, there exists some $\delta > 0$ such that
$$ |x-c| < \delta \implies \big||f(x)| - |f(c)|\big| < \frac{\varepsilon}{2}. $$
It then follows that
$$ -\frac{\varepsilon}{2} < f(x) - f(c)
\implies |f(x)| > |f(x)| - \frac{\varepsilon}{2} > \frac{\varepsilon}{2}, $$
with the last inequality following from the assumption that $|f(c)| > \varepsilon$.  From the monotonicity of the integral, we have
$$ \|f\|
= \int_{a}^{b} |f(x)|\,\mathrm{d}x
\ge \int_{c-\delta}^{c+\delta} |f(x)|\,\mathrm{d}x
\ge \int_{c-\delta}^{c+\delta} \frac{\varepsilon}{2}\, \mathrm{d}x
= \frac{\varepsilon}{2}\big[ (c+\delta) - (c-\delta) \big]
= \varepsilon
> 0.
$$
In particular, we have shown that if $f\not\equiv 0$, then $\|f\| > 0$.  Therefore positive definiteness holds.
Triangle Inequality
The triangle inequality follows fairly directly from the triangle inequality for real numbers.  To wit, suppose that $f,g\in C[a,b]$.  Then we have
\begin{align*}
\|f+g\|
&= \int_{a}^{b} |f(x) + g(x)|\,\mathrm{d}x \\
&\le \int_{a}^{b} \big( |f(x)| + |g(x)| \big)\,\mathrm{d}x && \text{(triangle inequality in $\mathbb{R}$)} \\
&= \int_{a}^{b} |f(x)|\,\mathrm{d}x + \int_{a}^{b} |g(x)|\,\mathrm{d}x && \text{(linearity of the integral)} \\
&= \|f\| + \|g\|,
\end{align*}
which is the desired result.
Homogeneity
This also follows pretty quickly from the homogeneity of the absolute value.  That is, if $\lambda,\mu\in\mathbb{R}$, then $|\lambda\mu| = |\lambda||\mu|$.  From this, it follows that if $\lambda\in\mathbb{R}$ and $f\in C[a,b]$, we have
$$ \|\lambda f\|
= \int_{a}^{b} |\lambda f(x)|\,\mathrm{d}x
= \int_{a}^{b} |\lambda| |f(x)|\,\mathrm{d}x
= |\lambda| \int_{a}^{b} |f(x)|\,\mathrm{d}x
= |\lambda|\|f\|,
$$
which is what we wanted.
