Prove $\int _a^b |f(t)| \, dt $ is a norm Let $a < b$ be real numbers and $X = C[a,b]$ be the space of continuous functions $f : [a,b] → \mathbb R$. Prove that $$ \|f \|_1 =\int _a^b |f(t)|\,dt $$indeed deﬁnes a norm on $X$.
Struggling on $\|f\|_1=0 \iff f=0$.
For the backward direction, it seems easy but the forward is not really obvious.
 A: Suppose $\left\Vert f\right\Vert =0$ but $f\neq0$. Recall that compositions
of continuous functions (e.g. $|\cdot|\circ f$) are continuous. Use
the continuity of $t\mapsto|f(t)|$ to establish the existence of
an interval $I\subset[a,b]$ on which $|f(t)|\geq M$ for some $M>0$.
This yields the desired contradiction since
$$
\left\Vert f\right\Vert =\int_{a}^{b}|f(t)|dt\geq\int_{I}|f(t)|dt\geq\int_{I}Mdt>0.
$$
A: Suppose $\int_a^b |f(x)|dx = 0$, then $\phi(t) = \int_a^t |f(x)|dx = 0$ for all
$t$, and so $\phi'(t) = 0 = |f(t)|$.
To see why $\phi$ is zero, note that the integral of a non negative quantity must be non negative. Hence $\phi(t) \ge 0$ for all $t$. Also, if $a,b$ are
non negative quantities, then if $a+b = 0$ we must have $a=b=0$.
Since $\phi(b) = \int_a^b |f(x)|dx = 0 = \phi(t) + \int_t^b |f(x)|dx $, we see
that we must have $\phi(t) = 0$.
A: I assume that you already know how to show homogeneity, triangle inequality and $f = 0 \implies \lVert f \rVert_1$.
Observe that if $\lVert f \rVert_1 = 0$, then $f$ must be $0$ almost everywhere. Then $f = 0$ since continuous functions agreeing a.e. are equal.
