Can anyone prove the second property of a the following metric? $d: C([a,b])\times C([a,b]) \to R ; \ \ \ d(f,g)=\int_{a}^{b}|f(t)-g(t)|dt$ $$d: C([a,b])\times C([a,b]) \to R ; \ \ \ d(f,g)=\int_{a}^{b}|f(t)-g(t)|dt$$
$2.) d(f,g)=0 \iff f \equiv g$
Now in my notebook some lemma is called upon, concerning integrals, but it is unclearly written, so I cannot understand how..
 A: You are essentially asking for a proof that for continuous $h$: $[a, b]\longrightarrow\mathbb{R}$ the fact that $\int_a^b \left|h(t)\right| dt=0$ implies $h\equiv 0$. There is an elementary proof, but it's not too obvious:
Suppose $h(t_0)\neq 0$ for some $t_0\in [a, b]$. Then $\left|h(t_0)\right|>0$ and hence, since $\left|h\right|$ is continuous, there is a subinterval $t_0\in [c, d]\subseteq [a, b]$ such that $\left|h\right|>0$ on $[c, d]$. Now a continuous function on a compact interval attains a minimum, so $\min\limits_{[c, d]}\left|h\right|:=m >0$. But then
$\int_a^b \left|h(t)\right| dt\geq\int_c^d \left|h(t)\right| dt\geq\int_c^d m dt=(d-c)m>0$,
where the last integral inequality holds since $\int$ is monotonic. This is a contradiction.
A: $(\Rightarrow)$ Suppose that $d(f,g)=0$. Then $$\int_a^b |f(t)-g(t)|dt=0.$$ I will call $|f(t)-g(t)|$ the (continuous) function $h(t)$. Suppose that there exists $t_0$ such that $$h(t_0)=|f(t_0)-g(t_0)|=y>0.$$ Letting $\varepsilon=y/2$, there exists $\delta>0$ (also, $\delta<b-a$) such that if $d(t,t_0)<\delta$, then $d(h(t),h(t_0))<\varepsilon$. So $h(t)>y/2$ for all $t$ such that $d(t,t_0)<\delta$. Thus, $$\int_a^b|f(t)-g(t)|dt\geq \frac{\delta y}{2},$$ a contradiction.
The other direction is trivial.
A: Suppose $\int_{a}^{b}|f(t)-g(t)|dt = 0$. For any $a \le x \le b$,
\begin{align}
          0 \le \int_{a}^{x}|f(t)-g(t)|dt & \le \int_{a}^{x}|f(t)-g(t)|dt+\int_{x}^{b}|f(t)-g(t)|dt \\ & = \int_{a}^{b}|f-g|dt=0.
\end{align}
Then, the fundamental theorem of Calculus applies because $|f(t)-g(t)|$ is continuous:
$$
         |f(x)-g(x)| = \frac{d}{dx}\int_{a}^{x}|f(t)-g(t)|dt.
$$
