In the book Linear functional analysis by Rynne and Youngson, Second edition:
Definition 1.17 If d is a metric the following conditions hold:
(...)
(b) $d(x, y) = 0 \iff x = y$
(...)
Theorem 1.52
Let $(X, Σ, d\mu )$ be a measure space and let $f \in \mathcal{L}^1(X)$.
(a) If $f(x) = 0$ a.e., then $f \in \mathcal{L}^1(X)$ and $\int_X f d\mu = 0$.
(c) If $f, g \in \mathcal{L}^1(X)$ and $f(x) ≤ g(x)$ for all $x \in X$, then $\int_X f d\mu ≤ \int_X g d\mu $. If, in addition, $f(x) < g(x)$ for all $x \in S$, with $µ(S) > 0$, then $\int_X f d\mu < \int_X g d\mu $.
Corollary 1.56
b) The function
$d_p(f, g) =( \int_X |f-g|^p d\mu)^{1/p}$, for $1\ge p < \infty$ and
$d_p(f, g) =\text{ess sup |f-g|}$, for $ p = \infty$
is a metric on $L^p(X)$ (condition (b) in Definition 1.17 follows from properties (a) and (c) in Theorem 1.52, together with the construction of the spaces $L^p(X)$)
I have a doubt with this, I 'dont see why property (c) is needed to proof $d_p(f, g) = 0 \iff f = g$
My proof:
- for $ p\in [1,\infty) )$
If $( \int_X |f-g|^p d\mu)^{1/p} =0 \implies |f-g|^p =0 \text{ a.e} \implies f=g \text{ a.e} \implies f \equiv g$ , since they belong to the same class
If $f\equiv g \implies |f-g|^p =0 \text{ a.e}$, so$\implies$ (using (1.52 a))$( \int_X |f-g|^p d\mu)^{1/p} =0$
- for $p=\infty$
If $0={\text{ess sup} |f-g| } = \text{inf}\{C\ge 0 ; |f(x)-g(x)|\le 0 \text{ a.e} \}$ $\implies |f(x)-g(x)|\le 0 \text{ a.e} \implies f=g \text{ a.e} \implies f \equiv g $
If $f=g \implies \text{ess sup}(|f-g|) =\text{ess sup}(0) =0$ since the smallest $C \ge 0$ such that $0 \le C$ is $C=0$
So where is (c) needed? Is my proof correct? Is there is anything that is not written preciselly Or where property (a) should have been called please tell me as well