Proposition 3

If f and g are two equivalent mappings of X into a Banach space F, then $N_p(f - g) = 0$ for $1 \leq p < +\infty$; conversely, if $N_p(f - g) = 0$ for a value of $p \geq 1$ then f and g are equivalent.

The proposition follows at once from Th. 1 of §2, No. 3.

If f and g are two equivalent mappings of X into F, then $N_p(f) = N_p(g)$ for all $p \geq 1$ (§ 2, No. 3, Prop. 6); thus, $N_p(f)$ depends only on the class $\tilde{f}$ of f, and one sets, by definition, $N_p(\tilde{f}) = N_p(f)$. Since the classes of mappings of X into F form a vector space (§ 2, No. 4), the relations (3) and (4) may also be written

One defines similarly $N_p(\tilde{f})$ for every class of equivalent numerical functions (finite or not).

One can then define $N_p(f)$ for a function with values in F (resp. in $\overline{\mathbf{R}}$) defined almost everywhere in X, by setting $N_p(f) = N_p(\tilde{f})$; it is then clear that the relations (3) and (4) again hold (assuming $\alpha \neq 0$ and $f + g$ defined almost everywhere, in the case of numerical functions, finite or not).

If $0 < p < 1$, one again sets $N_p(f) = (\int^* |f|^p d|\mu|)^{1/p}$, but the inequalities (4) and (5) are no longer valid (cf. Ch. I, Exer. 6 and Ch. IV, §6, Exer. 13).